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[ [ "\nGlassy carbon may be a good choice. Looking at [this](http://onlinelibrary.wiley.com/doi/10.1002/elan.1140071207/abstract), electrooxidation of methanol doesn't seem to be possible at reasonable potentials on glassy carbon, and it's a common working electrode material, relatively inexpensive, durable, and available in many shapes and sizes. Graphite could also work, though it may need to be replaced periodically as graphite electrodes tend to be less robust.\n\n\n", "2" ] ]

[ [ "\n\"[A] factory worker who spilled a gallon of methanol down his trouser leg, was dizzy on the following day, took a short nap, and woke up total blind\" [Biochemical aspects of Methanol Poisoning](http://thetruthaboutstuff.com/pdf/(116)%20Cooper%201962%20Biochemical%20Aspects%20of%20Methano%20Poisoning.pdf), Biochemical Pharmacology, vol. 11, page 405.\n\n\nSo spilling methanol on your skin is serious, but the amount in your case is relatively small. \n\n\nThe article also explains that ingestion of as little as \"a teaspoon\" (5 mL) has caused permanent blindness.\n\n\nIn any case, the internet is no place to get medical advice. If you are concerned you should call your local poison control center.\n\n\n", "7" ], [ "\nFor any poisonous material we have special terms defined. In your case biochemists defined [PEL](http://en.wikipedia.org/wiki/Permissible_exposure_limit) (permissible exposure limit). \n\n\nPEL is mainly important when it comes to deciding about for example \"will the mineral water of that source be applicable for everyday-use of humans?\" Many things affect the PEL for humans; such as age (kid, adult, etc.), type of exposure (ingestion, dermal exposure, breathing etc.), the medical situation of the exposed (other ailments or symptoms that can aggravate the harmful effect of a special kind of chemical) and so forth.\n\n\nAs your profile says, you're in your thirties, and thus are considered an adult. According to [here](http://www.methanol.org/Health-And-Safety/Safe-Handling/Methanol-Health-Effects.aspx) an ingestion of up to 500 milligrams in an adults diet is no problem, among other trivia. And according to [here](http://www.cdc.gov/niosh/docs/81-123/pdfs/0397.pdf), (in which you're considered to be supposedly exposed short-termly) if you didn't have headaches there couldn't be any problem.\n\n\n**Addendum** :\n\n\nYou were exposed to methanol in the dermal way, not ingestion. As a result, as the second reference mentions, the liver *will* be able to turn methanol into formaldehyde and then formate. Formate is toxic. As some of this methanol will evaporate and the rest will eventually be absorbed by the skin you might not be in serious trouble. However, a specialist's advice is the best in this case. Human biology isn't really something anyone will be able to give statements about without empirical observation.\n\n\n", "1" ] ]

[ [ "\nMy first guess was that it was putrescine and other polyamines---but the smell of putrefying potatoes actually comes from methyl mercaptan, dimethyl sulfide, and dimethyl trisulfide. Check it:\n\n\n\n\n\nSource: A. Kamiya, Y. Ose, [Study on offensive odor (Report IV): A Consideration of Putrefaction and Offensive Odor of Solid Waste](https://www.jstage.jst.go.jp/article/taiki1978/18/5/18_5_453/_pdf), Journal of Japan Society of Air Pollution, 18(5), 1983, pp 453-463.\n\n\nNotice that the headspace composition changes quite a bit with time. Most of the headspace chromatography studies I found dealt with early detection of disease organisms in infected potatoes, rather than potatoes in full putrefaction mode. You'll also find many references to solanine poisoning from potatoes; [solanine is a toxic glycoalkaloid](http://en.wikipedia.org/wiki/Solanine), is nonvolatile, and has nothing at all to do with the foul smell and toxic gas produced by putrid potatoes.\n\n\n[Methyl mercaptan](http://en.wikipedia.org/wiki/Methanethiol) ($\\rm CH\\_3SH$) has an odor described as \"rotting cabbage\" by ATDSR; it's one of the major contributors to the smell of farts (oh, sorry, \"flatus\" if we're being polite). It has an odor detection threshold as low as 1 ppb. \n\n\n[Dimethyl sulfide](http://en.wikipedia.org/wiki/Dimethyl_sulfide) ($\\rm (CH\\_3)\\_2S$) is responsible for the smell of the sea (in low concentrations); it too has a cabbagy smell. \n\n\n[Dimethyl trisulfide](http://en.wikipedia.org/wiki/Dimethyl_trisulfide) ($\\rm CH\\_3SSSCH\\_3$) is present in relatively smaller amounts but it has an even stronger odor. The detection threshold is around 1 part per trillion, and it is apparently a strong insect attractant. \n\n\nThe gases produced by rotting potatoes are quite toxic (see [Rotting potatoes in basement kill four members of Russian family](http://www.upi.com/Top_News/World-News/2013/08/23/Rotting-potatoes-in-basement-kill-four-members-of-Russian-family/54831377268770/)). \n\n\n(To keep the notes below in context, in my original answer I said that a plot point in \"The Walking Dead\" was that zombies couldn't smell delicious humans if they were wearing coats smeared with rotting flesh. I suggested that when the zombie apocalypse arrives, packing your pockets with putrid potatoes might work, too. Now I'm not so sure. Can zombies distinguish between polyamines and sulfur compounds? Perhaps they're stench connoisseurs.) \n\n\n\n\n\n", "22" ], [ "\nIt is a combination of potato and a fungus or bacterium that causes a potato to rot and produce the offensive odor. A major contributor to the odor is dimethyl disulfide which has been identified as a key component of the emitted volatiles ([see here](http://onlinelibrary.wiley.com/doi/10.1111/j.1745-4549.1990.tb00134.x/abstract))\n\n\n\n\n\nAlthough the above reference doesn't mention it (at least not in the abstract) dimethyl disulfide is often accompanied by hydrogen sulfide, dimethyl sulfide and dimethyl trisulfide, [the trisulfide being particularly odiferous](http://www.ncbi.nlm.nih.gov/pubmed/21150089). \n\n\nNot only is the odor of rotting potatoes extremely unpleasant, but it has been [reported to be lethal](http://www.parentdish.co.uk/2013/09/04/girl-8-orphaned-after-gas-from-rotting-potatoes-kills-her-entire-family/). Certainly [eating rotten potatoes can be lethal](http://www.smithsonianmag.com/arts-culture/horrific-tales-of-potatoes-that-caused-mass-sickness-and-even-death-3162870/?no-ist) (solanine poisoning).\n\n\n", "5" ] ]

[ [ "\nYou can prove that the Gibbs free energy is equal to the maximum amount of work that a certain reaction (or any physical process in general) can perform, i.e. $\\Delta G = w\\_{max}$. Such a proof can be found in any decent Physical Chemistry textbook (such as Atkins'). The remaining energy must be \"lost\" as heat to basically account for the observation that is essentially stated with the Clausius inequality $\\Delta S \\geq 0$ (i.e. the entropy of the universe is always increasing) . Of course the definition of entropy $\\Delta S = \\frac{q\\_{rev}}{T}$ and the Clausius inequality that is derived from it (and hence the Gibbs free energy) are inherently connected to Kelvins (and Clausius') statement of the second law. You can see from the definition (don't forget the change of entropy of the surroundings!) that (since $\\Delta U = q + w$) the entropy is decreasing when heat is completely converted to work.\n\n\nAsk yourself the following: if heat could be completely converted to work, then the collision of an inelastic ball with the earth (or any other object) would be reversible, have you ever seen a tennis ball suddenly jumping back from the ground?\n\n\n", "6" ] ]

[ [ "\nThis is impossible to answer. Usually you have to assume that when no calorimeter heat capacity is given, then it negligible (i.e. you only use the heat capacity of the 70g $\\ce{H\\_2O}$). You know the temperature drop of the metal and the energy increase of the water, combine both to obtain the heat capacity of the metal.\n\n\n", "2" ], [ "\nYou know the heat capacity of the water given by:\n$$C=4.148\\ \\mathrm{\\frac{J}{g\\ K}}\\ \\times 70\\ \\mathrm g = 290.36\\ \\mathrm{J/K}$$ Since the water gained $3807.44\\ \\mathrm J$, you know that the metal loss $3807.44\\ \\mathrm J$. Using $Q = mc\\Delta T$ you get: $$c = \\frac{Q}{m\\Delta T} = \\frac{{-3807.44\\ \\mathrm J}}{180.45\\ \\mathrm g\\times -66\\ \\mathrm K} = 0.320\\ \\mathrm{\\frac{J}{g~K}}$$\n\n\nUsing the Dulong–Petit law, which states that the molar heat capacity of a metal is approximately $3R$, we get:\n$$\\frac{3R}{c} = \\frac{3 \\times 8.314\\ \\mathrm{\\frac{J}{mol~K}}}{0.320\\ \\mathrm{\\frac{J}{g~K}}} = 78.01\\ \\mathrm{g/mol}$$\n $M=78.01\\ \\mathrm{g/mol}$ is closest to copper so I would guess that is the answer.\n\n\n", "1" ], [ "\nq=m X c X (T2-T1)\n\n\nset heat lost by metal equal to heat gained by water. any difference is due to the calorimeter absorbing heat. hopefully you have values for the specific heats (c) of your various metalS?? if not, hopefully other answers get you where you need to be!\n\n\n", "-1" ] ]

[ [ "\nThere are a couple of things wrong.\n\n\nFirst, the mass of the volatile gas is not $0.8384\\ \\mathrm g$. It's the difference between $12.4528\\ \\mathrm g$ and $12.0468\\ \\mathrm g$.\n\n\nSecond, the volume is incorrect. $11.6144\\ \\mathrm g + 999.97\\ \\mathrm{kg/m^3} \\times 0.339012\\ \\mathrm{m^3} \\ne 350.6264\\ \\mathrm g$\n\n\n", "1" ] ]

[ [ "\nThe maximum temperature for water boiling is around 100 Celsius and there is a great deal of water present. Oil can reach greater temperature and the absence of water allows other reactions, such as the [Maillard reaction](http://en.wikipedia.org/wiki/Maillard_reaction), to occur. That's why you never see browning in boiled food.\n\n\n", "8" ], [ "\nBrinnb is correct, but the other big difference between boiling and frying is that vegetables already have a lot of water in them. Boiling in water doesn't really change that because everything is already surrounded in water, but frying in oil is hot enough to boil the water inside the food, so in addition to the Maillard reaction, deep frying tends to dehydrate food. This is one of the reasons the outsides of french fries get crispy, and taken to the extreme, this effect is used to dehydrate ramen noodles in the production of instant noodles.\n\n\n", "6" ] ]

[ [ "\nDish soap traditionally contained foaming agents such as sodium laureth sulfate which is chemically stable over long periods of time. But more recently environmentally conscious soap companies are turning to foaming agents that have a shorter half-life. These chemicals tend to break down sooner - perhaps just by being exposed to air. This helps reduce foaming in the settling basins of wasterwater treatment plants, or storm drains that enter the ocean and it takes less water to rinse away the soap.\n\n\nRead the ingredients of the soap you are using and research to see which are the foaming agents, and what the stable half-life is.\n\n\n", "6" ] ]

[ [ "\nNitric acid corrodes copper. Drop a penny into some nitric acid (under a hood!) and you'll see that your power supplies really don't stand much of a chance in that environment.\nThis happens with dilute nitric acid:\n\n\n$$\\ce{3 Cu + 8 HNO3 -> 3 Cu^{2+} + 2 NO + 4 H2O + 6 NO3^{−}}$$\n\n\nWhen you open up one of the failed power supplies, do you see bluish crusty white stuff anywhere? That's nitric acid corrosion.\n\n\n**Addendum**: Here are some quantitative corrosion rates: \n\n[Karl Hauffe and Roman Bender: Copper. In Corrosion Handbook. Wiley-VCH, 2008.](http://dx.doi.org/10.1002/9783527610433.chb02210) \n\n\n**Addendum 2**: Even though the concentration of nitric acid in the air is low, your power supplies are sucking a lot of air through them in 12 hours. \n\n\nLet's suppose you have a power supply that dissipates 300 W of heat and you can tolerate a 10 degree Celsius temperature rise. Then according to [this reference](http://www.jmcproducts.com/wp-content/uploads/2012/02/comparative_analysis.pdf) you need an airflow rate of about 53 cubic feet per minute.\n\n\nThat's 1.5 cubic meters of air per minute.\n\n\nThe density of air in a warmish (300 K) room is about 1.177 kilograms per cubic meter. So you're pulling \n\n\n$$\\frac{1.5~\\mathrm{m^3}\\ \\text{air}}{1~\\mathrm{min}} \\frac{1.177~\\mathrm{kg}\\ \\text{air}}{1~\\mathrm{m^3}\\ \\text{air}} \\frac{9~\\mathrm{mg}\\ \\ce{HNO3}}{1~\\mathrm{kg}\\ \\text{air}} = 15.9~\\frac{\\mathrm{mg}\\ \\ce{HNO3}}{1~\\mathrm{min}}$$\n\n\nwhich over 12 hours is \n\n\n$$\\frac{15.9~\\mathrm{mg}\\ \\ce{HNO3}}{1~\\mathrm{min}} \\frac{60~\\mathrm{min}}{1~\\mathrm{hour}} (12~\\mathrm{hours}) = 11.4~\\mathrm{g}\\ \\ce{HNO3}$$\n\n\nIf *all* of that reacted with copper in the power supply, this is how much copper would be converted into corrosion:\n\n\n$$11.4~\\mathrm{g}\\ \\ce{HNO3}\\frac{1~\\mathrm{mol}\\ \\ce{HNO3}}{63.01~\\mathrm{g}\\ \\ce{HNO3}}\\frac{3~\\mathrm{mol}\\ \\ce{Cu}}{8~\\mathrm{mol}\\ \\ce{HNO3}}\\frac{63.546~\\mathrm{g}\\ \\ce{Cu}}{1~\\mathrm{mol}\\ \\ce{Cu}} = \\fbox{4 g Cu }$$\n\n\n...that's a lot of copper to lose (a penny is 2.5 g!). Even if only 10% of all the $\\ce{HNO3}$ that's being sucked through the power supply reacts, it's still 0.4 grams of copper, certainly enough to significantly corrode electrical contact surfaces.\n\n\n", "23" ] ]

[ [ "\nYou need to investigate the safety and handling of every chemical you use in your experiments individually. The handling and safety precautions can be very different depending on what substances exactly you use.\n\n\nIn terms of fire, the one thing you must know before each experiment is which methods won't work or even fail catastrophically. Trying to extinguish a sodium fire with water is a really, really bad idea, and that is something you must know in advance. \n\n\nMetal fires are the most common kind of fire where you have to be careful about choosing the right method. Some metal fires can't even be extinguished by carbon dioxide. There are special fire extinguishers for this (class D), other methods might not work.\n\n\nA bucket of sand is the most useful method for small fires in a lab. It works for almost everything and doesn't cause a big mess. But it is obviously useless if you cause a bigger fire and can't reliably cover it with sand anymore.\n\n\nPowder extinguishers cause quite a mess, you'll spend a lot of time cleaning up the room afterwards. Obviously this is a lot better than burning down your home, but I wouldn't use them for small fires you could stop with some sand.\n\n\nFoam and powder are probably the most useful general-purpose extinguishers. Carbon dioxide is a bit trickier to use from what I heard, I probably wouldn't recommend it in your case.\n\n\nThinking about a fire extinguisher is a good first step. Another thing I very strongly recommend is that you read up on general safety procedures in the lab. Simple things like not storing large amounts of solvent directly in your fume hood (or workplace) might prevent a small fire from becoming a dangerously large one. Knowing that you need to inactivate any reactive reagents before disposing them might save you from throwing away some oxidizing agent and accidentally setting your waste on fire. There is a lot more to chemical safety than just choosing the right fire extinguisher.\n\n\n", "5" ], [ "\n*Admittedly, I'm not a big fan of experimenting at home - particularly when (larger volumes of) flammable liquids are involved.*\n\n\nHome labs typically lack huge sinks with faucets high and huge enough to get a face under it. Neither do they have a fire shower at the exits, which is standard in German university labs. This aside and adding to the answer given by **Mad Scientist**: \n\n\nFoam extinguishers work well even in the rare case when a person is on fire. This does however require a second person to operate it! With respect to this and other scenarios, where pain and panic prevents you from treating a mishap reasonably (acid and alkaline spills in the face, severe cuts due to breaking glas, strong burns, etc.) the golden rule for students in university labs therefore is: **NEVER work alone!** \n\n\n", "1" ] ]

[ [ "\nLitmus is a weak acid, with a $\\mathrm{p}K\\_\\mathrm{a}$ of about 6.5. \n\n\nAlcohols are *extremely* weak acids, with $\\mathrm{p}K\\_\\mathrm{a}$'s (typically) around 17. \n\n\nThe litmus itself is a much stronger acid than the alcohol. Dissociation of the alcohol won't produce enough protons to shift the equilibrium between the two colored forms of the indicator, so you'll see no color change.\n\n\n", "14" ], [ "\nAlcohol in water solution is not acidic enough to change the solution pH. So, it doesn't turn blue litmus red.\n\n\n", "2" ], [ "\nThe pH of ethanol ($\\ce{C2H5OH}$) or pure alcohol is $7.33$. This means that ethanol is slightly basic. And the colour of litmus paper change at $\\mathrm{pH} =7$.\nConsider the ionization of indicator in solution represented by this equation: \n\n\n$$\\ce{HIn <--> H+ + In-}$$\n\n\nAnd the acidic colour of $\\ce{HIn}$ is red, the basic colour of $\\ce{In-}$ is blue.\nAccording to Le Chatelier’s principle, if we put indicator in basic solution, $\\ce{H+}$ reacts with base, more $\\ce{HIn}$ ionized, more $\\ce{In-}$ formed, so the colour of blue litmus unchanged.\n\n\n", "-3" ] ]

