weiming1998

Although HF is a weak acid in aqueous solution, as it becomes more and more concentrated (approaching 100% hydrogen fluoride) , it becomes a lot more acidic, dissociating into H+ and HF2- ions instead, and doing so far more readily than when it is diluted in water. This could mean that it can donate a proton to the H2SO4, forming the H3SO4+ ion and the stable HF2- ion. Read this (http://en.wikipedia.org/wiki/Hydrofluoric_acid).

There are several ways you can concentrate 3% H2O2. One way is by freezing it. Since H2O2 has a lower freezing point than water, it will stay liquid while all the water freezes. It is possible to get the H2O2 to at least 15% with freezing. With fractional crystallisation, a 60% concentrated solution could be made, but this would be difficult to set up. Another way is by evaporation/boiling. For your purposes, you do not need to fear about combustible vapours (H2O2 vapours are only significantly dangerous/explosive if your H2O2 is above 70% in concentration), but cranking the heat all the way up will only result in the decomposition of the H2O2. Ideally, a vacuum would be used to remove all the water, leaving behind extremely concentrated H2O2 (more than 90% concentration) since H2O2 is less volatile than water, but you can let the 3% H2O2 evaporate on the stove at low heat. Make sure the glassware is clean and free from any sort of metal ions (which catalyses the decomposition of the H2O2).

Because an alcohol, after all, is just any molecule with an OH group. The rest of the molecule could be anything from just a simple saturated chain of carbon and hydrogen to ring/rings made of carbon and hydrogen to branching structures containing both rings, carbon hydrogen chains, double/triple bonds, and other functional groups. It is like the difference between a frog and an elephant. They both have four legs, but they are drastically different in most other aspects, and thus doesn't even look remotely similar.

Add a couple of ml of the unknown acid that you're testing to a dilute solution of potassium permanganate or dichromate. If it is sulfurous acid, the bright orange/purple colour will change. If it is sulfuric acid, the colour won't change. Carefully smell the acid. Sulfuric acid hardly has a smell at room temperature, while sulfurous acid (which is just SO2 in water) has an odour resembling burnt matches (when dilute) and stings your nose (when more concentrated) Sulfurous acid has a bleaching effect on some dyes, while sulfuric acid doesn't.

Question 1: (a), The ratio of NH4NO3:Na3PO4 is 3:1. 3 moles of NH4NO3 is approximately 240 grams while 1 mole of Na3PO4 is approximately 164 grams. 30 is 1/8 of 240. 1/8 of 164 is 20.5 grams. Clearly the limiting reagent is NH4NO3. (b), As NH4NO3 is the limiting reagent, the maximum amount of moles of product is 1/8 of the amount of moles of each product in the equation. Thus, The maximum amount of (NH4)3PO4 formed is 1/8 moles and the maximum amount of NaNO3 formed is 3/8 moles. You can work out the rest from here. ©, This can be worked out using the information I gave you in (a). Question 2: (a), The ratio of CaCO3:FePO4 is 3:2. 3 moles of CaCO3 is 300 grams while 2 moles of FePO4 is 302 grams. 100 grams is 1/3 of 300. 1/3 of 302 is 100.6 grams. FePO4 is the limiting reagent. (b) 45 grams of FePO4 is 14.9% of 302. Thus, 14.9% of 300 grams of CaCO3 reacts to form 14.9% of a mole of Ca3(PO4)2 and 14.9% of a mole of Fe2(CO3)3. © This can also be worked out using the information I gave you in (a). Question 3: 1 mole of Mg reacts to form 1 mole of H2. One mole of Mg is 24.3 grams. Thus, 40 grams is 1.65 moles of Mg. This can form 1.65 moles of H2. Question 4: This can be worked out by simply dividing the maximum amount of H2 that can form by the amount of hydrogen that actually formed, then using the resulting number to divide 100.

No, carbon dioxide is not polar. Although the C=O bond is somewhat polar, the molecule's symmetry balances out the polarity, resulting in a non-polar molecule. Thus, liquid CO2 is non-polar.

Stainless steel contains considerable amounts of chromium, which is in solution as trivalent Cr3+ ions. Iron is also dissolved in solution. However, it is in the form of Fe2+ ions that are a light grass-green in colour. The colour of Cr3+ is a much more intense deep green. Thus, stainless steel dissolved in hydrochloric acid yields a much deeper colour than iron dissolved in hydrochloric acid. I'm not sure why the solution looked greener after the addition of aluminium (optic effects due to precipitate of metal?), but the carbon-like particles are iron particles with traces of chromium in it. The aluminium displaced the Fe2+ and some Cr3+ out in a single displacement reaction.

