I have 3 chemistry questions that I can not figure out?

Question

First one is:How many moles of Cr(OH)3 will dissolve in 1.00 L of a solution with a pH of 7.63. (Assume the pH does not change as the Cr(OH)3 is placed in the water.) Second one is:Calculate the standard entropy in J/mol*K for the reaction: CaCO3(s) ® CaO(s) + CO2(g) Given:S°(CaCO3(s)) = 92.9 J/mol*K,S°(CaO(s)) = 40J /mol*K DS°(CO2(g)) = 213.6 J/mol*K. I have tried to calculate the standard entropy and it keeps on tellin me I 'm wrong. Can you tell me where I went wrong on this question. I took 40 J/mol*K because it is CaO entropy number and added it to 213.6 J/mol*K because that it CO2 entropy number then I subtracted 92.9 J/mol*K from CaO and CO2 put together and I got 160.7 kJ/mol*K and it told me I was wrong could please explain to me where I went wrong and tell me the correct answer Third one is: Estimate the value of the equilibrium constant, K°, at 1200 K for the reaction: 2 SO2(g) + O2(g) ® 2 SO3(g) SO2(g)H° -296.83 S°248.22, SO3(g) H° -395.72 S°256.7 O2(g) H° 0 S°205.14

whatever I put on the values is what they gave me. They didn't give me any ksp values for any questions.
The first one only had 1.0L and the pH of 7.63.
The second one is the one that it keeps on telling me I'm wrong
The third one it just gave me the balanced equation, the delta H and delta S for each of the species and it gave me the temperature.

Zor Prime,
For the second problem I put in that answer and it told me that I was wrong.

Answer 1

Lancenigo di Villorba (TV), Italy

FIRST CHEMISTRY's QUESTION)
You have to dissolve Chromium(III) Hydroxide in aqueous media. The chemical process involved

Cr(OH)3(s) <---> Cr+++(aq) + 3 OH-(aq)

is related to the Product's Solubility as defined for this compound at 25°C

6.0E-31 = Ksp = |Cr+++| * |OH-|^3

You wrote that the aqueous medium is characterized from
pH = 7.63, so you have defined |OH-|

|H+| = ALOG(0 - pH) = 2.3E-8 M
|OH-| = Kw / |H+| = 1.0E-14 / 2.3E-8 = 4.3E-7 M
|Cr+++| = Ksp / (|OH-|^3) = 6.0E-31 / (4.3E-7)^3 = 7.5E-12 M

This concentration results like 7.5E-12 mol in one liter.

SECOND CHEMISTRY's QUESTION
The reaction involved follows

CaCO3(s) ---> CaO(s) + CO2(g)

You gave me some informations, e.g. the Standard Formation's Entropy as evaluated at 25°C

S°f,CaCO3 = +92.9 J/(mol * K)
S°f,CaO = + 40.0 J/(mol * K)
S°f,CO2 = +213.6 J/(mol * K)

DeltaS° = (S°f,CO2 + S°f,CaO) - (S°f,CaCO3) =
= (213.6 + 40.0) - (92.9) = +160.7 J/(mol * K)

THIRD CHEMISTRY's QUESTION
The reaction involved follows

2 SO2(g) + O2(g) ---> 2 SO3(g)

You gave me some informations, e.g. the Standard Formation's Enthalpy as evaluated at 25°C

H°f,SO2 = -296.83 J/(mol * K)
H°f,O2 = 0.0 J/(mol * K)
H°f,SO3 = -395.72 J/(mol * K)

DeltaH° = (2 * H°f,SO3) - (2 * H°f,SO2 + H°f,O2) =
= (2 * (-395.72)) - (2 * (-296.83) + 0.0) = -197.78 kJ/mol

You gave me some informations, e.g. the Standard Formation's Entropy as evaluated at 25°C

S°f,SO2 = +248.22 J/(mol * K)
S°f,O2 = + 205.14 J/(mol * K)
S°f,SO3 = +256.70 J/(mol * K)

DeltaS° = (2 * S°f,SO3) - (2 * S°f,SO2 + S°f,O2) =
= (2 * 256.70) - (2 * 248.22 + 205.14) = -188.18 J/(mol * K)

Starting from the Undefined Gibb's Formula

DeltaG° = DeltaH° - T * DeltaS°
DeltaG° = -197,780 - 298.16 * (-188.18) = -141,672 J/mol

I will be able to apply the Differential Isobaric van't Hoff's Formula

(DeltaG / T) - (DeltaG° / T°) + (DeltaH° / (T°)^2) * (T - T°) = 0
DeltaG = T * [(DeltaG° / T°) - (DeltaH° / (T°)^2) * (T - T°)]
DeltaG = 1,200 * [(-141,672 / 298) - (-197,780 / (298)^2) * (1,200 - 298)] = +1,840,178 J/mol

Finally, I will apply the Isothermal van't Hoff relation

DeltaG / (R * T) + LN(Keq) = 0
Keq = EXP(0 - DeltaG / (R * T))
Keq = EXP(0 - (+1,840,178) / (8.31 * 1,200)) = 1.2E-80

I hope this helps you.

Answer 2

ok - you seriously is not abl=e to comprehend the dissociation of acetic acid in water in case you write the acid as: HCH3O2. Write it as ethanoic acid, that's its propername: CH3COOH + H2O ? CH3COO - + H3O+. Writing the acid this form you are able to very extremely see the ionisable H. This acid is a susceptible acid, it producesd a low share of H3O ions, so it really is a susceptible electrolyte. The concentrayted, glacial acetic acid does no longer dissociate. CH3COOH ? no dissociation, no ions, non-conductor of electricity. Dilute the acid: You get the equilibrium equation above. by technique of use of acceptable inert ( platinum) electrodes you receives oxygen and hydrogen liberated. notwithstanding the acid is any such susceptible electrolyte, the production of gas is sluggish. you receives hydrogen come off on the adverse electrode (4 H2O(l) + 4e- = 2 H2(g) + 4 OH-(aq)) and oxygen on the precious electrode (2 H2O(l) = O2(g) + 4 H+(aq) + 4e-), The acid acts in elementary words as an electrolyte and it really is the water that identity separated into hydrogen and oxygen,

Answer 3

Please supply Ksp with solubility problems.

Sometimes what ever is telling you that your answer is wrong is wrong. Refer it to your teacher.