[ [ "\nIn general, the \"alpha\" hydrogen is a hydrogen attached to a carbon that is [\"alpha\" (adjacent to) the substituent](http://en.wikipedia.org/wiki/Alpha_and_beta_carbon). \n\n\n\n\n\n[image source](http://www.meritnation.com/img/shared/discuss_editlive/4255118/2013_10_30_17_50_50/zhcgfjcy4021270914596835552.png)\n\n\nSo in this case, an \"alpha\" hydrogen is a hydrogen attached to the carbon alpha to the carbonyl.\n\n\n\n\n\n[image source](http://en.wikipedia.org/wiki/Alpha_and_beta_carbon)\n\n\nThe following drawing represents a typical [keto-enol equilibrium](http://en.wikipedia.org/wiki/Keto%E2%80%93enol_tautomerism). If we removed the last alpha hydrogen and replaced it with a substituent (alkyl group, phenyl group, bromine, etc.) then we could not form a double bond with that alpha carbon, it would have 4 non-removable substituents. If we can't form a double bond that means we can't form an enol.\n\n\n\n\n\n[image source](http://en.wikipedia.org/wiki/Keto%E2%80%93enol_tautomerism)\n\n\n", "3" ], [ "\nAn $\\alpha$ hydrogen atom is bound to a carbon next to some functional group, e.g. a ketone or an aldehyde. Without this hydrogen there is no hydrogen that can be removed by enolization. Other hydrogen atoms on the carbon chain are much less acidic due to the lack of an electron stabilizing group such as a keto-group. Benzaldehyde contain no $\\alpha$ hydrogen atoms because of the $\\ce{sp^2}$ hybridized carbon atom. On the other hand, acetone for example has 6 $\\alpha$ hydrogen atoms.\n\n\n", "0" ] ]

[ [ "\nThe graphite itself can adhere to paper fibers with [London forces](http://en.wikipedia.org/wiki/London_dispersion_force); it's not a strong adhesion because the marks can be erased easily. Some of the graphite is just mechanically stuck between the paper fibers; rubbing a mark from a soft pencil can easily blur it. Enough graphite is deposited that [a line drawn by a pencil can conduct electricity](http://pubs.acs.org/cen/whatstuff/print/7942sci4.html).\n\n\nThe clay gives the lead structural integrity; an \"H\" pencil has a harder lead, and deposits less graphite; a \"B\" pencil is softer, and an \"F\" pencil can be sharpened to a finer point.\n\n\nIn my time as an art major, I made a lot of sketches using vine charcoal on newsprint. Vine charcoal has no binding clay, and the marks were easily smeared. On paper without much \"bite\", it was easy to lift the marks off completely with a piece of kneaded rubber, and if you wanted to keep a charcoal drawing (or a soft pencil drawing) you really had to spray it with a fixative. \n\n\nSo I'd say that it's possible that the binding clay also plays some role in binding the graphite or charcoal to the paper. \n\n\n\n> \n> In the manufacture of graphite pencils, air-classified grades of Volclay sodium \n> bentonite and micronized hectorite are regularly used to bind the graphite \n> compound in pencil lead. --- [American Colloid Company, Industrial and Household Applications](http://www.colloid.com/isg/Application-IndustrialAndHousehold.aspx)\n> \n> \n> \n\n\nBentonite and hectorite both have ionic surfaces that probably help bind it to OH groups on the surface of cellulose fibers.\n\n\n", "7" ] ]

[ [ "\nI can see how this question is troubling. \n\n\nIt don't think there is a solution. \n\n\n$K\\_b = \\frac{RT\\_b^2M\\_w}{1000\\Delta H\\_v}$\n\n\nWhere $T\\_b$ is boiling point, $\\Delta H\\_v$ is the molar heat of vaporization, and $M\\_w$ is molecular weight of the solvent.\n\n\n(see [Colligative Properties by W. R. Salzman](http://www.chem.arizona.edu/~salzmanr/480a/480ants/colprop/colprop.html))\n\n\nSo you can see that knowing $K\\_b$ and $T\\_b$ is insufficient to find $M\\_w$, knowledge of $\\Delta H\\_v$ is also required. \n\n\nExperimentally, from [Table of cryoscopic and ebullioscopic constants](http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf) you can also see there is no solution. \n\n\n", "5" ] ]

[ [ "\nLet's imagine that your transport box has a decently closing lid. It you place the herbs in the box, close the lid , stuff the box into a vacuum sealer bag, **remove the air from the bag until it is shrunk around the box** and finally seal it, you may consider it sealed. \n\n\nThis means:\n\n\n1. According to \"household standards\", it is air-tight to the outside. No air from the environment will enter the bag, unless it is damaged.\n2. There is however still air in the box. Whatever degradation, loss of flavour, etc. may happen to the herbs during transport still can, since there is oxygen.\n\n\nHowever, if you **vaccum-seal the box**, or replace the air in the box with nitrogen, you will minimize (or fully prevent) degradation of the herbs.\n\n\n", "3" ], [ "\nI know this is an old post but I hope I can still help. It would be a very bad idea to vacuum pack fresh herbs for shipping. From a chemical stand point, the living herb are respiring and so releasing CO2. Cold shipment will be needed to lower the rate of respiration, check the Q10 for the herbs you are shipping, or a not air-tight container. Look at the packaging in a grocery store, they are not air-tight.\nFrom a microbiological view, you want to place fresh herbs (may have some soil still on them), that most likely have not been irradiated (best way to kill microbes on food), into an anaerobic environment. Research *Clostridium botulinum* and you will see why this is a bad idea.\n\n\n", "2" ], [ "\nAfter vacuum cleaning of goods which you are mentioning.. there is still a chance of entry of gases during the packing time. If you are good enough to manage the goods vacuum protected until packing then your goods will be vacuum protected if you vacuum seal the box perfectly.. \n\n\n", "1" ] ]

[ [ "\nAtoms with one outer electron (e.g. sodium), called alkali metals, or two outer electrons (e.g. calcium, strontium), called alkaline earths, and even the aluminum-scandium metals are not good at sharing -- they tend to lose their outer electrons given the opportunity, e.g. when a halogen is near. These form *ionic* compounds, such as table salt, in which case the Na+ has given up one electron, which stays close to the Cl- ion. \n\n\nElements in columns closer together don't get away with outright theft, and share their outer electrons more equitably in a covalent bond. \n\n\nThat said, it's an oversimplification, in that a bond is not necessarily fully ionic or fully covalent and may fall somewhere in between. See <https://www.youtube.com/watch?v=OY2_q-1t2Q8> for a video on this. \n\n\n", "1" ] ]

[ [ "\nBriefly, since ethyl alcohol expands more rapidly with increasing temperature than does water, then measuring the mix by volume at a warmer temperature would give you less alcohol and more water, because the alcohol would be less by weight for the same volume.\n\n\n**Thermal Coefficient of Expansion (approximate, measured at room temperature)**\n\n\nAlcohol, ethyl (ethanol): 0.00109 / K\n\n\nWater:                            0.00021 / K\n\n\nUsing these constants, you can figure out the answer to your question. But note the wording: given a **mixture**. If the EtOH/H2O is **already mixed**, changing the temperature does **not** change the concentration or proof, and you would have to separate the ingredients to measure their respective volumes. BTW the volume of the mix is slightly less than the sum of the volumes of its ingredients.\n\n\nMoral: if your measuring your drinks by shot-glass rather than with a scale, you'll get a weaker drink in a warm bar ;-)\n\n\n", "1" ] ]

[ [ "\nCertainly the are many different types of melamine resin. \n\n\nIs there a way to make a melamine resin without using formaldehyde?\n\n\nYes, a different aldehyde could be used such as acetylaldehyde. Glucose, benzaldehyde, acrolein and other alternatives to formaldehyde have been studied also (see page 33 of Polymer Synthesis by Sandler et al.). \n\n\nIs there another way to make an abrasive foam sponge out of melamine resin that is not formaldehyde-melamine-sodium bisulfite copolymer?\n\n\nGood sources of information about formaldehyde-melamine-sodium bisulfite copolymer and possible alternatives are the patent [Resilient foam based on a melamine-formaldehyde condensate](http://www.google.com/patents/US4540717) and related patents.\n\n\nFirst the introductory portion of the patent explains:\n\n\n\n> \n> It is true that foams based on melamine-formaldehyde condensates have been described in various patent publications, but they have not hitherto been employed industrially for heat or sound insulation in the construction industry. They are produced by stirring air into an aqueous melamine resin solution which contains an emulsifier and a curing agent. Such as described, for example, in German Pat. No. 870,027, have the serious disadvantage that they are very hard and brittle, and easily break or crumble when handled. It is alleged in French Pat. No. 1,073,642 that melamine resin foams can be produced by heating a resin powder in a mold under reduced pressure. This process, however, does not give useful resilient foams. U.S. Pat. No. 3,093,600 describes melamine resin foams which are said to have improved resilience and resistance to crazing as a result of the incorporation of triols, e.g. trimethylolpropane. However, it has been found that the resilience, and especially the recovery after compression, of such foams is inadequate for many applications.\n> \n> \n> \n\n\nThen in describing what is considered new about the invention, many different alternatives are included. Such as adding aldehydes other than formaldehyde, urea, and many other substances. \n\n\n", "2" ] ]

[ [ "\nWhether trying to figure out if a double bond is \"E\" or \"Z\", or if a chiral center is \"R\" or \"S\", the [Cahn–Ingold–Prelog priority rules](http://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules) (see the section \"Assignment of priorities\") are used to prioritize the groups. In the case of a double bond there are 2 substituents (a lone pair of electrons can be a substituent) at each end of the double bond. Priorities (1 and 2) are assigned to the substituents at each end using the Cahn–Ingold–Prelog rules provided in the link above. The first rule used to set priorities is based on the atomic number of the atoms attached to the double bond. If there is a tie (the 2 attached atoms at one end of the double bond have the same atomic number), then we move out to the next atoms and compare these. The process is continued until the tie is broken (eventually the tie must be broken if the molecule being analyzed is really capable of being described as \"E\" or \"Z\"). After the priorities have been assigned, if the two priority-1 substituents at opposite ends of the double bond are on the same side of the double bond, then the molecule is the \"Z\" (zusammen, together) isomer, otherwise it is the \"E\" (entgegen, opposite) isomer.\n\n\nExample:\n\n\n\n\n\n[image source](http://driverlayer.com/img/e%20z%20nomenclature/20/image?tab=1)\n\n\n", "2" ] ]

[ [ "\nYes, the fast reaction is a syn-periplanar E2 reaction. If we are ranking the rates of E2 based on the orientation of the C-H and C-LG bonds, anti-periplanar would be the fastest, syn-periplanar would be slower, and perpendicular would be the slowest. The anti-periplanar arrangement is the fastest because it provides the best overlap of the (filled) bonding C-H orbital with the (empty) anti-bonding C-LG orbital. The syn-periplanar arrangement is slower because there is much less productive overlap between these orbitals. However, the small lobe of the bonding C-H molecular orbital does provide some overlap with the anti-bonding C-LG orbital. A perpendicular arrangement results in no productive overlap.\n\n\n\n\n\nThe slow reaction must occur by some other mechanism, such as an E1 or E2-conjugate base mechanism.\n\n\n\n\n\n", "2" ] ]

[ [ "\nSince green is the *absorption* of orange light, you need to supply the orange light to measure how much is absorbed. As the alcohols are oxidized to the acids, the dichromate is reduced, so you can measure how much acid has been formed. BTW, what alcohol was used? Butyl alcohol oxidation has a distinct aroma that can be detected without spectroscopy.\n\n\n", "1" ] ]

[ [ "\n\n> \n> [...] ethanol could be used in a solvent extraction of two solids, but it also says that isopropyl alcohol could also be used, but must be used in higher concentrations. Why? [...]\n> \n> \n> \n\n\nThey hopefully didn't write about the \"concentration of isopropanol\" but mentioned that a larger amount of the solvent has to be used.\n\n\nIt is not uncommon that organic compounds with polar substituents are less soluble in higher alkanols.\n\n\nVanillin is a typical example, the solubilities are in the range of 4.15 mol/L in methanol, 2.5 mol/L in ethanol and 1.8 mol/L in 1-propanol. (Data are taken from a freely accessible collection of experiments performed by students: [OpenNotebook Science Challenge](http://onschallenge.wikispaces.com/list+of+experiments)). I couldn't find any data for 2-propanol. \n\n\n\n> \n> So would this suggest that if one were to use IPA as a solvent, one could use the same amount of IPA as they would Acetone?\n> \n> \n> \n\n\nNo, not really. A ketone and a secondary alkanol, although looking similar on first sight, may have behave completely different as solvents. Polystyrene (foam), as an example is pretty much insoluble in isopropanol but highly soluble in acetone!\n\n\n", "3" ] ]

[ [ "\nHere are some things to focus on:\n\n\n* How quickly\\slowly did you add the diazonium salt to the NaI solution?\n* Did you maintain the reaction temperature below 10°C during the entire addition?\n* Did you also keep the diazonium salt cooled while adding it?\n* After addition was complete, did you let the reaction stir at 10°C until nitrogen evolution ceased and then let the reaction slowly warm to room temp?\n\n\n[Here](http://myweb.brooklyn.liu.edu/swatson/Site/Laboratory_Manuals_files/Exp18.pdf) is a tested lab procedure for the preparation of 4-iodonitrobenzene. Compare it to your notes and see where you differed.\n\n\nKeeping the reaction cold during the addition is key. If the diazonium salt warms up too much, the diazonium can react with itself ([coupling](http://en.wikipedia.org/wiki/Azo_coupling)) to give colored by-products. The diazonium ion can also eliminate nitrogen to generate the extremely reactive phenyl cation which will react with any nucleophile around, water for example (see page 6, the Mechanism section in this [reference](http://nptel.ac.in/courses/104103022/download/module7.pdf)). Both pathways can contaminate and reduce the yield of the desired product.\n\n\n", "8" ], [ "\nWash the product (sludge) with 95% ethanol when filtering. The sludge is due to impurities.Your pure product should be light brown.\n\n\n", "0" ] ]

[ [ "\nElimination from vinyl halides occurs faster when the halide and proton have a trans relationship. When the halide and proton are located on the same side of the double bond (cis), the elimination to form an alkyne is much slower. The faster elimination from the trans arrangement is consistent with a preferred anti-periplanar arrangement in the E2 reaction. \n\n\nThe fact that elimination still occurs, albeit at a slower rate, when the groups being eliminated are arranged cis to one another suggests that a different mechanism is involved in the \"cis\" case. Two possible mechanistic alternatives are\n\n\n* cis elimination from a higher energy (higher energy, therefore less preferred, slower) syn-periplanar conformation, or\n* an [E1CB mechanism](http://chem.ucr.edu/documents/faculty_neuman/chapter%209.pdf) where the proton is removed in a first step and a carbanion is generated, and this is followed by halide ejection in a second step.\n\n\nExperiments in analogous molecules suggest that the E1CB pathway is most often followed in \"cis\" elimination from vinyl halides.\n\n\n", "4" ] ]

[ [ "\nAt constant temperature, only changes to the concentration (technically [activity](http://en.wikipedia.org/wiki/Activity_(chemistry))) of a reactant or product, shift the equilibrium. \n\n\nAdding inert gas does not change the concentration of any reactant or product, so it does not affect equilibrium. \n\n\n", "1" ], [ "\nThe gas added works in both directions and the fraction of pressure is the same becaussed the total pressure increses but the inert gas do not affect the features of the others and exerts the same effect. If the number of molecules do not change, the concentration remains constant, and the principle of Le chatelier sais that in case of increase the pressure the reaction take the direction where there are less molecules because the pressure force them to agregate.\n\n\n", "0" ] ]

[ [ "\n\n> \n> How can hydrogen form two bonds?\n> \n> \n> \n\n\nIn the case of an enol, hydrogen does not form two full bonds. What the drawing below is trying to show with the enol hydrogen is that one bond (the O-H with the solid line) is elongated, partially broken, while the other bond (the O..H with the dotted bond) is also elongated and only partially formed. It's as if the partially broken OH bond and the partially made OH bond are each \"half a bond\", so all together the electron sharing is such that there is still just one full bond to hydrogen\n\n\n\n\n\n\n> \n> How does the steric repulsion destabilize the enol form?\n> \n> \n> \n\n\nLook at the bottom part of the drawing where we have inserted an alpha-methyl group into the 1,3-diketone. Notice the steric interactions indicated by the arrows. The situation looks similar to that seen with *cis*-2-butene. The methyl groups are large enough (larger than a hydrogen) and close enough to one another that their electron clouds \"bump\" into one another causing what we often call \"steric destabilization.\" Remember that *trans*-2-butene is more stable than *cis*-2-butene because the *trans* isomer does not have the adverse steric interaction present in the *cis* isomer. It is that same type of steric interaction that occurs here with the enol when an alpha-methyl group is present, and just like in *cis*-2-butene, it destabilizes the enol. Therefore less of the enol will be present at equilibrium when an alpha methyl group is added to the enol.\n\n\n", "8" ] ]

