# Tag Info

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The equilibrium constant for combustion of organic matter in air with oxygen is not small, but extremely large ($K_\mathrm{eq} \gg 1$), as is expected from a reaction that is simultaneously very exothermic and (usually) increases entropy due to the formation of more gaseous molecules than the input oxygen. The major reason carbon-based life can exist at ...

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There's another question related to salt bridges on this site. The purpose of a salt bridge is not to move electrons from the electrolyte, rather it's to maintain charge balance because the electrons are moving from one-half cell to the other. The electrons flow from the anode to the cathode. The oxidation reaction that occurs at the anode generates ...

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XKCD's source material is an article from the blog of one of the funnier chemists I've read, Dr. Derek Lowe. The chemical in question made his "Things I Won't Work With" list, and the article is found here. Dioxygen difluoride, $\ce{O_2F_2}$, sometimes evocatively labeled according to its atomic arrangement $\ce{FOOF}$, is first and foremost a vigorous ...

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Indeed, $\ce{Zn}$ is lower than $\ce{Na}$ in activity series of metals, so the following reaction won't take place $$\require{cancel}\ce{Zn + 2NaOH \cancel{→} Zn(OH)2 + 2Na}$$ Remember, however, that $\ce{Zn}$ is amphoteric, so it can reacts with a strong base such as $\ce{NaOH}$ as an acid forming sodium zincate $$\ce{Zn + 2 H2O + 2 NaOH -> Na2Zn(OH)4 ... 29 Since I will deal with all of the alkali metals in this answer, I think the question should also be broadened. There is no point in covering one single metal (sodium) without touching the others since it is the trend going down the group that we are interested in. All thermodynamic data is taken from Prof. M. Hayward's lecture notes at Oxford. So, firstly, ... 28 Ivan's answer is indeed thought-provoking. But let's have some fun. IUPAC defines oxidation as: The complete, net removal of one or more electrons from a molecular entity. My humble query is thus - what better way is there to remove an electron than combining it with a literal anti-electron? Yes, my friends, we shall seek to transcend the problem ... 25 There are many, here are some examples Ion interchange reactions of various types, such as precipitation \ce{NaCl + AgNO3 -> AgCl \downarrow + NaNO3 } Polar molecule insert/ejection. The most common case is hydrolysis and reverse reactions. \ce{PCl5 + 4 H2O -> H3PO4 + 5 HCl} Reorganization of bonds between atoms of same type, such as catalytic ... 23 We could explain it from an electrolytic point of view. Iron has a certain tendency to dissolve in water according to the following equation: \ce{Fe (s) -> Fe^2+ (aq) + 2e^-} The release of electrons causes a small current to flow in the metal (Remember that metals conduct electricity). This turns the point where iron dissolves into an anode, and the ... 22 \ce {O_2F_2} doesn't spontaneously combust. It is a supporter of combustion, which means that it's basically a better version of oxygen when it comes to supporting fires. Basically, when placed in contact with \ce{O_2F_2}, other materials spontaneously combust. Here's an analogy: Substances such as nitroglycerin and TNT are like a person with a short ... 22 This is actually related to the thermite reaction, where the rust on the balls reacts with the pure aluminum in the foil. Here is a pdf detailing the experiment. The activation energy for the reaction is reached through the friction and pressure, and the rust on the balls is actually very important, since it is the exchange of oxygen between the iron oxide ... 20 According to the source mentioned in the comments to your question, the first step is indeed nucleophilic substitution of the OH group by \ce{I-}, faciliated by protonation of the alcohol. For the second step (\ce{HI} reduction), a radical species was found as an intermediate, and therefore a reduction by single electron transfer (SET) with oxidation of ... 20 First, let's think about what makes a substance coloured: a substance will appear to be coloured if it absorbs light in the visible spectrum. Beta carotene: absorbs blue/green light (400–500 nm), so it looks red-orange when white light is shined on it as the blue/green light is subtracted from the white and what remains is what reaches our eyes. The ... 20 Your assertions are correct, for the most part. \ce{BH3} is a Lewis acid, and it does not reduce carbonyl groups by directly donating a hydride group like \ce{NaBH4} does. The \ce{NaBH4} reduction mechanism is fairly short and involves a direct transfer of the a hydride ion to an electron deficient carbonyl carbon: \hspace{2.4cm} \ce{NaBH4} is ... 20 Aldehydes, including aldoses, are oxidized to their respective carboxylic acids in the presence of \ce{Br2} in \ce{H2O}. The reason this reaction is often discussed with carbohydrates is that it is useful for differentiating aldoses from from ketoses, which cannot be further oxidized. A solution of \ce{Br2} and \ce{H2O} will lose its characteristic ... 