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Electrode Potential : The amount of electric potential difference through a half cell reaction.

It would be like that if you added only ions and not water with it. stargaze is right in that the conductivity it proportional to the sum of the absolute charges. And since the conductivity is taken over a specific volume it is proportional to the concentration of the ions as well. To...

Um, how about unsaturation? CH3Cl ----> -(-CH2-CHCl-)-n PS. What is vinyl chloride? What is PVC?

Might as well be bored and productive!

EDTA, ethylenediamine, 1,10-phenanthroline, DEDTC, acetylacetanato, oxalato, dimethylglyoxime and glycinato. I think... it was last session so it is a tad distant now!

It means it is monoprotic.

NH3 + HNO3 -> NH4+ + NO3- For HNO3; c = 0.151 M V = 0.0272 L n = 0.151 * 0.0272 = 0.00411 mol For diluted NH3; n =1/1 * 0.00411 mol V = 0.025 L c = 0.00411/0.025 = 0.164 M dillution factor is 500/25 = 20 For undiluted NH3; c = 0.164 * 20 = 3.29 M

H(C8H4O4)- + OH- -> C8H4O4 2- + H20 For KH(C8H4O4); m = 0.917 g M = 204.2 g/mol n = 0.00449 mol For NaOH; n = 1/1 * 0.00449 mol V = 0.0272 L c = 0.00449/0.0272 = 0.165 M

Hmmm ... I've always wanted to use a hydrolic press to make acid pellets and slip 'em into someone's [insert fizzy drink]. Though I fear people may get the wrong impression and I'll end up in a cell! You know you are a chem nerd when the morning/afternoon after a cocktail party you find...

The formal oxidation numbers (ON) are; 1/2 H2 + OH- --> H2O + e- E = 0.83V .......0.....-2.1........1..-2 So the ON of one H is increased by one; so this is an oxidation half reaction. To be an actual reaction we need to pair it with a reduction half equation; Cu 2+ + 2e- --> Cu(s)...

It says the beakers are labelled in order of collection AND that pure substances were collected. Since Methyl Propanoate has the second lowest BP it would have been collected in beaker 2.

Hence the assumption. I would hate a multiple choice q. without the assumption:: [H+] = 0.0005 + y [HSO4-] = 0.0005 - y [SO4 2-] = y 0.0005^2 - y^2 = y * 10^-1.99 0 = y^2 + y * 10^-1.99 - 0.00000025 2y = -10^-1.99 + &sqrt;(10^-3.98 + 0.000001) y = 2.437 * 10 ^-5 [H+] =...

Hi dchait, NaOH completely dissociates in water. So [OH-] = [NaOH] [OH-] = 0.13 M The equation for the ionic product of water is; [H+][OH-] = Kw = 1.008*10^-14 @ 25C [H+] = 0.77 fM (that's femtomolar!) pH = - log [H+] pH = -log 0.7753846...*10^-15 pH = 13.1

Q27 a BaSO4: M = 233.4 g/mol m = 1.8g n = 0.00771 mol SO4 2-: M = 96.1 g/mo n = 0.00771 mol m = 0.741 g Percentage Sulfate = 0.741/1 *100% = 74.1% b This reaction has several limitations. The scale of the reaction and the slight solubility of BaSO4 makes it useless for...

It is important to understand that the second proton is more tightly bound, though to put it into perspective; A neutralisation reaction involving sulfuric acid reaches an end point at ~pH 7. The solution contains 2000000 more SO4 2- ions than HSO4 - ions. This may seem to condradict my...

Yes sulfuric acid is diprotic, though in solution bisulfate ions aren't present in any great quantity (unless you want to add a load of HCl as well). This is a limiting reagents question. In excess NaOH it is certainly a). Though consider; 2H2SO4(aq) + 2NaOH(aq) → 2Na+ 2SO4 2- + 2H+ +...

Volatile Liquid : A liquid with a relatively low B.P., such that the substance vapourises with a significant partial pressure within the range of 'room temperatures'.

^^ Doesn't necessitate being used for ozone levels. I am certainly glad the last one I used wasn't as you describe it!

Q1. The solubilities depend on the amount/type of dissolved molecules in the solvent and temperature. Silver acetate will dissolve 1.11% (w/w) in deionized water at standard conditions. It is something like an additional 10K doubles reaction rates, and since partial solubility is an eq reaction...

Also Q.2 is a redox reaction so an applied voltage would favour one side or the other, that is assuming you had the system setup correctly. Though that isn't really a concern of yr 11 is it? :D