[ [ "\nBriefly: valence orbitals on adjacent metal atoms interfere to form \"bands\" of closely spaced molecular orbitals that spread throughout the metal. \n\n\nIf you have N metal atoms with occupied s orbitals, they interfere to form an \"s band\" with N molecular orbitals. The lowest energy orbital is fully bonding; the orbital on the top of the band is fully antibonding. The spread of the band will be finite, and since N is a huge number, the molecular orbital energies will be *very* closely spaced, practically continuous. \n\n\nIf you have two valence s electrons in your metal, the s band will be full (each orbital can hold up to 2 electrons, you have 2N electrons, and N orbitals in the band). If you have only one valence s electron, the s band will be half full (the first N/2 orbitals will be filled, and the remaining orbitals will be empty). \n\n\nAbove the s band you'll have a similar p band. In gold, the p band is unoccupied, while the s band is half full (gold has 6s$^1$). However, IF the bottom of the p band overlaps with the top of the s band, the s electrons can easily move into the bonding MO's at the bottom of the p band. They can also move into any vacant orbitals in the s band that are just above the highest occupied orbital. *In either case, the s electron will be delocalized*. \n\n\nTo figure out whether the valence electrons will delocalize, then, you'll have to decide what bands are present, where the highest occupied molecular orbital is, and what the width and the spacing of the bands are (does the unoccupied band overlap with the occupied band, or is there a gap between the two?)\n\n\n", "3" ], [ "\nValance electron of s and p orbital take part in delocalisation over the positively charged gold atoms\n\n\n", "-1" ] ]

[ [ "\nThe orbital and geometrical symmetry are closely related. You know that $\\ce{XeF\\_4}$ is square planar, therefore $\\ce{D\\_{4h}}$ symmetric. That also means that the four fluorine atoms are indistinguishable.\n\n\nSo if you perform some manipulation of the molecule, e.g. rotation by $\\ce{90^{\\circ}}$, you must end up with the same picture (or just with opposite sign), therefore ruling out our suggestion of 3:1 different orbitals. This holds for all orbitals, not only bonding.\n\n\nOne could possibly argue, what is the physical meaning of the unoccupied orbitals, but this is out of scope of the question.\n\n\n", "2" ], [ "\nIt's not so much that these are arbitrary arrangements or that symmetry is \"conserved.\"\n\n\nWe take the 4 fluorine atoms as a *set* and consider the reproducible representation of all fluorine 2p orbitals.\n\n\nUnder the $D\\_{4h}$ point group of $\\ce{XeF4}$, the 4 fluorine orbitals consist of an $a\\_{1g}$, $b\\_{1g}$ and $e\\_u$ representations. (There are other representations of the out-of-plane 2p orbitals too.) Four atomic orbitals, 2x1D 1x2D representation.\n\n\nNow we try combining the representations of the fluorine orbitals.\n\n\nIt's a matter of *math*... the reason $b\\_1g$ is non-bonding is because there is no Xe atomic orbital with the same symmetry. If you consider the direct product of the fluorine $b\\_{1g}$ with any Xe orbital, you get zero - they are orthogonal and have no overlap.\n\n\nNow, you ask why can't you have 3:1.. well, you have 4 identical fluorines. If you had $\\ce{XeF3Cl}$ or something like this, you'd then have different symmetry and different orbitals.\n\n\n", "2" ], [ "\nSymmetry reduces the level of energy of any thing. Most of the things of nature has inbuilt symmetry with them. It is the property of nature to have different kinds of symmetry. So that it reduces the energy of the orbital or molecule\n\n\n", "0" ] ]

[ [ "\nIn *The Preparatory Manual of Explosives* ([paperback at amazon](http://rads.stackoverflow.com/amzn/click/0578142813)), the author, Jared Ledgard, provides a procedure for the preparation of copper fulminate that can be summarized as follows:\n\n\n\n> \n> To 100 mL of 70% $\\ce{HNO3}$, 30 g of anhydrous $\\ce{Cu(NO3)2}$ is added under stirring. If the copper nitrate doesn't dissolve properly, add water dropwise until it does. (**NOTE**: *Adding water to concentrated acids usually isn't the best idea!*) The solution is stirred for 24 hrs at room temperature and then heated to roughly 80 °C. Under stirring, 120 g of 95% ethanol are added. After addition of the ethanol, the heat source is removed and the mixture is alllowed to reach room temperature. The precipitate is rapidly filtered off, washed with several hundred mL od cold water and dried in a desiccator over magnesium sulfate. The product, copper fulminate, should be stored in a desiccator over sodium sulfate in a refrigerator.\n> \n> \n> \n\n\n**NOTE** \n\n\n1. To my opinion, the procedure leaves out significant information, such as the **exact** temperature of the aqueous solution while adding the ethanol. It is not clear whether and how much ethanol is supposed to evaporate. While the author suggests to carry out the reaction in a beaker, a setup with a two-neck flask, reflux condenser and a dropping funnel might be a better idea.\n2. Handling fairly concentrated nitric acid requires safety precautions: A fume hood, eye protection, gloves and a lab coat are mandatory!\n3. The reaction product is a **primary explosive**! While less shock sensitive than mercury fulminate, it still will go off on impact (or upon heating above ~200 °C).\n\n\nIn the case of these or similar sensitive materials, there is a good rule of thumb: **If you have to ask on a Q&A site - Don't touch it!**\n\n\n", "8" ] ]

[ [ "\nYou are right! There is no molecular geometry in any chart that has two lone pairs and one bond, because the only possible geometry with one neighboring atom is linear.\nThe central atom in a chemical moiety **must have at least two neighboring atoms** in order to predict the geometry around this central atom (according to VSEPR model).\nLet's take the case of oxygen atom of the hydroxyl group (in Ibuprofen). The oxygen atom here, as a central atom, has the steric number: 4 ( It has two lone pairs and two simple bonds with carbon and hydrogen atoms). So, The molecular geometry around oxygen atom is bent. \n\n\n", "3" ] ]

[ [ "\nFor ice I melting pressure as a function of temperature:\n\n\n$$\\pi = 1- 0.626000 \\times 10^6 (1- \\theta^{-3}) + 0.197135 \\times 10^6 (1- \\theta^{21.2})$$ \n\n\nwhere\n\n\n$\\pi =$ (pressure in MPa)/0.000611657 MPa\n\n\n$\\theta =$ (temperature in kelvins)/273.16K\n\n\nSource is [International Equation of the Pressure along the Melting and along the Sublimation Curve of Ordinary Water substance](http://www.nist.gov/data/PDFfiles/jpcrd477.pdf) (no paywall).\n\n\nFor carbon dioxide:\n\n\n$$\\frac{p\\_m}{p\\_n} = 1 + a\\_1 (\\frac{T}{T\\_t}) + a\\_2 (\\frac{T}{T\\_t})^2$$\n\n\nWhere:\n\n\n$T$ is temperature in kelvins\n\n\n$T\\_t =$ 216.592K\n\n\n$p\\_m$ is the melting pressure in MPa\n\n\n$p\\_n =$ 0.51795 MPa\n\n\n$a\\_1 =$ 1955.5390\n\n\n$a\\_2 =$ 2055.4593\n\n\nSource is [A New Equation of State for Carbon Dioxide...](http://www.nist.gov/data/PDFfiles/jpcrd516.pdf) (no paywall)\n\n\nCheck original sources before using equations in case I made typo(s). \n\n\n", "4" ] ]

[]

[ [ "\nIt seems you were on right path - for example common [feldspars](https://en.wikipedia.org/wiki/Feldspar) are [solid solutions](http://%20en.wikipedia.org/wiki/Solid_solution) and they are mixed because they share cations, while they easily substitute one another, otherwise they exsolve - become separate phases. Also you're right that [van der Waals forces](https://en.wikipedia.org/wiki/Van_der_Waals_force) aren't important here as they are ionic compounds.\n\n\n", "2" ] ]

[ [ "\nBe sure you have really obtained TS, that means stationary point, i.e. all gradients are zero. Than, you should have one imaginary frequency corresponding to the motion along the reaction coordinate. If this is true and you still get both IRCs going to same minimum, your assumptions regarding the shape of Potential Energy Surface (PES) could be wrong.\n\n\nIt is easier to find the minima first, so be sure you have well optimized reactant, product, and if by any means reasonable, also intermediates. It is possible that there exist some high energy intermediates, which complicate the whole story (if you are not aware of them).\n\n\n", "5" ], [ "\nIt could also be the case (impossible to tell, as you haven't told us what molecules you are using) that the two minima are equivalent by symmetry.\n\n\n", "2" ], [ "\nWhen I generate an IRC from a TS - using GAMESS (US) software - I calculate the IRC for the forward direction and the IRC for the backwards direction using MacMolPlt. Is it possible you've stitched together two forward (or two backward) direction IRCs by accident?\n\n\nAlso, I once tried to calculate and IRC from a TS but both the forward and backward IRCs went in the same direction. I don't know why ... I got around it by cribbing someone else's code.\n\n\nA good place for code is the GAMESS (US) user group on google groups.\n\n\nHope the above helps...\n\n\n", "1" ] ]

[ [ "\nTake $\\pu{200 mL}$ of the $30\\%$ solution, add water to a final volume of $1$ liter. Then you have a $6\\%$ solution. Next, take $\\pu{59 mL}$ of the $6\\%$ solution and add water until it is $1$ liter.\n\n\nAfter the first dilution it's not exactly $6\\%$, because the density of $30\\%$ $\\ce{H2O2}$ is $\\pu{1.135 g/cm3}$. It's more like $6.6\\%$, but the instructions are referring to it as $6\\%$. If you want exactly $6\\%$ you should dilute by weight instead of volume.\n\n\n", "10" ] ]

[ [ "\nPreamble\n========\n\n\nFor a reaction with no intermediate steps, you could derive the [rate law](http://en.wikipedia.org/wiki/Rate_equation) from the reaction equation. For a generic equation\n$$\\ce{$a$A + $b$B -> C}$$\nthe rate is given by\n$$r = k\\cdot c^{a}(\\ce{A})\\cdot c^{b}(\\ce{B}).$$\nRigorously, this is only possible for [elementary reactions](http://goldbook.iupac.org/E02035.html). A prominent example for a strictly unimolecular reaction is radioactive decay. Also quite commonly known is the photolytic or thermal cleavage of hydrogen or bromine:\n\\begin{align}\n\\ce{H2 &->[\\Delta T] 2H. }\\\\\n\\ce{Br2 &->[h\\nu] 2Br. }\n\\end{align}\nAn example for a bimolecular reaction would be\n$$\\ce{2NO2 -> NO3 + NO}$$\n\n\nReaction mechanism of the combustion of hydrogen bromide\n========================================================\n\n\nThis seems to be a reaction with a fairly complicated mechanism, involving multiple species, most of them radical. I would assume, that it is somewhat similar to the combustion of hydrogen, that I have described [here](https://chemistry.stackexchange.com/q/14704/4945).\n\n\nA quick search on some forums will turn out a reaction mechanism, which I think is wrong. But I will show it here anyways, since it gives you a first clue, about how complicated this reaction might be. *(Note: These are supposed to be gas phase reactions.)*\n\\begin{align}\\ce{\nHBr + O2 &-> HOOBr\\\\\nHOOBr + HBr &-> 2 HOBr\\\\\nHOBr + HBr &-> H2O + Br\\\\\n}\\end{align}\nThe following mechanism is taken from [Lilian G. S. Shum and Sidney W. Benson *Int. J. Chem. Kin.*, **2004**, *15* (4), 341-380.](http://dx.doi.org/10.1002/kin.550150404) I think this is a much more suitable and credible approach to the problem. *(Note: Most of the reactions are in equilibrium.)*\n\\begin{align}\\ce{\nHBr + O2 &~<=> HO2 + Br\\\\\nHO2 + HBr &~<=> H2O2 + Br\\\\\nH2O2 + 2HBr &~->T[wall] 2H2O + Br2\\\\\nBr + Br + M &~<=> Br2 + M\\\\\\hline\n4HBr + O2 &~<=> 2H2O + 2Br2\\\\\n}\\end{align}\n\n\nUnfortunately I was unable to determine what $\\ce{M}$ refers to in this case. It might be any inert gas used as a solvent in the reaction or the wall.\n\n\nRate law of the combustion of hydrogen bromide\n==============================================\n\n\nAs a consequence of the mechanism you cannot find the rate law from theory, but only from experiments. It is important to realize, that most of the reactions are in equilibrium and the whole system is dependent on the concentrations of each of the involved species. Therefore the kinetics of this reaction may be expressed as a complicated system of coupled differential equations.\n\n\nHowever, Shum and Benson conclude, that the first step will be the rate-determining and the rate expression is given through\n\n\n\n> \n> $$\\frac{\\mathrm{d}[\\ce{H2O}]}{\\mathrm{d}t} = 2k\\_g[\\ce{HBr}][\\ce{O2}].$$\n> \n> \n> \n\n\nTherefore it can formally be regarded as a bimolecular reaction.\n\n\n", "3" ] ]

[ [ "\nElimination via a E2 mechanism cannot take place at the bridgehead carbon of a bicyclic compound due to the lack of the required trans geometry between the leaving group at the bridgehead carbon and any of the six β-hydrogens. \n\n\nOn the other hand, as you have correctly observed, a E1 mechanism has to be ruled out as well due to the strain at the bridgehead carbocation. In fact bicyclic structures prevent the tertiary bridgehead carbon becoming planar, which implies that is very high in energy because the nonplanar structure forces the cation to be an empty \"sp3 orbital\" instead of an empty \"p orbital\". This observation was made for the first time by Bredt, therefore is generally known as \"Bredt's rule\".\n\n\n", "1" ] ]

[ [ "\nA container that is not at ambient temperature will generate air currents around it. If you place such a non-ambient container in a balance, air currents will develop around the container as it heats or cools to ambient. These air currents will cause the balance to read incorrectly.\n\n\n", "19" ], [ "\nThe other two answers here are fine; it's true that as the hot object you're trying to weigh warms the air around it, the air will rise, creating air currents that will give you an unsteady reading. \n\n\nI'd like to add three things they didn't mention:\n\n\n* The balance pan will heat up, too, causing metal parts in the balance to expand. This will also contribute to an error in your reading, and as the metal cools, the reading will change.\n* Another contribution to the error is that the density of a hot sample is less than the density of a cold sample, so its buoyancy is different. The object will appear to gain a tiny amount of weight as it cools!\n* Don't worry about $E=mc^2$ in chemistry unless you've got some serious nuclear reactions going on.\n\n\nAnd a fourth thing: if you put a hot object into an air-tight dessicator, the air inside will contract as the object cools---and you might have some trouble getting the lid off again! \n\n\n", "10" ], [ "\nYour gut feeling is right: **Very hot crucibles *can* cause damage to the balance**. However, this isn't the main reason for why chemists cool their samples down before measurement.\n\n\nWe are all familiar with [convection currents](http://en.wikipedia.org/wiki/Convection_current): When air is heated the molecules will collide more than its previous state; resulting in an increase in either volume or pressure depending on the \"flexibility\" of the environment of effect. Thus, the density of hot air decreases, and it goes up and \"replaces\" the cold air. If the process of heating the air continues, hot air will continuously go up in what we call a \"convection current\".\n\n\nThe system of the crucible is thermodynamically open, so in contact with air it will lose some of its heat to its surrounding gas, which is normally atmosphere air. If we put the hot crucible for measuring its mass, convection currents will occur around the crucible, where it's in contact with the surrounding air. The motion of the hot air upwards forces your crucible upwards, but in small amounts.\n\n\nThe balance has to be precise enough to detect these differences, though. Most balances in the labs can measure up to 0.01 or 0.001 gram, and *will sense* the difference. Since you usually need (or are made to write) exact measurement results up to 1 miligram, these differences are going to be irking pests in measurement. \n\n\nThis is a very useful [source](http://content.bfwpub.com/webroot_pubcontent/Content/BCS_5/Harris_Exploring%20Chemical%20Analysis%205e/Lab%20Experiments/5eExpts%20for%20Web%20Nov%202011.pdf) from which the answer came from. It also has useful recommendations on practical problems in the lab.\n\n\n", "5" ], [ "\nWhile these are all very good answers, I have a much simpler one. These far overthink the question. The crucible has to cool in the desiccator to avoid adding moisture to it. The experiment you're doing deals with hydrates, therefore getting as much moisture out as possible is important as it will reduce your percent error.\n\n\n", "0" ], [ "\nYou must let the crucible cool before measuring it because the heat from the crucible warms the surrounding air, which rises, then that air cools down and falls. This rise and fall of surrounding air is called a convection current and will give you an unsteady reading that is rising and falling. These currents will make it hard to find the weight of the anhydrate(CuSO4), thus making it hard to find *n*, the moles of water per mole of CuSO4.\n\n\n", "-1" ] ]

[ [ "\n*Note that like most of the colleagues on this site with practical lab experience, I've used activated charcoal once in a while, but I never bothered to make it myself.* \n\n\nSupposed that you live on the countryside with enough space around, you can make your own charcoal in a earth-covered pile or an oven - unless local environmental regulations ban that.\n\n\nActivation, even when starting from ordinary charcoal is different. The **gas activation** requires heating (and partial burning) of the charcoal at up to 1000 °C in a stream of air, carbon dioxide or water vapour. I'd rather not do it in my backyard.\n\n\n", "2" ], [ "\nSorry for reviving this question, but I just saw it today.\nSince you already have charcoal, I guess you don't need the process required for making charcoal. But, you know, just in case:\n\n\n[making charcoal](http://www.survivalnewsonline.com/index.php/2013/12/making-charcoal/)\n\n\n[activated charcoal from charcoal](http://www.survivalnewsonline.com/index.php/2014/01/activated-carbon-from-homemade-charcoal/)\n\n\nThe only extra thing you need is calcium chloride. There's a nice video explaining all the steps in great detail, if you need it. I used these sites myself for making activated charcoal and they worked pretty well.\n\n\nNote: this will not produce high quality grade charcoal but it's effective.\n\n\n", "0" ] ]