Dilute nitric acid (3.0 mol/L would be dilute) reacts with copper in a different equation as the NO2 formed would redissolve back/oxidize the copper in solution: 3Cu(s)+8HNO3(aq)----->3Cu(NO3)2(aq)+2NO(g)+4H2O(l) 2.5L of the solution contain 7.5 moles of nitric acid (3x2.5). Multiply that by 3/8 (every 8 moles of HNO3 reacts with 3 moles of Cu) and you get 2.8125. 2.8125 moles of copper had reacted. Multiply that by 64 (atomic weight of copper) and you get 180. 180 grams of copper had reacted. 500-180=320.

Salt, or sodium chloride, is an ionic compound. That means that salt actually exists as Na+ and Cl- ions stacked together. The Na+ and the Cl- ions are bonded together with ionic bonds. As salt crystals dissolve in water, these bonds break, releasing Na+ and Cl- ions. Breaking bonds requires energy, so this is endothermic. But it doesn't stop here. The Na+ ion doesn't exist as free ions, as the negatively charged ends of water is attracted to and bonds to it, forming the coordination complex [Na(H2O)6}+, amongst other things. Bond-forming is exothermic, so this releases energy. I'm not sure about what happens to the Cl- ion, as it is a rather inert anion. The energy released and the energy absorbed balances each other out, resulting in a (very slight) release of heat when salt dissolves in water. As for why a salt-water mix freezes at a lower temperature, that is because in order for water to freeze, the molecules of water must crystallize in a certain structure. Salt interferes with the crystallisation of water, resulting in a lower freezing point. Other substances like antifreeze or various alcohols also interferes with the crystallisation of water and lowers the freezing point.

The ferric nitrate definitely decomposed. At such high temperatures (180 degrees), a solution of ferric nitrate will decompose (hydrolyse), producing ferric oxides, ferric basic nitrates, and nitrogen oxides. A lot of transitional metal nitrates (copper, iron, etc) are very sensitive to the combination of water and high temperatures, and will decompose. The correct way to prepare hydrated iron (III) nitrate is by combining hydrogen peroxide (dilute nitric acid will not oxidise iron (II) to iron (III) easily), excess dilute nitric acid, and iron without heating, then evaporate (not boil) the resulting solution. The anhydrous nitrate is a strong nitrating agent, and prepared by combining liquid N2O4 with iron. You don't need iron (III) nitrate to etch silver. Nitric acid works much better. Hot, concentrated nitric acid attacks silver vigorously. Even dilute nitric acid will etch silver slowly when heated. Other things (such as a mixture of hydrogen peroxide and vinegar) etches silver as well.

The gelatinous blue mass sounds like copper (II) hydroxide to me. That's strange, because it should have dissolved already in the nitric acid. Or maybe it is precipitated hydrated copper (II) nitrate, which makes more sense. Anyway, just add water, and the mass should dissolve if it is the nitrate. If not, then add nitric acid, and it should dissolve.

1, I deducted it from the chemistry of silicon. Silicon dioxide dissolves in HF to form hexafluorosilicic acid. Zirconium dioxide acts very similarly to silicon dioxide, and as hexafluorozirconic acid exists, I deducted that it would form. To actually practically test it, heat the resulting solution of H2ZrF6 to dryness. If fumes of HF in addition to water (careful!) is emitted, then the compound is almost definitely H2ZrF6. 2, Yes, but the formation of these ions are negligible and won't affect anything in a 48% solution. H2F2 doesn't exist, as far as I know.

An Arrhenius base is anything that increases the concentration of OH- in aqueous solution. In simpler terms, anything that causes the pH to rise above 7 when dissolved/mixed in water.

Some HCl will be formed by adding salt to vinegar, but in very small amounts (an equilibrium), I think a mix of salt and vinegar cleans pennies due to the complexing ability of Cl-. Usually, the reaction between vinegar and copper oxide/sulfide is very slow due to the weak acidity of the vinegar. The vinegar dissolves the layer of oxides, forming [Cu(H2O)6]2+ and CH3COO-. What the chloride does, however, is allowing the formation of CuCl4(2-) within the solution. Thus, the equation can be written as: CuO(s)+2H+(aq)+4Cl-(aq)---->CuCl4(2-)+H2O. The chloride ions can also penetrate the oxide layer easily, as it penetrates the layer of Al2O3 that usually surrounds aluminium metal, causing it to react with water.

Ammonia has an extremely weak tendency to donate protons outside aqueous solution, forming amides and nitrides. Only extremely strong bases, like butyllithium (I think) can deprotonate ammonia. The pKa of ammonia is about 40, while the pKa of water is about 15-16. Arrhenius acids donates protons in AQUEOUS SOLUTION, while if a substance can donate protons at all, it is a Bronsted-lowry acid. Now it's your turn to think.