19 There is no definitive answer; if you think you have one, you are wrong. See, this is much like asking "what is the northernmost big city". Depending on where you draw the line for being "big", the answer may be Moscow (latitude 55^\circN, population 13M), St. Petersburg (60^\circN, 5M), Murmansk (68^\circN, 300K), and quite a few others. There is no ... 18 This is a rather interesting question because these names actually refer to classes of reactions (specific to certain reagents and products), and aren't constrained by specific proportions of substances or even the identity of these substances. \hspace{4cm} A Rosenmund catalyst is used to reduce acyl chlorides to their corresponding aldehydes, and is ... 17 This post deals with the mechanism that is observed in the gas phase. It is of course not as simple as the equation might suggest and you did suspect that already.$$\ce{2H2 + O2 -> 2H2O}$$This will be divided into many different elementary sub reactions. Any mixture of oxygen and hydrogen is metastable (stable as long as you do not change the ... 15 The migratory aptitude list for the Baeyer-Villager oxidation is as it is because that is how the molecules undergoing the reaction behave. When we are learning about new reactions from a textbook, we often read about the mechanism and theoretical explanations of chemo-, regio-, and stereoselectivity. It is easy to forget that the experimental observations ... 15 Above a 1000\ \mathrm K, the \Delta_\mathrm fG of \ce {CO} from \ce C is more negative than that of \ce{CO2} formation from \ce C. Therefore, during smelting, when coke (\ce C) reacts with \ce {SnO2}, the formation of \ce {CO} rather than \ce {CO2} is thermodynamically preferable. Below that temperature, the reduction with carbon is not ... 15 Here are the two half reactions:$$\begin{align} \ce{[Ag(NH3)2]+ + e- &-> Ag^0 + 2NH3} \\ \ce{RCHO + 3OH- &-> RCO2- + 2H2O + 2e-} \end{align}$$which together yield the overall reaction$$\ce{2[Ag(NH3)2]+ + RCHO + 3OH- -> 2Ag^0 + RCO2- + 4NH3 + 2H2O}$$Here is a diagram of the reaction mechanism. The carbonyl group is oxidized in the ... 15 If we examine the Nernst equation: E = E^\circ-\frac{RT}{zF}\ln Q, the logarithmic term is what changes with concentration. The reaction quotient Q, gets smaller as the ratio of reactants to products increases, meaning that the log term will decrease (and become negative below Q = 1), so both equations will become more favourable as the ratio of ... 15 In the first equation, \ce{Fe} is oxidised to \ce{Fe^2+} by \ce{H+}; the \ce{SO4^2-} ions play no role in the reaction. You could replace \ce{H2SO4} with \ce{HCl} and get the same reaction. In the second equation, the same oxidation of \ce{Fe} to \ce{Fe^2+} by \ce{H+} also occurs. However, on top of that, some of the \ce{SO4^2-} ions are ... 15 You hit it right on the nose. The real key piece of information is that given enough time, all the unsaturated bonds will be reduced. This tells you that though the reduction is thermodynamically favorable, it is the difference in the energy barriers (\ce{\Delta \mathrm{G^{‡}}}) that prevents the carbonyl reduction from occurring at the same rate as the ... 15 Thermal decomposition of potassium chlorate is not disproportionation, just a redox reaction. Disproportionation refers to the same element acting both as oxidizing agent and a reducing agent, resulting in compounds which contain the same element in different oxidation states. On the other hand, if you consider preliminary decomposition stage involving ... 14 As noted in the cited paper by Sharpless and Gordon^{[1]}, the problem with the Corey-Schaefer mechanism is the existence of the selenous ester (7) which, due to its instability, would undergo hydrolysis prior to any elimination reaction. A study by Qingjiang Li and Tochtrop^{[2]} supports Sharpless and Gordon's counterargument. The former pair give a ... 14 I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements The preface of the book says: The approach: These reactions represent important facets of the elements involved but are typically presented as no more than ... 13 Without the salt bridge, the solution in the anode compartment would become positively charged and the solution in the cathode compartment would become negatively charged, because of the charge imbalance, the electrode reaction would quickly come to a halt. It helps to maintain the flow of electrons from the oxidation half-cell to a reduction half cell, ... 13 Permanganate as oxidizing agent works most efficiently in acidic solution, because it is reduced to the greatest extent in this medium, from oxidation state +VII in \ce{MnO4-} to +II in \ce{Mn^2+}.$$\ce{8H+ + MnO4- + 5e- ->~ Mn^2+ + 4H2O} Therefore, the number of electrons transferred from oxidized species per mole $\ce{KMnO4}$ (5 electrons) is ...

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This is based off my memory, as far as I can remember what my teacher taught me. Note, instead of $\ce{Pt}$, you can go for $\ce{Ni}$ (or Raney Nickel) or even $\ce{Pd}$ as a catalyst. Functional groups such as acid halide, cyanide, nitro, aldehyde, ketone, alkene, and alkyne are easily reduced by $\ce{H2/Ni}$. Some other groups can be reduced with \$\ce{H2/...

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