[ [ "\n\n> \n> does water's polarity help explain its unusual phase properties?\n> \n> \n> \n\n\nNot just its polarity, but **its ability to form [hydrogen bonds](http://en.wikipedia.org/wiki/Hydrogen_bond)**, together with the size and shape of its molecules. \n\n\nHydrogen bonds give water an unusually high boiling point for a molecule its size; many molecules of similar size that can't hydrogen-bond are gases at room temperature. They also give water a relatively high heat capacity and high enthalpies of fusion and vaporization. These latter two properties mean that ice has to absorb more heat to melt and liquid water has to absorb more heat to boil than would otherwise be necessary without hydrogen bonding. \n\n\nWater's shape lets it form 4 tetrahedrally arranged hydrogen bonds. This lets it hydrogen-bond [into networks of chair-shaped hexagonal rings in ice](http://www1.lsbu.ac.uk/water/hexagonal_ice.html). The voids in the center of the rings make ice less dense than water. This is unusual; the solid form of a compound is most often denser than the liquid form. \n\n\n", "4" ], [ "\n[Vapor pressure of water](http://en.wikipedia.org/wiki/Vapour_pressure_of_water) is quite small, thanks to H-bonding in the liquid. Therefore more energy is needed to evaporate (break H-bonds with neighboring molecules) a water molecule.\n\n\nIf you compare water to \"nearest\" non-H-bonded analog, dimethylether, $\\ce{CH\\_3-O-CH\\_3}$ you'll find, that [vapor pressure of ether](http://encyclopedia.airliquide.com/encyclopedia.asp?GasID=80) is much higher (500 kPa) than water (2.3 kPa).\n\n\n", "2" ] ]

[ [ "\nAs mentioned in the comments, safety is the main priority. Take particular heed to the warning labels and information on the chemical label - if unsure, look up the chemical details MSDS (or SDS) = (Materials) Safety Data Sheet.\n\n\nOne of the most important thing to do (and is often overlooked) is to keep an inventory of what chemicals you have. The safety data sheets can be obtained from the manufacturers or by looking up the chemical SDS itself, or from general websites such as [MSDS.com](http://www.msds.com/) - an example is [this one for mineral turpentine](http://www.illawarrasurf.com/msds/turps.pdf) - these contain considerable information about storage, disposal and first aid. (It is always a good idea to have this readily available and updated).\n\n\nPersonal protective gear is a must - gloves, safety glasses etc - usually the label or the SDS will have this information.\n\n\nAccording to the document [NERC Guidance on Safe Storage of Laboratory Chemicals](http://www.nerc.ac.uk/about/policy/safety/procedures/guidance_chemical_storage.pdf) (2010), has information that is pertinent to a workshop environment. There are 3 main principles in regard to storage of chemicals explained in the document:\n\n\n* **Segregation**, from the article:\n\n\n\n> \n> The key incompatibles to segregate from each other are strong acids from\n> strong bases and strong oxidisers from organic or flammable materials. \n> \n> \n> \n\n\nThe NERC document has a list in its appendix (too long for this reply)\n\n\n* **Separation**, related to the segregation,\n\n\n\n> \n> In a laboratory situation adequate separation can be achieved by means of\n> storage cupboards which physically divide incompatible classes of hazardous\n> chemicals. The cupboards may need specific properties or provide\n> separation by means of distance. They will also need to provide secondary\n> containment (eg spill trays or bunded shelves) and security (eg locks / bolted\n> to wall). \n> \n> \n> \n\n\n* **Ventilation** (from personal experience, this one is often overlooked).\n\n\n\n> \n> Ventilation is often an essential requirement for safe storage of hazardous\n> chemicals. Its main function is to allow dilution and extraction of vapours or\n> gases that may escape / seep out from containers during storage so they no\n> longer present problems from the viewpoint of noxious smell, hazardous\n> personal exposure or creation of an explosive atmosphere. \n> \n> \n> \n\n\nAlso consider an appropriate fire extinguisher, note that water is probably the worst one. [Here](http://www.wormald.com.au/fire-products/fire-extinguishers/extinguisher-selection-and-fire-classification) is an example of a fire extinguisher type guide - this guide suggests that powder based extinguishers may be best (but, make sure you study it and seek advice if unsure).\n\n\nSome of this may seem 'overkill', but it is better to have some preparedness for if anything goes wrong, than be unpleasantly surprised by it (from personal experience), also, depending on where you are, there could be a legal requirement for a degree of proper storage, handling and protection.\n\n\n", "10" ] ]

[ [ "\n$\\ce{H2O}$ does not absorb the radiation almost at all, so it will not interfere in the absorbance of the analyte we are analysing, thats why we use water almost always as a solvent in instrumental methods of analysis.Therefore $\\ce{MeCN}$ wil absorb some of the radiation so then of course the absorbance of the analyte will also change ( decrease ) and the spectrum will be different ( the peaks of the analyte will be lower ). \n\n\n", "1" ], [ "\nI haven't looked for the spectrums for the compound in the different solvents, so I can only theorize what the differences would be. Since the nickel ion has the energy levels for visible light, then the differences would be due to what surrounds the nickel ion. In water, the nickel ion would be surrounded by water, where as in acetonitrile, the molecule $\\ce {[NEt4]2[NiBr]}$ would stay together.\n\n\n\n> \n> would one of them have more peaks?\n> \n> \n> \n\n\nNo, since it is the nickel atom that influences the visible spectrum.\n\n\n\n> \n> Also how would the molar extinction coefficients differ and what do these numbers tell you?\n> \n> \n> \n\n\nI cannot predict the difference in the molar extinction coefficient. The coefficient would tell you how easy the transition took place in the atom. The larger the number, the easier the transition.\n\n\n", "0" ] ]

[ [ "\nIf the proton spectrum is unbroadened and not shifted out of the ~0-10ppm range, then I would have to say it is a diamagnetic compound.\n\n\nIt could be low spin square planar. Tetrahedral and octahedral d8 cannot be diamagnetic.\n\n\n\n\n\nImage source: <http://upload.wikimedia.org/wikipedia/commons/a/a6/Chem507f09sqvstet2.png>\n\n\nCould also be two nickel atoms in the complex that are antiferromagnetically\ncoupled for a net spin of zero.\n\n\n", "4" ] ]

[ [ "\nWhen burning a blunt, joint, doobie or whatever, THC evaporates (bp of tetrahydrocannabinoles is around 155 °C). Depending on \n\n\n* the length of the \"object\" , related to the temperature gradient between the front and the mouth piece, and\n* the type of the \"filling\" and its adsorptive properties\n\n\na part of the vaporized material might condense and/or get absorbed in a colder region of the filling. Similar effects, such as the enrichment of nicotine in the stump of cigarettes have examined.\n\n\nIt is however unlikely that any of the transported and reabsorbed components can be inhaled from a cold \"reactor\". The relevant components are no longer in the vapour phase and the vapour pressure of the tetrahydrocannabinoles will be too small.\n\n\n\n", "7" ] ]

[ [ "\nIn potentiometric analysis, there are two electrodes, an indicator electrode (platinum here) where reduction takes place and reference electrode (calomel electrode) where\noxidation takes place. \n\n\nSo here the reference is not $E\\_{STP}$ cell but the $E\\_{cathode}$ that is the calomel electrode, so there is drop to $-241mV$ after the equivalance point. Hence,\n $E\\_{cell} = E\\_{cathode}$\n\n\n", "1" ] ]

[ [ "\nMany spectroscopic experiments record analog data, which is subsequently Fourier transformed to give a frequency-domain spectrum. For example, in a nuclear magnetic resonance experiment, the signal emitted by an excited nucleus is picked up as a voltage in an RF coil for 1-2 sec. This time-domain signal is processed then converted to the frequency domain by FT. \n\n\nAs for 'popular', Richard Ernst won the 1991 Nobel in Chemistry for his contributions to this field of research. \n\n\n", "1" ], [ "\nHow about the [patch-clamp technique](http://en.wikipedia.org/wiki/Patch_clamp) in neurochemistry? This technique allowed the study of ionic currents through single molecules of a voltage-gated ion channel. It won the [Nobel prize](http://www.nobelprize.org/nobel_prizes/medicine/laureates/1991/press.html) in 1991. \n\n\nThe signal processing required to simply observe the single-molecule opening and closing events is fairly minor. However, processing single-molecule ion-current data for multiple experiments when inhibitors are titrated, or to compare ensemble-average kinetics to single- molecule kinetics, can be fairly complex.\n\n\nHere is a paper from 1998 on single-molecule enzymology; it's an extension of the patch-clamp experiments developed for the specific case of single-molecule ion channels: <http://www.sciencemag.org/content/282/5395/1877.full> Read the paper for a better description of some of the signal processing involved.\n\n\n", "1" ] ]

[ [ "\nLet's consider the following redox couples:\n\n\n$\\ce{Cu^{2+} + e^- ->Cu^+\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,}{ E^o\\_1}(\\ce{Cu^{2+}/Cu^+}=0.17\\, \\ce{V})$\n\n\n$\\ce{Cu^{2+} + e^- + I^- ->CuI\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,}{ E^o\\_2}(\\ce{Cu^{2+}/CuI}=?)$\n\n\nAs the chemical species $\\ce{Cu^{2+}}$, $\\ce{Cu^+}$ and $\\ce{CuI}$ are together in the aqueous solution, the two redox couples $\\ce{Cu^{2+}/CuI}$ and $\\ce{Cu^{2+}/Cu^+}$ are in equilibrium and that means: $$ E\\_1(\\ce{Cu^{2+}/Cu^+})=E\\_2(\\ce{Cu^{2+}/CuI})$$ Using Nernst equation for each couple:$$ E\\_1(\\ce{Cu^{2+}/Cu^+})=E^0\\_1(\\ce{Cu^{2+}/Cu^+})+\\frac{RT}{F}\\ln{\\frac{[\\ce{Cu^2+}]}{[\\ce{Cu^+}]}}$$ \n$$ E\\_2(\\ce{Cu^{2+}/CuI})=E^0\\_2(\\ce{Cu^{2+}/CuI})+\\frac{RT}{F}\\ln{[\\ce{I^-}][\\ce{Cu^2+}]}$$\nWe can write then:\n$$E^0\\_2(\\ce{Cu^{2+}/CuI})= E\\_1^0(\\ce{Cu^{2+}/Cu^+})+\\frac{RT}{F}\\ln{\\frac{[\\ce{Cu^2+}]}{[\\ce{Cu^+}][\\ce{Cu^2+}][\\ce{I^-}]}}$$ We rewrite the above equation after simplification:$$E^0\\_2(\\ce{Cu^{2+}/CuI})= E\\_1^0(\\ce{Cu^{2+}/Cu^+})+\\frac{RT}{F}\\ln{\\frac{1}{[\\ce{Cu^+}][\\ce{I^-}]}}$$\n$$E^0\\_2(\\ce{Cu^{2+}/CuI})= E\\_1^0(\\ce{Cu^{2+}/Cu^+})+\\frac{RT}{F}\\ln{\\frac{1}{K\\_{sp}(\\ce{CuI})}}$$\n\n\n$$E^0\\_2(\\ce{Cu^{2+}/CuI})= 0.17+\\frac{8.314 \\times 298}{96500}\\ln{\\frac{1}{10^{-12}}}=0.88 \\ce{V}$$\n\n\n", "4" ] ]

[ [ "\nToothpaste is what is called a [**non-newtonian fluid**](http://en.wikipedia.org/wiki/Non-Newtonian_fluid), more specifically toothpaste is a [**Bingham plastic**](http://en.wikipedia.org/wiki/Bingham_plastic). This means that the viscosity of the fluid is linearly dependent on the shear stress, but with an offset called the yield stress (see figure below). This yield stress is what makes it hard to say whether it is liquid or solid. The fact that toothpaste is viscous *alone* is not sufficient to explain this, because water is also viscous, but doesn't behave like a solid (unless frozen, but that's another phenomenon).\n\n\n\n\n\nWhat the yield stress does is the following. Below a certain shear threshold the fluid responds as if it were a solid, as you can see happening when you have put toothpaste on your toothbrush, it just sits there without flowing away. A highly viscous but newtonian fluid would flow away (although slowly as pointed out by @ron in his comment to the answer of @freddy).\n\n\nNow if you put sufficient shear stress on the toothpaste, when you squeeze the tube of paste, it will start flowing and respond as a liquid.\n\n\nOther examples, as mentioned in the Wikipedia link in my first sentence, are e.g. mayonnaise and mustard. Another example is silly putty.\n\n\n", "106" ], [ "\nHere's a genchem-level answer for a genchem-level question about the classification of matter:\n\n\nToothpaste is a *sol*: a stable suspension of tiny solid particles in a liquid. When the toothpaste dries out you can see what the solid part alone looks like.\n\n\nMixtures with more than one phase often have interesting properties and behaviors that the components alone don't have; the other answers to your question touch on some of these. \n\n\nOther examples of sols are paints and solid-pigment inks. Again, you can see the solid part when they dry.\n\n\n", "44" ], [ "\nHere's the boring answer: Toothpaste is a mixture of some solids and some liquids.\n\n\nThe question \"is it solid or liquid?\" makes sense when you're talking about a substance or a mixture with a single [phase](https://en.wikipedia.org/wiki/Phase_(matter))—that is, a substance or mixture that's pretty much completely uniform throughout space. Examples of single-phase materials include pure water (a liquid), a chunk of copper (a solid), and sugar water (which is a liquid—including the sugar).\n\n\nBut some materials consist of multiple phases, and in this case, we have to ask the question \"is it solid or liquid?\" separately for each phase. Examples of multiple-phase materials include ice water (a mixture containing both a solid (ice) and a liquid (water)), vinaigrette (a mixture of oil (a liquid) and vinegar (another liquid)), and toothpaste (a mixture of several kinds of solids and liquids).\n\n\n", "22" ], [ "\nIt is **Viscous.**\n\n\nViscous mean \"having a thick, sticky consistency between solid and liquid\"(dictionary meaning)\n\n\nThere are many more examples like tomato ketchup, honey, wax, toothpaste, etc. \n\n\n\n\n---\n\n\nTo know more check out [Wikipedia](http://en.wikipedia.org/wiki/Viscosity) \n\n\n", "14" ], [ "\nToothpaste is a colloid. just like dust particles suspended in water form suspension, colloids are much finer particles suspended in a medium. for example jelly, it is solid particles suspended finely in a liquid....and foam; that is gas suspended in a liquid. As such is deodorant spray that is liquid suspended in air(gas)\n\n\n", "14" ], [ "\nActually, toothpaste is both. I'm no chemist, but I am pretty sure that it can be correctly classified as a **semisolid**, which means exactly what you'd think. Semisolids have properties of both solids and liquids. Slime would be another example of a semisolid.\n\n\n", "13" ], [ "\nToothpaste is a **colloid**, more specifically a type of *liquid dispersed in a solid dispersing medium*. The solid component is much more in proportion than the liquid one, so the paste is quite thick. You can't really call it a solid or a liquid, its a combination of both. As Michiel says, it may be considered as a *non-Newtonian fluid*. If you have seen a toothpase, you could observe these:\n\n\nHas no definite shape (liquid property)\nCannot flow (solid property)\nSlight tendency to decrease surface area (liquid property)\n\n\nSo you can see that it has properties similar to both solids and liquids. Its a colloid of a liquid in solid.\n\n\n", "6" ] ]

[ [ "\nThere's no measuring the radius of a single atom, mainly because electrons are around the nuclei: **We can't define a radius for any single atom.** Take a loom at the uncertainty principle:\n\n\n\n> \n> Introduced first in 1927, by the German physicist Werner Heisenberg, it states that the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa. The formal inequality relating the standard deviation of position $\\sigma \\_x$ and the standard deviation of momentum $\\sigma \\_p$ was derived by Earle Hesse Kennard later that year and by Hermann Weyl in 1928: $$\\sigma \\_x \\sigma \\_p \\geq \\frac{\\hbar}{2}$$ [Wikipedia](http://en.wikipedia.org/wiki/Uncertainty_principle)\n> \n> \n> \n\n\nThus, if you want to measure an atom's radius, you have to measure the distance between two nuclei, and then divide it by two. This results in different types of radius defined for atoms. The most common are [covalent radius](http://en.wikipedia.org/wiki/Covalent_radius), [ionic radius](http://en.wikipedia.org/wiki/Ionic_radius), and [Van der Waals radius](http://en.wikipedia.org/wiki/Van_der_Waals_radius). Radii trends are what that are mostly studied in undergrad chemistry, rather than the how of measuring them with advanced techniques.\n\n\nChemguide is pretty informative when it gets to teaching the trends:\n\n\n[chemguide: atomic radius](http://www.chemguide.co.uk/atoms/properties/atradius.html)\n\n\n", "3" ] ]

[]

[ [ "\n[The IUPAC rules for punctuation marks](http://old.iupac.org/reports/provisional/abstract04/RB-prs310804/Chap2-3.04.pdf) in chemical formulas don't mention [|], so that notation must be peculiar to geology. They *need* different notation; chemists write formulas for molecules, while geologists focus more on complicated superstructures built from smaller structures that can be arranged in different ways, even within the same superstructure. \n\n\n[This paper](http://www.academia.edu/7540967/Brochantite_Cu4SO4_OH_6_OD_character_polytypism_and_crystal_structures) describes the structure of brochantite as layers and chains of copper(II) ions octahedrally coordinated with hydroxides. The octahedra in different layers and chains are crosslinked by sulfate ions. \n\n\nPerhaps the notation $\\rm [(OH)\\_6|SO\\_4]$ is an attempt to indicate this arrangement. The brackets denote the electronegative part of the structure, with what comes before the vertical bar representing directly coordinated ions, with what comes after representing ions that aren't directly coordinated. \n\n\nI'm speculating, though. Sorry. I would have put this into a comment, but it's too long and I wanted to give you the links to think about.\n\n\n", "3" ] ]

[ [ "\nWhen some number of atomic orbitals are mixed together to produce hybrid atomic orbitals, the same number of hybrid atomic orbitals must result. So, when N atomic orbitals are mixed together, N hybrid atomic orbitals are produced. In the case you describe, mixing one s atomic orbital with two p atomic orbitals means that three hybrid atomic orbitals must result.\n\n\nThese three hybrid orbitals will each contain 33.333% s character and 66.666% p character because that is the ratio of the one s orbital to two p orbitals that we are mixing together. We can describe such an orbital as $\\ce{s^{33.333}p^{66.666}}$, but if we reduce that notation to the lowest common denominator we get $\\ce{s^1p^2}$, or the commonly used description $\\ce{sp^2}$. This simply means that this hybrid orbital is composed of one part s character to two parts p character.\n\n\n**To summarize if we mix 3 atomic orbitals, one s and two p orbitals, then 3 $\\ce{sp^2}$ atomic hybrid orbitals will result.**\n\n\n", "3" ] ]

[ [ "\n1. Under acidic conditions, a blue colour is observed. This results from the formation of N,N'-diphenylbenzidine. In the course of the reaction, diphenylamine is oxidized to a quinoid radical which subsequently adds to another molecule of diphenylamine.\n2. Under slightly acidic conditions, bleaching of the solution is observed. Sulfite is oxidized to sulfate while $\\ce{MnO4-}$ is reduced to $\\ce{Mn^{2+}}$.\n\n\n", "2" ] ]

[ [ "\nMy advice is to use copper and protect it with a sacrificial anode. Copper would be the choice in a non corrosive environment and if you can protect it go for it.\n\n\nHowever, if the coils will contact parts of the boat that are important than the coils should be aluminum. Copper would cause the other metal parts to corrode faster.\n\n\nWhy not use rubber or plastic, something like the pipes used in drip irrigation?\n\n\n", "1" ], [ "\nhave you thought about using cupro-nickle, which is used exensively in marine engineering. I am aware it is expensive, but further research should reveal its suitability. I hope that you have found this helpfull. \n\n\n", "0" ] ]

[ [ "\nSpray it as a fine mist into a drying chamber - the solvent will evaporate and you are left with a powder. \nNo chemical involved though, but still process chemistry. \n\n\n", "6" ], [ "\n1. If you decrease the solvent volume, the dissolved material will start to precipitate.\n2. Solubility most often increases with the temperature of the solvent.\n3. Slow decrease of the temperature or slow evaporation of the solvent at constant temperature mostly leads to slow crystallization, starting from a small(er) number of nucleation centres and typically results in the formation of less but larger, well-formed crystals.\n\n\nRapid decrease of the temperature with additional scratching the surface of the crystallisation vessel (beaker, etc.) with a glas rod, stirring, or sonification will typically result in precipitation of the solute in the form of smaller particles.\n\n\n", "2" ] ]

[ [ "\nAll complex anions you noted that include only non-metal atoms are always considered covalent. Except for a few extreme examples including fluorine and hydrogen, non-metals only form predominantly covalent bonds between themselves.\n\n\nIt gets a little more tricky when considering those anions whose central atom is a transition metal. In this case at first approximation, we can consider the anion a coordination compound and the coordinate bonds within coordination compounds are typically considered more covalent. Moving on to a second approximation clarifies the picture: the formal oxido-ligands are in fact very strongly attached to the metal and do not undergo ligand substitution. This means that they are even *more* covalent than other coordination complexes.\n\n\nTherefore, it is typically safe to consider the bonds within complex anions to be predominantly covalent.\n\n\n", "6" ], [ "\n\"Electronegativity\" is a measure of the power of an atom in a molecule to attract shared electrons to itself. If you look up the electronegativities for the atoms on either end of a bond, there's a quick and dirty way to decide for yourself how much covalent character the bonds have.\n\n\nCalculate the electronegativity difference between the central atom and oxygen. The smaller the difference, the more covalent character the bond has. For example, for Si-O, Si and O have Pauling electronegativities of 1.8 and 3.5, respectively, so the electronegativity difference is 1.7.\n\n\nSome textbooks set (rather arbitrary) thresholds for electronegativity differences to help general chemistry students classify bonds as non-polar covalent, polar covalent, or ionic. Pauling's original text says that if the electronegativity difference is 1.7, the bond has about 50% ionic character, so you can say an Si-O bond is about 50% ionic, 50% covalent. If the electronegativity difference is greater than 1.7, you can call it \"predominantly ionic\"; if it's less than 1.7, you can call it \"predominantly covalent\". \n\n\n", "3" ] ]

[ [ "\nHomologation methods exist, but typically not for unbranched alkanes.\nA typical example is the Arndt-Eistert homologation of carboxylic acids.\n\n\nFor simple **unbranched** alkanes, stepwise chain elongation is not a technical relevant process. On the contrary, petrochemical processes reach for the opposite (formation of smaller alkanes) by thermal or catalytical cracking.\n\n\nFormation of long chain alkanes, although not via subsequent step up reactions, is possible by catalytic oxidative dimerization of terminal alkynes (Glaser, Hay or Eglinton coupling) and subsequent hydrogenation or by Kolbe dimerization of carboxylic acids.\n\n\n**Branched** alkanes, such as triptane (2,2,3-trimethylbutane), which are interesting as 'anti-knocking' fuel additives, are another league. Here, the source of the methyl group to be transfered is methanol, catalysts are either zeolites or indium(III)-iodide ($\\ce{InI3}$) ([DOI](http://dx.doi.org/10.1021/ja803029s)).\n\n\n", "9" ], [ "\nOne method could be metathesis of alkanes. There is a nice paper about it published by [J. M. Basset et al., SCIENCE 1997, 276, 99-102, \"Metathesis of Alkanes Catalyzed by Silica-Supported Transition Metal Hydrides\"](https://dx.doi.org/10.1126/science.276.5309.99) and for sure more recent ones have been published since then. \n\n\n\n> \n> **Abstract**\n> The silica-supported transition metal hydrides $\\ce{(≡Si-O-Si≡)(≡Si-O-)2Ta-H}$ and $\\ce{(≡Si-O-)xM-H (M, chromium or tungsten)}$ catalyze the metathesis reaction of linear or branched alkanes into the next higher and lower alkanes at moderate temperature ($\\pu{25^\\circ}$ to $\\pu{200^\\circ C}$). With $\\ce{(≡Si-O-Si≡)(≡Si-O-)2Ta-H}$, ethane was transformed at room temperature into an equimolar mixture of propane and methane. Higher and lower homologs were obtained from propane, butane, and pentane as well as from branched alkanes such as isobutane and isopentane. The mechanism of the step leading to carbon-carbon bond cleavage and formation likely involves a four-centered transition state between a tantalum-alkyl intermediate and a carbon-carbon bond of a second molecule of alkane.\n> \n> \n> \n\n\n", "9" ], [ "\nCarbenes, like $\\ce{CH2}$, are known to insert into everything, including, in case of absence of better substrate, carbon-hydrogen bonds. However, the process is not clean, as carbene will insert into any suitable position it can find, with no selectivity. Carbenes are usually produced by decomposition of precursors, usually ketene ($\\ce{CH2=C=O}$) or diazomethane ($\\ce{CH2=N=N}$). \n\n\nIn case there is some group in the compounds, it is often possible to selectively substitute hydrogen, activated by the group. For example, many electronegative groups, like carbonyl group, activate nearby hydrogenes, making them quite acidic by organic chemistry standards. Produced anion can react with equivalent of carbocation. Carbonyl group may be later eliminated by Wolff–Kishner reduction (heating with hydrazine and alkaly metal hydroxide)\n\n\nHowever, there is no general way to increase chain of alkanes cleanly and selectively. \n\n\n", "4" ] ]

[ [ "\nTo measure the total chlorophyll from leaves, probably not, since you need access to methyl isobutyl ether or pyridine.\n\n\nBut you can separate the different pigments in the leaf. From [The Naked Scientist](http://www.thenakedscientists.com/HTML/content/kitchenscience/exp/chlorophyll-chromatography/): Cut a rectangle 1 cm wide and long enough to stand up in a small jar from a coffee filter. Place your leaf over the top of the strip and roll the coin across the strip about 2 cm from the bottom. Now add a little nail varnish remover to the bottom of your jar, half a cm is plenty. Suspend your strip of paper so the bottom end is sticking a few mm into the acetone. you can hold it up by folding the top end of the strip over forming a hook which will rest over the lip of the jar. Wait a few minutes, with any luck you should see an interesting effect. Try the experiment with a strongly colored leaf like a purple one. \n\n\n", "1" ] ]

[ [ "\nIt is used to make the conjugate base of the imide which is far more nucleophilic than the imide itself, but not too much to attack another molecule of the alkyl halide.\n\n\n", "3" ], [ "\nThe goal is to make a primary alkylamine without causing polyalkylation. If you do not activate the pthalimide nitrogen with a base, you have a highly delocaled lone pair on the nitrogen. At best, you get an Sn1 reaction, however that will not give a primary amine as loss of the halide (if you can force it to go) will undergo Wagner-Meerwein rearrangement (hydride shift) to give a secondary or tertiary amine. \nThis is the point of the Gabriel phthalimide synthesis.\n\n\n", "1" ] ]

[ [ "\nThe required HLB doesn't depend on the concentration of water. The required HLB of mineral oil is ca. from 9 to 12. You have to find out the required HLB of your mineral oil at first.\n\n\n", "0" ] ]

[ [ "\nI don't know what this standard or the \"first area\" are, but the hazard quotient is simply the ratio of potential exposure to the maximum exposure without adverse effects. You can't take the ratio of the two values you have because 1300 mg/L is a concentration and 837 µg is an absolute quantity. You need to first decide how much water your potential exposure will involve and multiply that by 1300 mg/L.\n\n\nIncidentally, 1300 mg/L of copper shouldn't be any kind of standard level in drinking water. That's more like a standard solution for AAS or ICP, is definitely bright blue, and probably contains several percent of nitric acid.\n\n\n", "1" ] ]

[ [ "\nOne mechanism of glycation is the [Maillard Reaction](http://en.wikipedia.org/wiki/Maillard_reaction).\n\n\nThe mechanism starts with nucleophilic attack of the sugar carbonyl group by a side chain amino group from the protein.\n\n\nThe Milliard Reaction occurs in the human body and has been implicated in various adverse health conditions.\n\n\nThe cornea, lens and retina are known to be damaged by this reaction.\n\n\nSee [The Maillard reaction in the human body. The main discoveries and factors that affect glycation](http://linkinghub.elsevier.com/retrieve/pii/S0369811409001928?via=sd&cc=y) Pathologie Biologie, Volume 58, Issue 3, Pages 214-219 for a recent review.\n\n\n", "1" ] ]

[ [ "\nHome chemistry can be difficult sometimes due to the expense of the essential working materials and instruments required. As a mathematician all I need is a sharp pen (which is sometimes still quite difficult to obtain!) and blank paper. I can give you the following general (personal) advice regarding purchasing chemistry lab equipment:\n\n\n* Determine your budget and the needs for the experiments you want to perform. Do you really need a scale that has to weight with an accuracy of 0.001g or is 0.01g sufficient (or even 0.1g)? Ask yourself what you REALLY need for an experiment BEFORE searching for instruments. If you do it in the converse order you often buy something way more fancier (and pricier) than what you actually needed in the first place.\n* Look at specialized chemistry equipment supply websites to orientate on what is available and in what price range certain products are (think of scales, chemicals, glassware, etc.)\n* A lot of stuff can be bought incredibly cheap on websites where you not expect (at least I did not) people to sell lab supplies/instruments. Think of [ebay.com](http://ebay.com) and [amazon.com](http://amazon.com) as examples. They sometimes list items at a fifth of the usual price that would have to be paid at a specialized store.\n* Take a look at local non-specialized stores like the construction market (you know, the place where they sell all kinds of construction material). They sometimes hide hidden (cheap) gems! As an example I was able to obtain a vacuum source (membrane pump) for < 100\\$, which would have cost me easily > 400\\$ at a chemical supply website.\n\n\nAbout your particular question: I really recommend a 0.01g precision 200g (or 600g) scale. It brings the benefit of having relatively accurate weighting for even small samples, while still being able to weight bulk mass. Furthermore, they are most common in the lab/home environment and hence relatively cheap in comparison with professional analytical scales. If you plan to do a lot of microscale experiments (yielding a product < 200 mg) the analytical scale (precision of 0.001g) may suit your needs better.\n\n\nI wish you much success, think and buy wisely!\n\n\n", "2" ], [ "\nLet me first say that @Jori's answer is excellent. I would only add two things.\n\n\n1) A range of equipment is usually desirable if it is also affordable. So measures across all 3 accuracies (0.1, 0.01, and 0.001) is what you want ideally, particularly given that precision tends to decrease with absolute mass.\n\n\n2) There is a trend towards green chemistry, which emphasizes minimal reactants, products, and waste whenever possible. Thus, there will be a trend towards microchemistry where small amounts of reactants are the norm, and this will necessitate higher precision balances (i.e 0.001g precision).\n\n\n", "1" ] ]

[ [ "\nIt's conceivable that somewhere out there, someone with a completely fresh point of view could have an *annus mirabilis* in an area like theoretical chemistry, working with just a paper and pencil (or a laptop) and home-brewed equipment and resources. Yes, for most of us, all of the low-hanging fruit is gone---but for a real giant, the whole tree might be full of low-hanging fruit. \n\n\nMost of us aren't giants, but as Newton said, we can stand on the shoulders those who are. Doing that requires some ability to search the extant literature, even for \"small to moderate contributions\". \n\n\n**In addition to the obstacles you've listed above, there is a third impediment: paywalls.** \n\n\nThere was a time when you could walk into a university library and sit down with the Science Citation Index and do a good job of finding research related to yours. Nowadays, you need a faculty or student university ID to do such a search and retrieve the papers you need. Without an ID at a good university, our amateur, hobbyist, or science buff will hit paywall after paywall. Amateurs might have to pay half a day's wages or more for the pdf of *a single paper*! \n\n\nPaywalls are not only a major obstacle for amateurs---they discourage professionals in the third world, and they shut out young people who are beginning their careers outside of university.\n\n\nConsider Jack Andraka's story, here: <http://blogs.plos.org/thestudentblog/2013/02/18/why-science-journal-paywalls-have-to-go/> \n\n\n", "4" ], [ "\nEric Betzig, one of this year's Nobel laureates in chemistry, did much of the work for which he won the prize (superresolved fluorescence microscopy) in his living room .\n\n\n", "3" ], [ "\nBack in the early days of chemistry, it was all done by \"amateurs\". And, as far as what costly equipment is needed--go back and look at an old pre 1900 textbook like Fownes' -- it is amazing how much chemistry they got right without NMR, IR, MS, HPLC, GC, etc.\n\n\n", "3" ], [ "\nI'm a mathematician by training, and I think math is a decent analogy to your question. The answer is that yes, amateurs could certainly produce new insights, but this rather low probability. The reason is that humanity has been applying the scientific method to all sciences for a long time, resulting in many new results. But this also means that the \"low hanging fruit\" in all sciences have been heavily picked. It's simply the case that most of the remaining problems are hard in any scientific discipline. You may still be able to find some non-researched problem, but this is increasingly unlikely. And if it is non-researched, it is likely **very** hard. So this does not rule out amateurs, but it does set a high bar.\n\n\n", "1" ], [ "\nWhat about the Mpemba effect? Does that count as chemistry? Maybe it's more physics or chemical engineering. \n\n\nErasto Mpemba fits the definition of a total amateur. He was in high school back then. \n\n\nOTOH, 1963 was half a century ago now. \n\n\nAt least in Chemical Engineering, a fair bit of innovation happens still by amateurs. Ok, not total amateurs, but plant operators, entrepreneurs, businessman, scrap dealer, contractors, foreman, et cetra. i.e. someone outside the conventional BS-MS-PhD academic system. \n\n\n", "1" ] ]

[ [ "\nGenerally speaking, there is no protection unless the sacrificial anode is immersed in the electrolyte. It might be better to anodize and seal the aluminum, which should protect it against most acids. There are many aluminum anti-corrosion treatments, but in any case the area under attack would have to be covered. See <http://en.wikipedia.org/wiki/Anodizing>.\n\n\n", "2" ] ]

[ [ "\nSome higher-molecular weight alkanes definitely have an odor, but they aren't as intense as most of the compound types you'll find stinking up an organic lab. \n\n\n* \"The odor threshold for hexane is 130 parts per million (ppm), with a faint peculiar odor reported.\" ---<http://www.epa.gov/ttnatw01/hlthef/hexane.html>\n* \"[Heptane is a] colorless liquid with a mild gasoline-like odor.\" ---<http://www.cdc.gov/niosh/docs/81-123/pdfs/0312.pdf>\n* \"Alkanes have relatively high odor detection threshold (37 to 50 mg/m$^3$ for dodecane) as compared to other VOC groups such as sulphides (0.0003 to 0.16 mg/m$^3$ for\ndimethylsulphide), phenolics (0.022 to 4 mg/m$^3$ for phenol), and nitrogen heterocycles (0.0004 to 0.0008 mg/m$^3$ for skatole) (O'Neill and Phillips, 1992)\" ---<http://www.bae.uky.edu/jbicudo/odorVOC_02a.pdf>\n\n\nAccording to the following paper, there are apparently at least two olfactory receptors in rats that respond to alkanes; one accepts alkanes with areas around 3.5 nm$^2$, and another responds to hydrocarbons with areas of about 5 nm$^2$. The authors note that response to cycloalkanes is higher than for normal alkanes, so the shape matters too.\n\n\n* \"Concanavalin A reveals olfactory receptors which discriminate between alkane odorants on the basis of size.\" ---<http://www.ncbi.nlm.nih.gov/pubmed/2803264>\n\n\n", "3" ] ]

[ [ "\nLow-lying, empty antibonding orbitals will suffice. The reaction of carbon dioxide and water to form carbonic acid serves as a parallel example. Water donates one of the lone pairs on oxygen towards bond formation, water is the Lewis base in this case. Carbon dioxide has empty, low-lying $\\ce{\\pi}$\\* orbitals that serve to accept these electrons; carbon dioxide is a Lewis acid in this example. \n\n\n", "2" ] ]

[ [ "\nI think the reagent you're talking about is ketene (R=H in the figure below). It is a linear molecule (the central carbon is $\\ce{sp}$ hybridized) and very reactive. \n\n\n\n\n\nWhen mixed with an alkane and heated it dimerizes to form diketene rather than react with the alkane.\n\n\n\n\n\nHowever, when irradiated with light, ketene will produce the reactive carbene \"methlylene\" ($\\ce{CH2:}$). When this photolysis is performed in the presence of an alkane, the methylene will insert itself into the various $\\ce{C-H}$ bonds like you've drawn. However the reaction is not synthetically useful for a number of reasons. For example, the $\\ce{C-H}$ insertion reaction is very indiscriminate and a variety of products will be formed if different $\\ce{C-H}$ bonds exist in the molecule. \n\n\n", "1" ], [ "\nI would describe the reaction as a C-H insertion reaction of the carbene generated in situ from ketene. The reaction starts with an initial release of carbon monoxide to generate the highly electrophilic 6 electrons carbene species, which is then involved in a C-H insertion process. The C-H insertion can proceed through two mechanism: via singlet state or triplet state of the carbene. The carbene is a very electrophilic sp2 hybridized species with 6 electrons: a lone pair and two pairs in bond orbitals (with H atoms). Singlet carbenes have the lone pair of electrons in a nonbonding sp2 orbital and have an empty p orbital, while triplet carbenes have two unpaired electrons, one in a sp2 and the other in a p orbital. C-H insertion mediated by singlet carbenes is believed to proceed through a concerted mechanism, with the side-on approach of the carbene which provide constructive orbitals overlapping. Triplet carbene insertions should follow a two-step radical pathway, but it is generally accepted that the reaction proceed through singlet state carbenes (supported by retention of stereochemistry observed when if the C-H bond is at a stereogenic center).\n\n\n", "0" ] ]

[ [ "\nThe relative stability of your two compounds is not determined by arguments based on hyperconjugation. Rather the relative stabilities have to do with **ring strain**. Placing a double bond in smaller and smaller rings increases the angle strain in the ring system. A double bond, being roughly $\\ce{sp^2}$ hybridized, would prefer to have bond angles around 120°. This becomes difficult in 3- and 4-membered rings where the internuclear angle is 60° and 90° respectively. Therefore molecules containing double bonds in 3- and 4-membered rings will be more strained than molecules with the double bond in a larger ring.\n\n\n**Edit:** Response to OP's comment\n\n\nNo, the given solution is not the correct answer to the problem. It is correct that structure B has more alpha (alpha to the double bond) hydrogens (4) than structure A (only 2 alpha hydrogens). It is also true that hyperconjugation can be used to explain why alkyl groups attached to a double bond act to stabilize a double bond (see [this earlier answer](https://chemistry.stackexchange.com/questions/19027/inductive-effect-and-hyperconjugation-one-elephant-different-parts/19032#19032) for an explanation), and that the double bond in B is more stabilized by hyperconjugation than the double bond in A. However, hyperconjugation has a much smaller effect on the stability of these molecules than the ring strain effect.\n\n\n", "7" ] ]

[ [ "\nYes, there would be a difference. The acids would be weaker. Why? Because the heavier isotope lowers the zero-point energy (the lowest possible energy) in the acid molecule, making the acid's bond to the D stronger than it would have been to the H. Acids in $\\rm D\\_2O$ are weaker than they are in ordinary water, too.\n\n\nAny reaction that involves dissociation of these acids will run more slowly, too. This is called \"the kinetic isotope effect\". You can read more about that here: <http://en.wikipedia.org/wiki/Kinetic_isotope_effect>\n\n\nYou might want to ask further questions: *why* is the zero point energy lower for a bond to D than a bond to H? How exactly does the lower zero point energy make the bond stronger? How exactly does it affect the rates of reactions that include acid dissociation?\n\n\nAsk away!\n\n\nIf you'd like to see how substituting a D for an H in acids affects the acid dissociation constant for particular acids, [here's a good reference](http://nvlpubs.nist.gov/nistpubs/jres/73A/jresv73An3p299_A1b.pdf): \n\n\nR. A. Robinson, M. Paabo, R. G. Bates, \"Deuterium Isotope Effect on the Dissociation of Weak Acids in Water and Deuterium Oxide\", *J. Res. Nat. Bureau. Standards*, **73A(3)**, 1969, pp 299-308.\n\n\n", "8" ] ]

[ [ "\nIf you compare the [magnesium price](http://www.metalprices.com/p/MagnesiumFreeChart) (about 2$ per pound)\n\n\nto the [aluminum price](http://www.metalprices.com/p/AluminumFreeChart) (about $0.80 per pound)\n\n\nyou can see it is unprofitable. \n\n\nEven considering the elements' relative atomic weight, you would still be losing money if you did that. \n\n\n", "4" ] ]

[ [ "\nYes it is true. \n\n\nConditions are temperature in the 650 to 850 degrees C range. \n\n\nThere is a [cool looking youtube video of the reaction](http://www.google.com/url?sa=t&rct=j&q=&esrc=s&frm=1&source=web&cd=6&cad=rja&uact=8&ved=0CEcQtwIwBQ&url=http%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DMJ3oCT_HMoE&ei=7Qq5VPqNJo3ksATKxYLYCg&usg=AFQjCNEY1LEmWs661zxDe8Mmz9uD4EyADA), not that that makes it true.\n\n\nFor a more serious discussion see [Production and Purification of Silicon by Magnesiothermic Reduction of Silica Fume](https://tspace.library.utoronto.ca/bitstream/1807/25788/3/Sadique_Sarder_E_201011_MASc_thesis.pdf)\n\n\nand [Ordered Mesoporous Silicon through Magnesium Reduction of Polymer Templated Silica Thin Films](http://pubs.acs.org/doi/abs/10.1021/nl801759x)\n\n\n", "3" ], [ "\nYes, it is true below $\\pu{1683 K}$, the melting point of silicon, the $\\Delta\\_\\mathrm{f}G^\\circ$ curve for the formation of $\\ce{SiO2}$ lies above the $\\Delta\\_\\mathrm{f}G^\\circ$ curve for $\\ce{MgO}$, so, at temperature below $\\pu{1683 K}$, $\\ce{Mg}$ can reduce $\\ce{SiO2}$ to $\\ce{Si}$.\n\n\nOn other hand, above $\\pu{1683 K}$ the $\\Delta\\_\\mathrm{r}G^\\circ$ curve for $\\ce{MgO}$ lies above for $\\ce{SiO2}$, hence at temperatures above $\\pu{1683 K}$, $\\ce{Si}$ can reduce $\\ce{MgO}$ to $\\ce{Mg}.$\n\n\n", "1" ] ]

[ [ "\nHmmm. I don't know, but I'd say that if you *significantly* lowered the percentage of $\\rm O\\_2$ in the air, the average molecular weight of the molecules in the bubble would drop (there'd be more low-molecular weight nitrogen and less higher-molecular weight oxygen), so the rate at which gas would diffuse out of the bubble would increase, and the foam would flatten faster. \n\n\nTry blowing a foam with nitrogen, and compare it with the results of blowing a foam with air or oxygen---if I'm right, the foam blown with nitrogen will flatten a bit faster.\n\n\nThere'll be a bit more $\\rm CO\\_2$ in the air from the candle, and that would diffuse more slowly out of the bubble (in fact, $\\rm CO\\_2$ is used as a \"blowing agent\" to make foams, partially for that reason). But I'd think in this case that the small amount of $\\rm CO\\_2$ produced by the candle would dissolve in the liquid, with gas dissolution rates enhanced by the lower surface tension and higher surface area of the foam.\n\n\n", "3" ] ]

[ [ "\nYou are right! There is an approximation here in only considering the vapor pressure of the gas. But let's consider the value of vapor pressure of water at $298\\ce{K}$: it's $13.85\\,\\ce{mm Hg}$. So, if you neglect the vapor pressure of water, the relative uncertainty is $$\\frac{13.85}{760}\\times100=1.83\\%$$\nAs you can see, it's negligible.\n\n\nP.S. To calculate the vapor pressure of water, I used the first equation in <http://en.wikipedia.org/wiki/Vapour_pressure_of_water>\n\n\n", "2" ] ]

[ [ "\n$\\pu{0.1 M}~\\ce{NaOH}$ is potentially corrosive when in contact with metals or human skin (see [here](http://www.chemicals.co.uk/uploads/documents/SODIUM-HYDROXIDE-0.1M-MSDS.pdf)), so I advise precautions such as wearing protective standard gloves and keep metals away from feasible damage.\n\n\n$\\ce{HCl}$ and $\\ce{H2SO4}$, especially when they're hot and are in higher concentrations, are very corrosive acids that are dangerous in contact with many materials. They can also release noxious gases in the reactions, but their weakness is glass. Due to reasons irrelevant, these acids can't corrode glass. I recommend extreme precaution. You can work under a \"glassy\" protection and as Yomen mentions, under a fume hood. Gloves are very important here. [Here](http://kni.caltech.edu/facilities/msds/HCl.pdf)'s what Caltech thinks about precautions using $\\ce{HCl}$ and [here](http://www.cchem.berkeley.edu/rsgrp/SOPs2013/SulfuricAcid_Sarpong.pdf)'s what Berekley does about sulfuric acid.\n\n\n$\\ce{KMNO4}$ is a very strong oxidizer. It may cause fire in contact with different materials (we had a volcano experiment in elementary school where we mixed it with glycerine and the result was very exothermic), and it has biohazards such as kidney damage. [Here](http://avogadro.chem.iastate.edu/MSDS/KMnO4.htm) it's explained more comprehensively.\n\n\n", "5" ], [ "\nYou must be careful when working with hot and concentrated acid solutions as they are corrosive. Otherwise, the chemicals you mentioned are common ones.\n\n\nBecause you are not used to manipulate chemicals, I advise you to wear goggles and protective gloves while preparing solutions. Work under fume hood when refluxing $\\ce{HCl}$. \n\n\n", "0" ] ]

[ [ "\nThe Wikipedia article for [ammonium perchlorate](http://en.wikipedia.org/wiki/Ammonium_perchlorate) gives a hint at how this can be done. The hint is that ammonium perchlorate is curiously less soluble than sodium perchlorate, and thus with the right concentrations, you would get ammonium perchlorate to precipitate by mixing solutions of sodium perchlorate and and, say, ammonium chloride.\n\n\n$$\\ce{NH4Cl(aq) + NaClO4(aq) -> NH4ClO4(s) + NaCl(aq)}$$\n\n\nThat being said, the best advice that can be given on how to make ammonium perchlorate is \n\n\n\n> \n> **Don't. Especially do not use heat!**\n> \n> \n> \n\n\nAmmonium perchlorate is fairly dangerous explosive that can decompose with not much provocation into a whole heck of a lot of gas for just a little bit of solid.\n\n\n$$\\ce{2NH4ClO4(s) -> N2(g) + Cl2(g) + O2(g) + 4H2O(g)}$$\n\n\nThat this decomposition produces toxic [chlorine](http://en.wikipedia.org/wiki/Chlorine#Health_effects_of_the_free_element_and_hazards) gas should only make you want to prepare ammonium perchlorate less. \n\n\nSafely producing, storing, and transporting this compound is going to require a lot of specialized equipment and training (which is expensive). The danger here is not of the type that may lead to funny smells and dizziness in poorly ventilated rooms. The hazards associated with the compound involve loss of life or limb and serious structural damage if you look at it funny. **Don't.** For more information into the sort of destructive power ammonium perchlorate has, consider that this is the rocket fuel that got the space shuttle into low earth orbit. This stuff has the power to [level industrial complexes with the forces of an earthquake](http://en.wikipedia.org/wiki/PEPCON_disaster). \n\n\n", "8" ] ]

[ [ "\n$\\rm O\\_2$ is the most common form of oxygen---the periodic table won't really tell you that. \n\n\nAvogadro's number tells you how many particles there are in a mole. \n\n\nThere are $6.022\\times 10^{23}$ O atoms in a mole of O atoms.\n\n\nThere are $6.022\\times 10^{23}$ $\\rm O\\_2$ molecules in a mole of $\\rm O\\_2$. \n\n\nSince you have 2 oxygen atoms in one $\\rm O\\_2$ molecule, there are\n$2\\times 6.022\\times 10^{23}$ O atoms in a mole of $\\rm O\\_2$. \n\n\nDo you see the difference?\n\n\n", "5" ], [ "\nA 'mole' is not short for a 'molecule'.\n\n\nThe 'mole' is a specific quantity (number of objects) defined by Avogadro's constant $N\\_A=6.022\\times10^{23}mol^{-1}$. So a 'mole' of oxygen molecules has $6.022\\times10^{23}$ molecules.\n\n\nThe '2' in $O\\_2$ means there are two oxygen atoms in an oxygen molecule.\n\n\n", "3" ] ]

[ [ "\nSimply adding 4 mL of solution to 100 mg of enzyme is not an accurate way of preparing this solution. I am going to describe what to do from an analytical chemist's point of view. If you don't need this level of precision, you could do this following the same steps without volumetric glassware.\n\n\nThe first thing you need to do is decide what volume of your 25 mg/mL solution you need. Ideally, you will have a volumetric flask corresponding to this volume. For now, let's go with the 4 mL that you mentioned.\n\n\nNext you will need to make a stock solution of your buffer. This will include all of the components except for the enzyme:\n\n\n1. Decide what volume of buffer to make. You will need to make at least enough for your working solution, and again you will want to use a volume for which you have a volumetric flask. Also, you want to be weighing out reasonable amounts of your reagents (because weighing out <10 mg of something can really suck). Let's assume you're making 1 L. You're reagents are cheap anyway and the larger scale will make everything easier.\n2. Calculate how much $\\ce{TrisHCl}$ and $\\ce{CaCl2}$ you will need to obtain the desired concentrations in 1 L of solvent.\n3. Weight out the $\\ce{TrisHCl}$ and $\\ce{CaCl2}$ and add it to your volumetric flask.\n4. Add 300 mL glycerol to your flask using a graduated cylinder. 30% glycerol probably refers to v/v.\n5. Fill *nearly* to volume with water. You need everything to be in solution so you can adjust the pH, but you also need to leave some room to work with to do the actual adjustment. I'd shoot for approximately 80% full.\n6. Adjust the pH by adding concentrated $\\ce{HCl}$ dropwise.\n7. Fill to volume with water and thoroughly mix the solution.\n\n\nNow that you have your buffer solution, you just need to weigh out the appropriate amount of your enzyme and dissolve it in your buffer. From the example above, you would weigh out 100 mg of the enzyme (don't just assume you have 100 mg, even if the bottle says so) and dissolve it in your buffer using a 4 mL vol. flask.\\*\n\n\n\\*\nIf you don't have a 4 mL vol. flask and you actually need all 100 mg of your enzyme in this solution, you will need to mess around with dilutions. Start by by creating a 50 mg/mL (2 mL vol. flask) or 100 mg/mL (1 mL vol. flask) solution.\n\n\n", "0" ] ]

[ [ "\nMore than likely the case of your laptop became charged and was attracting the salt. Yes salt crystals can be coerced by an electrostatic field.\n\n\nHere is a post on a teaching site that actually uses salt (and pepper) to demonstrate electrostatic attraction.\n\n\n<https://www.teachervision.com/electricity/lesson-plan/5787.html>\n\n\n", "0" ] ]

[ [ "\n[Effervescence](http://en.wikipedia.org/wiki/Effervescence) if there are bubbles and \n\n\n[Degasification](http://en.wikipedia.org/wiki/Degasification) more generally speaking.\n\n\nThere is also the term [outgassing](http://en.wikipedia.org/wiki/Outgassing), but that term is broader than just gas coming out of liquid solution.\n\n\n", "7" ] ]

[ [ "\nExact statements are:\n\n\n\n> \n> Because of the instability of its $\\ce {Pb^4+}$ cation, lead dioxide reacts with **warm** acids, converting to the more stable $\\ce {Pb^2+}$ state and liberating oxygen\n> $$\\ce{2 PbO2 + 4 HNO3 → 2 Pb(NO3)2 + 2 H2O + O2}$$\n> \n> \n> \n\n\nand \n\n\n\n> \n> Lead dioxide is produced commercially by several methods, which include ... or reacting $\\ce {Pb3O4}$ with **dilute** nitric acid:\n> $$\\ce{Pb3O4 + 4 HNO3 → PbO2 + 2 Pb(NO3)2 + 2 H2O}$$\n> \n> \n> \n\n\nSince nitric acid in warmer in first reaction, the following reaction occurs easily, which is actually a decomposition reaction:\n\n\n\n> \n> When heated, lead(II) nitrate crystals decompose to lead(II) oxide, oxygen and nitrogen dioxide.\n> $$\\ce{Pb(NO3)2->PbO +NO2 +O2}$$\n> \n> \n> \n\n\nHence the oxygen might be liberated with decomposition-cum-oxidation reaction.\n\n\n", "3" ], [ "\nYou need to consider the reaction conditions. \n\n\nThe Wikipedia [Lead Dioxide article](http://en.wikipedia.org/wiki/Lead_dioxide) is saying that in **dilute** nitric acid the first reaction occurs, while in **warm** nitric acid the second reaction occurs. \n\n\nSo the first reaction would only proceed to the second reaction under harsher conditions, heat and more concentrated acid.\n\n\n", "3" ] ]

[ [ "\nPolymers follow a slightly different nomenclature system than you are used to in organic chemistry. Instead of naming the polymer after the monomers *as they are*, we name them after the monomers *as they originally were*. In other words, a polymer made from ethylene monomers is called polyethylene. Since PTFE and FEP are made from fluorinated ethylene and propylene, we just put \"poly\" in front of the monomer names. This is despite the fact that after the reaction, the bond order is reduced by one. \n\n\n", "9" ], [ "\nThe compounds are polymers. The unit structure of PTFE is tetrafluoroethene. PTFE is named in accordance with the subunits from which they are polymerized. You can apply the same logic to the second one (you apply the same logic to all polymers). There are no common names for polymers just abbreviations. IUPAC rules do have polymer rules too, so you will necessarily see them adhering to the IUPAC rules. It is, however, a matter of the unit structure in brackets, not the monomer that determines the name.\n\n\n", "0" ] ]

[ [ "\nReason 1: \n\n\nwhile $T\\,\\mathrm dS=\\mathrm dQ$ for a reversible process, this does not mean $T\\Delta S = Q$\n\n\nReason 2: \n\n\nThe term \"reversible reaction\" is unrelated to \"reversible process\".\n\n\nLame Reason 3:\n\n\nBecause your \"always\" statement does not repeat the \"constant pressure\" requirement assumed for $Q=\\Delta H$\n\n\n", "3" ], [ "\nThe problem is that you are equating \"reversible reaction\" with \"reversible process.\"\n\n\nWhat you are seeing when you plug equations 2 and 1 into the Gibbs Free Energy equation is that at equilibrium, \n\n\n$\\Delta G = 0$\n\n\nThe reason that this works is because a reversible process is one that progresses by very small movements away from equilibrium states - when we say a process is \"reversible\" what we really mean is that we are making an approximation that allows us to ignore entropy generation. When you make this approximation and apply it to the Gibbs Free Energy equation, you are forcing it to stay at equilibrium - hence, $\\Delta G = 0$.\n\n\nI think your confusion is coming from the similar names of \"reversible reaction\" and \"reversible process.\" A [reversible proces](http://en.wikipedia.org/wiki/Reversible_process_(thermodynamics))s is one in which the entropy generation term is zero, but a [reversible reaction](http://en.wikipedia.org/wiki/Reversible_reaction) is one in which the forward and reverse rates of reaction result in a mixture of reactants and products at equilibrium.\n\n\nIn other words, a reversible **reaction** can be in an equilibrium ($\\Delta G = 0$) or non-equilibrium state ($\\Delta G \\neq 0$), but a reversible **process** is always in an equlibrium state.\n\n\n", "2" ], [ "\nYou are missing the subscripts. Heat transfer only makes sense if you define a system and its surrounding. Your equations become:\n\n\n$$\\Delta G\\_\\text{sys}=\\Delta H\\_\\text{sys}-T\\,\\Delta S\\_\\text{sys}$$\n\n\nAt constant pressure, \n\n\n$$q = \\Delta H\\_\\text{sys}\\tag{1}$$\n\n\nand we know that $T \\Delta S=q$ (reversible)\nso finally \n\n\n$$T\\,\\Delta S\\_\\text{surr}=-q\\tag 2$$\n\n\nPutting values of Equation $\\text{(1)}$ and $\\text{(2)}$ in Gibbs energy equation ,\nwe get \n\n\n$$\\Delta G\\_\\text{sys} = -T\\,\\Delta S\\_\\text{surr}-T\\,\\Delta S\\_\\text{sys} = -T\\,\\Delta S\\_\\text{univ}$$\n\n\nSo the reaction will proceed if $\\Delta G\\_\\text{sys}$ is negative or $\\Delta S\\_\\text{univ}$ is positive. This is how the textbooks tell it.\n\n\n", "0" ] ]

[ [ "\nIt doesn't matter if there is a container. I'm picturing bubbling the nitrogen through the water, for example the water is in a flask, and there is a hose going down into the water, maybe a frit at the end of the hose.\n\n\nThe important assumption is that nitrogen reaches equillibrium with the water, which is a big assumption. \n\n\nBut adding 1.2 grams of water would increase the total volume by about 1 L. I would use the total volume to answer the question, not 48L, but it will only change the answer by about 2%. \n\n\n", "2" ], [ "\nThis is an example of using partial pressures for a mixture of ideal gases.\n\n\nWe are making a big assumption - that water is an ideal gas - but once we make that assumption, the way to proceed is to treat each gas in the mixture *as if it were by itself* in the container. As you pointed out, it isn't very obvious from the way the question is worded that we would assume that both would occupy 48 L, but, since the volume of a gas depends either on the container volume or on temperature and pressure when there is no container, it does make sense. The idea is that whatever is constraining the $\\ce{N2}$ to 48 L would also constrain the $\\ce{H2O}$ to the same volume, since it is a mixture of gases.\n\n\nSince the partial pressure equals the vapor pressure, the solution is given by calculating the pressure of this mass of water under these conditions. \n\n\n", "2" ] ]

[ [ "\nAccording to the way this question is worded, adding 1 atmosphere would be incorrect. The reason is that they specify an *empty* container - if you have a rigid container and it is truly empty (high vacuum), then the absolute pressure would be zero, and the gauge pressure would be -1 atm (see this [wikipedia article](http://en.wikipedia.org/wiki/Pressure_measurement#Absolute.2C_gauge_and_differential_pressures_-_zero_reference) for reference to these terms).\n\n\nOnce the $\\ce{CO2}$ sublimates, the final absolute pressure would be given by $PV=nRT$ - there is no need to add atmospheric pressure. To find the gauge pressure, you would actually *subtract* one atmosphere.\n\n\nIt is possible that the question wants you to assume that \"empty\" means \"filled with air at 1 atm,\" but that is a guess and I would argue that it's not safe to assume that based on the way the question is written.\n\n\n", "1" ] ]

[ [ "\n1) Most definitely. Adsorbed molecules often are very distorted, up to chemical reaction with support, especially when reactive surface is used. Despite common misconception, even noble metal surfaces are quite reactive.\n\n\n2) DFT, while being indeed inexact (as any calclation with limited size of basis set), catches major features of electronic density very well.\n\n\n3) AFM stands for atomic force microscopy, which is not connected to electronic density directly. Now, naphthalene molecule features large $\\pi$-system, which is very polarizable. Since nonvalent interatomic forces beyond electrostatic interactions usually increases with polarizability of interacting particles, I would expect $\\pi$-systems to be highlighted on the NC-AFM images.\n\n\n4) I'm not an expert on atomic-resolved microscopy, but I never read about this possibility in the articles I worked with, though the works I usually works with used STM. and SEM mostly.\n\n\n", "5" ], [ "\nThis is not really my field, but I've certainly followed it. I'd guess some combination of all of your reasons, plus one more.\n\n\n* I think usually the substrate is considered, but *imperfections* and defects in the surface may not be properly considered.\n* Perhaps, although I think this is probably less important than other causes.\n* Exactly. DFT-computed electron density is not exactly what the AFM measures.\n* The probe will certainly have different interactions with the molecule, and could be subtly asymmetric.\n\n\n**I'd guess the most likely culprit is different.**\n\n\nIn AFM (and all scanned-probe methods) the substrate is rarely perpendicular to the probe. Some minor image correction is performed to \"line-fit\" the substrate.\n\n\nConsider if you take a photo of a page on your desk. The page is probably slightly rotated in the plane of the desk, but the camera is almost never exactly parallel to the desk. This results in [\"keystoning\"](https://en.wikipedia.org/wiki/Keystone_effect).\n\n\nThe AFM software corrects for this effect, and users can fit a bit after data-collection, but it can cause subtle distortions.\n\n\n", "4" ] ]

[ [ "\nIn a barometer at equilibrium, the pressure of the mercury equals the pressure of the air (at the air/mercury interface).\n\n\nThe compressibility of mercury in principle does make it non-linear. The density would be [3.8 parts per million](http://hyperphysics.phy-astr.gsu.edu/hbase/tables/compress.html) higher at the air interface than at the vacuum interface. \n\n\nCertainly there are other greater sources of error: the vacuum isn't really a vacuum but contains mercury vapor, density varies with temperature, vapor pressure varies with temperature, the size of the tube varies with temperature, there are surface tension effects. \n\n\nJust like any experimental technique, you need to evaluate sources of error and report a measurement with appropriate uncertainty.\n\n\n", "1" ], [ "\n\n> \n> Why does a liquids height in a Barometer depend linearly on pressure?\n> \n> \n> \n\n\nSince: $$P=\\frac FA=\\frac {mg}{A}=\\frac {\\rho Vg}A=\\rho g h$$\n\n\n\n> \n> However, this derivation seems misguided, since P=F/A, and P is the pressure of the gas, not of the liquid, as the derivation later relates it to.\n> \n> \n> \n\n\nNo at equilibrium the pressure of both will be same.\n\n\n\n> \n> Furthermore, it doesn't make sense to me that the height of the liquid and the pressure are necessarily linearly related. Wouldn't this depend on the compressibility of the liquid?\n> \n> \n> \n\n\nSince we are assuming uniform density of liquid the mass at any place of same volume will be same, so if we add some water mass adds, so force adds.\n\n\n\n> \n> Or rather, it seems like depending upon the properties of the liquid, it may be very difficult to change the height past a certain point, for example. In this sense the added energy from increasing the pressure goes into bonds or is stored in electrical interactions.\n> \n> \n> \n\n\nSuch a situation never arrives, as far as theoretical concerns are made, since we assume water as an in compressible fluid.\n\n\n", "0" ] ]

[ [ "\nThe *average* rate $\\Delta[A]/\\Delta t$ over time interval $\\Delta t$ is only approximately equal to the true *instantaneous* rate $\\rm d[A]/dt$ at time $t$. The difference between the two *can* be significant if the instantaneous rate changes over the chosen time interval. \n\n\nHere's an analogy: suppose you're driving your car on the highway at a steady 55 mph. Your average and instantaneous rates of travel will be much the same even over a long time interval. On the other hand, if you're driving through a city with a lot of starts and stops at intersections, your instantaneous rate of travel won't match your average rate of travel unless you're looking at averages over very short time intervals.\n\n\nTo see how far off we'll be if we don't use a small time interval, let's compare the average and instantaneous rates for your reaction with time intervals of 0.1, 0.5, and 1.0 s:\n\n\n\n\n\nThe larger the time interval, the larger the difference between $\\Delta[A]/\\Delta t$ and $\\rm d[A]/dt$ and the larger the error in the rate constants estimated using your method.\n\n\nSo why not just use very small time intervals? Well, there's a limit to how small you can make $\\Delta t$ if you're computing $\\Delta[A]/\\Delta t$ from experimental measurements. You're taking differences between concentrations that are almost the same size. Any error in experimental measurements of $\\rm [A]$ may well swamp your rate estimates as you go to smaller and smaller time intervals. \n\n\n", "6" ], [ "\nThat rate constant '$\\kappa$' is for \"instantaneous rate of reaction\"(irr) which you wish to calculate from \"average rate of reaction\"(arr)\n$$\\frac{d[A]}{dt}=r\\_{irr}=\\lim\\_{\\Delta t\\to0 }r\\_{arr}=\\lim\\_{\\Delta t\\to0 }\\frac{\\Delta [A]}{\\Delta t}$$\nSo if the time duration is very small you would get a better approximation.\n\n\n", "0" ] ]

[ [ "\nThis conclusion assumes that the enthalpy of the water remains entirely constant, which is not the case. Given that the water is being uniformly heated, but no end to the heating was specified, the enthalpy of the water will continue to increase, and thus force more and more of the liquid water into vapor form. Additionally, as you specified a very large surrounding, any water that enters the gaseous form would not largely impact the air pressure of the environment, so there would be few things acting to maintain some kind of equilibrium here.\n\n\n", "4" ], [ "\nYou have a closed system, consisting of a certain volume and containing a certain amount of water. At a constant temperature it will find some equilibrium consisting of water vapor, liquid water, and ice (depending on temperature -- I'll ignore the ice for now). As you heat the pan, you are heating the entire system and thus changing the balance of water vapor to liquid water. Where that liquid water actually resides is a problem for the water to figure out - it does not have to reside long term in the pan. Eventually, as you keep heating, you will get to a temperature of the system where the liquid will be thermodynamically unstable, and the enclosure will contain water vapor only. \n\n\nRemember, for thermodynamic equilibrium, how fast you heat/cool does not enter into it - those are kinetic questions.\n\n\n", "1" ], [ "\nIf the water vapor you start out with is at 1 atm then that means it has to be at 100 C. It also means that the walls of the box have to be either extremely well insulated or externally heated. Otherwise, condensation will occur on the walls and the gas will cool down. Assuming that the box and pan are somehow held at 100 C then the system will be in equilibrium and nothing will happen (i.e. no boiling). However, if the pan heats up beyond 100 C or the box walls cool off then you will see boiling. In the former case, the water vapor pressure will increase to maintain equilibrium. In the latter case, the water will boil just enough to compensate for the vapor lost by condensation.\n\n\n", "1" ] ]

[ [ "\nYou are on the right track, but if you google [terephthalic acid](https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=terephthalic%20acid), you will know that your formula is not correct. \n\n\nOne mistake you are making is using the final mass of the $\\ce{CaCl2}$ and $\\ce{NaOH}$ and not the change in mass. For example, let's take the $\\ce{CaCl2}$:\n\n\nFirst, we don't know what the initial mass of $\\ce{CaCl2}$ is, so we cannot calculate the final mass the way you are trying to do. We are only given the mass of the terephthalic acid (0.4953 g). The calcium chloride and sodium hydroxide are not part of this mass. \n\n\nHowever, we know that the calcium chloride increased in mass by 0.1613 g, presumably by absorbing water (you are correct that the calcium chloride is absorbing water). So, the mass of water produced by the combustion reaction is 0.1613 g.\n\n\nSimilarly, we only know the change in mass of the sodium hydroxide (1.0495 g), which occurred by absorbing carbon dioxide. So the mass of carbon dioxide is 1.4095 g.\n\n\nThe second mistake is to assume that all of the mass of the carbon dioxide and water produced must come from the original mass of the terephthalic acid. Not so! We have excess oxygen added contributing mass to the reactant side, and this oxygen is also in the products. \n$$\\ce{C}\\_x\\ce{ H}\\_y\\ce{ O}\\_z\\ce{ +}\\left(\\frac{4x+2y-z}{4}\\right)\\ce{ O2 -> }x\\ce{CO2 +}\\frac{y}{2}\\ce{ H2O}$$\n\n\nWe need to do more maths to figure out how much of the leftover mass in the original sample is oxygen. Here is the beginning of the setup, now that you have some sense of the mass of water and carbon dioxide produced:\n\n\n1. Mass of water produced $\\ce{->}$ moles of water produced $\\ce{->}$ moles of hydrogen in the water produced (and in the original sample!) $\\ce{->}$ grams of hydrogen (in sample)\n2. Mass of carbon dioxide produced $\\ce{->}$ mass of carbon (in sample)\n3. Mass of oxygen in sample (by subtraction) $\\ce{->}$ moles of oxygen in sample\n\n\nYou are doing everything correctly after this.\n\n\n", "1" ], [ "\nI am afraid I could not follow your reasoning and unfortunately can not pinpoint you to your error therefore. The result you obtain, does not resemble the empirical formula of [terephthalic acid](http://en.wikipedia.org/wiki/Terephthalic_acid). As a general tip: These problems are of entirely mathematical nature, so it is important to use proper equations and keep the units at all times. I am going to tackle this problem from beginning to end.\n\n\n\n> \n> Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticisers. It contains only $\\ce{C}$, $\\ce{H}$, and $\\ce{O}$. In one experiment, $0.4953~\\mathrm{g}$ of the acid was burnt in excess oxygen. The gases produced were passed through $\\ce{CaCl2}$ and $\\ce{NaOH}$. It was found that the mass of the $\\ce{CaCl2}$ increased by $0.1613~\\mathrm{g}$ and the mass of the $\\ce{NaOH}$ increased by $1.0495~\\mathrm{g}$. What is the empirical formula of terephthalic acid?\n> \n> \n> \n\n\nFirst of all it is important to formulate the correct reaction equation:\n$$\\ce{C\\_{$x$}H\\_{$y$}O\\_{$z$} + $\\frac{\\left(2x+\\frac{y}{2}-z\\right)}{2}$\\,O2 -> $x$\\,CO2 + $\\frac{y}{2}$\\,H2O}$$\n\n\nThe information about calcium chloride and sodium hydroxide is only a way to tell you the mass of the product. It is completely irrelevant to the question itself. However, the correct equation would then be:\n$$\\ce{C\\_{$x$}H\\_{$y$}O\\_{$z$} + $\\frac{\\left(2x+\\frac{y}{2}-z\\right)}{2}$\\,O2 ->[\\ce{CaCl2, NaOH}] $x$\\,NaHCO3 + $\\frac{y}{2n}$\\,CaCl2.$n$\\,H2O}$$\n\n\nFirst of all, we need to gather all the information we are given in the question or that we know otherwise:\n\\begin{align}\nm(\\ce{C\\_{$x$}H\\_{$y$}O\\_{$z$}}) &= 0.4953~\\mathrm{g} &M(\\ce{C\\_{$x$}H\\_{$y$}O\\_{$z$}}) &= ?\\\\\nm(\\ce{H2O}) &= 0.1613~\\mathrm{g} & M(\\ce{H2O}) &= 18~\\mathrm{\\frac{g}{mol}}\\\\\nm(\\ce{CO2}) &= 1.0495~\\mathrm{g} & M(\\ce{CO2}) &= 44~\\mathrm{\\frac{g}{mol}}\\\\\n\\end{align}\n\n\nWe can go ahead and calculate the number of moles of carbon dioxide and water produced. We therefore also know the relative number of moles of carbon and hydrogen in terephthalic acid. The basic formula that needs to be applied here is $$n=\\frac{m}{M}.$$\n\n\nHow many moles of carbon are in carbon dioxide, what is the mathematical relationship? What is it for hydrogen and water?\n\n\n\n> \n> $n(\\ce{C}) : n(\\ce{CO2}) = 1$ \n> \n> $n(\\ce{H}) : n(\\ce{H2O}) = 2$\n> \n> \n> \n\n\nHow many moles of carbon dioxide and water have been produced? How does that relate to carbon and hydrogen?\n\n\n\n> \n> $n(\\ce{CO2})=\\frac{m(\\ce{CO2})}{M(\\ce{CO2})}= 0.0239~\\mathrm{mol} \\therefore n(\\ce{C})= 0.0239~\\mathrm{mol}$ \n> \n> $n(\\ce{H2O})=\\frac{m(\\ce{H2O})}{M(\\ce{H2O})}= 0.0090~\\mathrm{mol} \\therefore n(\\ce{H})= 0.0180~\\mathrm{mol}$ \n> \n> \n> \n\n\nCalculate the mass of hydrogen and carbon in the original sample with\n$$m=n\\cdot M$$ and use it to find the mass of oxygen in the sample.\n\n\n\n> \n> $m(\\ce{C}) = 0.2868~\\mathrm{g}$ \n> \n> $m(\\ce{H}) = 0.0180~\\mathrm{g}$ \n> \n> $m(\\ce{O}) = m(\\ce{C\\_{$x$}H\\_{$y$}O\\_{$z$}}) - m(\\ce{C}) - m(\\ce{H}) = 0.1906~\\mathrm{g}$\n> \n> \n> \n\n\nHow many moles of oxygen are present in the original sample?\n\n\n\n> \n> $n(\\ce{O}) = 0.119~\\mathrm{mol}$\n> \n> \n> \n\n\nNow you only have to relate the number of moles for each element to integer ratios and you will find the empirical formula.\n\n\n\n> \n> $x:y:z = n(\\ce{C}) : n(\\ce{H}) : n(\\ce{O}) = 0.0239~\\mathrm{mol} : 0.0180~\\mathrm{mol} : 0.119~\\mathrm{mol}$ \n> \n> $x:y:z = 4:3:2$ \n> \n> Therefore the empirical formula is $\\ce{C4H3O2}$, which is half of the sum formula $\\ce{C8H6O4}$, see wikipedia for more information.\n> \n> \n> \n\n\n", "1" ] ]

[ [ "\nIf you add an excess of cyanide ions to a iron(II) solution you could conceivably get precipitate of $\\ce{Fe(SCN)2}$ which would then dissolve to form a complex with four thiocyanate ions. However, I can't find any references to such a complex on the internet.\n\n\nThe equations would be: $$\\ce{Fe^{2+}\\_{(aq)} + 2SCN-\\_{(aq)} -> [Fe(SCN)2]\\_{(s)}}$$ $$\\ce{[Fe(SCN)2]\\_{(s)} + 2SCN^{-}\\_{(aq)} -> [Fe(SCN)4]^{2-}\\_{(aq)}}$$\n\n\n", "2" ], [ "\nAs best I can tell, your answer is correct. I can't find any reference for iron(II) thiocyanate in the form your teacher says is correct, and several references that say $\\ce{Fe(SCN)2}$ is the correct form.\n\n\nFor precipitation reactions, the charge will never change. You should always assume that the charges in the product will balance. There might be some cases where the product has an unexpected formula that you can't predict based on charges alone, but I can't think of any off the top of my head, and you certainly shouldn't be expected to know them at this stage.\n\n\nedit:\n\n\nThis is the best reference I could find, it's a google books link to the [CRC handbook](https://books.google.com/books?id=c1rNBQAAQBAJ&pg=SA4-PA69&lpg=SA4-PA69&dq=iron+ii+thiocyanate&source=bl&ots=nBHC5pzvcy&sig=Upbact8-zBtevG43BjYbEDuLsRo&hl=en&sa=X&ei=cAe3VIurF5CRyATf5oDADw&ved=0CDQQ6AEwBDge#v=onepage&q=iron%20ii%20thiocyanate&f=false)\n\n\n", "2" ] ]

[ [ "\nThis is a little tricky. I would argue that only the first proton dissociates completely and should be considered \"strong\" - using the rules for writing net ionic equations, where we only dissociate **strong electrolytes**, this would meant that we would write:\n\n\n$\\ce{H2SO4 + 2KOH -> 2H2O + K2SO4}$\n\n\n$\\ce{H+ + HSO4- + 2K+ + 2OH- -> 2H2O + 2K+ + SO\\_4^{2-}}$ - here only the first proton dissociates completely\n\n\nCanceling out spectator ions gives:\n\n\n$\\ce{H+ + HSO4- + 2OH- ->2H2O + SO\\_4^{2-}}$\n\n\nYour teacher (or whoever is giving you that answer) is dissociating the second proton as well, to give:\n\n\n$\\ce{2H+ + SO\\_4^{2-} + 2OH- ->2H2O + SO\\_4^{2-}}$\n\n\nWhich becomes:\n\n\n$\\ce{H+ + OH- ->H2O}$\n\n\nThere are a few reasons why they might do this, but I would argue that the 1st approach (which only dissociates the first proton) is best because it more accurately reflects what is happening, in the sense that $\\ce{HSO4-}$ is a weak electrolyte. This will become important later when you study acid-base equilibria.\n\n\nIn general, you dissociate protons from acids as if they were regular metal cations - just \"pull them off\" and don't forget the positive charge. For polyprotic acids, the first proton is strongly acidic, or dissociates completely, but the other protons are not. There might be exceptions to this rule but I can't think of any off the top of my head.\n\n\n", "3" ] ]

[ [ "\nI think you have it all right. The OH- present must come from the autolysis of the water present, which is far in excess of the HCl. If you figure the pOH of pure water would be 7, then it makes sense to think that adding 1E-06 M (a small concentration) of HCl would lower the basicity (raise the pOH from 7 to 8). \n\n\nThe reason a scale of 0-14 is used for pH is that when you get to either extreme end of the scale (close to 1M in either acid or base), the sheer concentration of ions present, relative to water, is a game changer and the system no longer behaves in a nice linear fashion. So the scale does not work well outside the 0-14 range. Conceptually, you can extend the pH scale down below zero using concentrated mineral acids or strong acids like fluorosulfonic acid, and empirically measure their effectiveness in protonating a very weak base as a proxy for pH.\n\n\n", "1" ] ]

[ [ "\nI believe you are referring to the Sabatier process. $$\\ce{CO2 + 4H2 -> CH4 + 2H2O}$$\n\n\nThe reaction uses a $\\ce{Ni}$ catalyst and is done at $300-400\\,^{\\circ}{\\rm C}$ and high pressures. The reaction is exothermic, $\\Delta H = −165.0~\\mathrm{kJ\\,mol}^{-1}$, but requires initial heating in order to start the reaction off due to the activation energy.\n\n\n[Read more on Wikipedia](http://en.wikipedia.org/wiki/Sabatier_reaction).\n\n\n", "5" ] ]

[ [ "\nYou need to be familiar with some of the compounds earlier, otherwise you can't tell if it exists. Remember facts come first before explanation. It also depends usually on the atom/ion to which it is bonded. for example since Fluorine is most electronegative it can form $\\ce{XeF6,XeF4}$,etc. but $\\ce{XeH3}$ doesn't form also oxygen forms large oxidation states as it is electronegative too and there is also more energy released as many bonds are formed (ususlly pi-bonds.)\n\n\n", "1" ] ]

[ [ "\nYou wouldn't need to supply electric energy to get oxygen and hydrogen to recombine into water, because the [free energy change](http://en.wikipedia.org/wiki/Gibbs_free_energy) for that reaction is negative. In other words, the products have a lower potential energy than the reactants, and so the reaction will proceed \"downhill\" on its own. You would need to overcome the very small activation energy, which you could do with electricity or with a flame or spark. But once it got started, you would not need to supply additional energy, and could in fact *generate* electricity from the reaction.\n\n\nThis is the basis of how hydrogen fuel cells work - hydrogen and oxygen react in a device that is set up to extract the energy that is produced in the form of an electric current.\n\n\nAn easy way to think about it is that if you need to supply energy to make a reaction happen, then the reverse reaction will produce the same amount of energy (if we ignore losses due to inefficiency) or less (if we don't). The reverse is true as well.\n\n\nOn a side note - electrolysis specifically refers to splitting a compound. The \"-lysis\" suffix means \"break apart,\" and \"electro-\" prefix means \"with electricty.\" In your case, you are thinking of \"electrochemistry\" - which just means \"using electricity with chemistry.\" Electrolysis is one thing you can do with electrochemistry.\n\n\n", "3" ] ]

[ [ "\nI think the problem here is related to what each of these terms means, in a fundamental sense.\n\n\n$\\mathrm{d}S$ is the entropy of the *system*\n\n\n$\\frac{-\\mathrm{d}q}{T}$ is the entropy of the *surroundings* - the heat change of the surroundings divided by the temperature.\n\n\n$\\mathrm{d}H$ is the enthalpy of the *system* - it is equal to $\\mathrm{d}q$ only when the internal energy change occurs at constant pressure, and when there is no form of work being done besides PV work. In an electrochemical cell, moving the electrons is work, and so we can't leave it out of the total energy balance.\n\n\nIn other words, you can't just use $\\mathrm{d}q = \\Delta H$ for the 1st question because $\\mathrm{d}q\\_p \\neq\\Delta H$ when you are doing any work besides PV work.\n\n\nInstead you are using the Clausius inequality, which is just another way of stating the second law: the entropy of the universe must increase for any process. In this case, the universe is the system plus the surroundings, and so:\n\n\n$$\\Delta S - \\frac{\\mathrm{d}q}{T} \\geq 0$$\n\n\nSolving that for dq gives you the heat absorbed by the *surroundings*, which is equal to temperature times the entropy change of the *system*.\n\n\nFor a process that is only (possibly) doing PV work, at constant pressure, $\\Delta H = \\mathrm{d}q\\_p$, and so we can just use that. This would be an example of using the 1st law at constant pressure, when there is no other type of work besides PV. \n\n\nIn your case, you used:\n\\begin{align}\n\\Delta G &= \\Delta H - T\\,\\mathrm{d}S\\\\\n\\Delta H &= \\Delta G + T\\,\\mathrm{d}S\\\\\n\\end{align}\n\n\nThis works because the equation for $\\Delta G$ is derived directly from the Clausius inequality at constant pressure. In other words, you are using the 2nd law to derive the same result.\n\n\nEdit\n====\n\n\nTo answer your last questions, and questions from the comments:\n\n\n\n> \n> why is $T\\,\\mathrm{d}S$ less than $\\mathrm{d}q$ in my example? I don't really get this bit\n> \n> \n> \n\n\nIt's not - but you are dropping the signs and that makes it hard to see. If you leave the signs in:\n\n\n\\begin{align}\nT\\,\\mathrm{d}S &\\geq \\mathrm{d}q\\\\\n(298~\\mathrm{K})\\left( -61.6~\\mathrm{\\frac{J}{K\\cdot mol}}\\right) &\\geq -70~\\mathrm{\\frac{kJ}{mol}}\\\\\n-18~\\mathrm{kJ} &\\geq -70~\\mathrm{kJ}\\\\\n\\end{align}\n\n\nWhich is true, although it is easier to see if you divide by -1:\n\n\n$$18~\\mathrm{kJ} \\leq 70~\\mathrm{kJ}$$\n\n\n\n> \n> The entropy divided by the temperature gives $234.9~\\mathrm{JK^{−1}mol^{−1}}$ which is not the value for entropy as it should be according to the Clausius inequality given that the change in entropy is equal to the heat given off in part (b). Why is this so?\n> \n> \n> \n\n\nHere I assume you meant \"enthalpy divided by temperature.\" The second reaction is not carried out in a reversible manner, so we have no guarantee that $\\mathrm{d}S = \\mathrm{d}q/T$. All we know is that it will be greater than or equal to $\\mathrm{d}q/T$, and if you plug in the numbers, it is.\n\n\nRemember that all of this is built into the $\\Delta G$ equation as well, which is why we use it when doing \"regular\" chemical reactions at constant T & P.\n\n\n", "4" ], [ "\nFor first part since system is operated reversibly at constant temperature:\n$$\\newcommand{\\d}{{\\rm d}}\n\\d S=\\frac{\\d q\\_{rev}}{T}\\implies \\Delta S=\\frac{Q\\_{rev}}{T}$$\nThis is what you have done.\nFor second part, Yes Gibbs energy is the maximum (in reversible process) non-expansion/additional work (here the electrical work)\n$$\\Delta G^\\circ=-\\nu FE^\\circ=w\\_{add,max}$$\nThis value would be same as $\\Delta H^\\circ$ because from rearrangement of definition of $$\\Delta G^\\circ=\\Delta H^\\circ-T\\Delta S^\\circ$$$$\\Delta H^\\circ=\\underbrace{\\Delta G^\\circ}\\_{\\text{non expansion work}}+\\underbrace{T\\Delta S^\\circ}\\_{\\text{work extracted as heat}}$$\nAlso it can be viewed in another way, since $\\Delta H^\\circ$ is \"the energy supplied as heat at constant pressure\".Another thing to keep in mind is $\\d H=q\\_p$ only under constant pressure and no additional work, both to be satisfied. \nAnd for the Classius Inequality, there's some calculation error by you:\n$$\\frac{Q}{T}=\\frac{-70000}{298}=-234.8\\le-61.6(\\Delta S^\\circ)$$\n\n\n", "0" ] ]

[ [ "\n\n> \n> But apparently the correct molecular equation is $\\ce{6KCN\\+AlBr3 \\to 3KBr\\+Al(CN)6}$\n> \n> \n> \n\n\nThat's not correct, what happenend to the other 3 potassium atoms?\n\n\nIt could form $\\ce{Al2(CN)6}$ but you can't exclude $\\ce{Al(CN)3}$ without further information beyond balancing of equations.\n\n\nThere are two main steps to balancing equation:\n\n\n1. Balancing of each type of atom\n2. Balancing number of electrons\n\n\nIf you've done both of those correctly, you've done all you can with balancing alone. \n\n\n\n> \n> $\\ce{Cl2\\+2NaOH \\to 2NaCl\\+Cl2O\\+H2}$\n> \n> \n> \n\n\nAbove is clearly wrong because number of oxygen atoms doesn't balance\n\n\n\n> \n> $\\ce{Cl2\\+2NaOH \\to 2NaOCl\\+NaCl\\+H2O}$\n> \n> \n> \n\n\nAbove is clearly wrong because number of oxygen atoms doesn't balance\n\n\n", "4" ] ]

[ [ "\nThe ionization energy is described as the energy required to remove an electron from an atom in the gaseous phase. Apart from some exceptions, it is possible to identify a general trend in ionization energies within the periodic table: moving from left to right in a period the ionization energy increases, whereas it decreases from top to bottom in a group.\n\n\n", "3" ] ]

[ [ "\nUnit cells are just a way of describing a given crystal. You can certainly draw $\\ce{NaCl}$ the other way, with $\\ce{Na+}$ in the corners and it's still correct. There are a few ways of thinking about this structure:\n\n\n1. as two interpenetrating FCC lattices\n2. as a $\\ce{Cl-}$ lattice with $\\ce{Na+}$ in the octahedral holes\n3. as a $\\ce{Na+}$ lattice with $\\ce{Cl-}$ in the octahedral holes\n\n\nThey're all equivalent. The figure above actually shows both types of octahedral holes. Some might prefer to visualize the structure as the second case since, as you say, the chloride ions are bigger, but going the other way isn't wrong, nor does it describe anything different.\n\n\n", "7" ], [ "\nThis is a more important question than one would think and not adequately answered before. Yes, there is an equivalence so you would wonder why we define one as the lattice and the other as in holes. However this is an important idea of solid state chemistry.\n\n\nThe important thing is generally (GENERALLY) anions are bigger than cations. Also more often the same species. In oxides, this is especially the case. So you can think of the anions as a close packed lattice (FCC or HCP) in the sense of balls with closest packing. and the cations in certain holes of the lattice. This is the same in NaCl where chloride is bigger than sodium.\n\n\nIt's also important\n\n\nOf course this is chemistry, so everything depends. But if you want a useful initial organizing idea for MANY structures, think of close packed anions with cations in certain holes (perovskite, spinel, corundum, etc.).\n\n\nSo yes, there IS a reason why we think of the chlorides as the base lattice, not the sodium ions and you are right on that the reason is size.\n\n\n", "0" ] ]