# chemistry

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When carbon dioxide is heated in a closed container, it decomposes into carbon monoxide and oxygen according to the following equilibrium equation:
2CO2(g) ---> 2CO(g) + O2(g)

When 2.0 mol of CO2(g) is placed in a 5.0-L closed container andheated to a particular temperature, the equilibrium concentration of CO2(g) is measured to be 0.039 mol/L. Use an ICE table to determine the equuilibrium concentrations of CO(g) and O2(g)

• chemistry - ,

It's difficult to do spacing on these boards but if I turn the equation down instead of sideways, we may be able to do it. First will be I, followed by C and E horizontally. Let me know if this is too confusing. (First, (CO2) = 2.0 mols/5 L = 0.4 mols/L.
2CO2(g)  0.4   -2x   0.039
|
|
v
2CO(g)     0   +2x
+
O2(g)     0   +x
If equilibrium for CO2 is 0.039 and it was 0.4 initially, that means it must have changed by 0.4 - 0.039 = 0.361 and that is 2x. Then CO must be the same and O2 must be 1/2 that.

• chemistry - ,

Calculate the solubility of zinc hydroxide at 25 °C. The Ksp of Zn(OH)2(s) is 4.5 x 10^-17 at 25 °C?

What did I do wrong?

Zn(OH)2 <----> Zn2+ + 2 OH-

let x = moles/L of Zn(OH)2 that dissolve. This gives us x moles/L of Zn2+ and 2x moles / L of OH-
[Zn2+] = x
[OH-] = 2x

Ksp = [Zn2+][OH-]^2 = x ( 2x)^2 = 4x^3

4.5 x 10^-17 = 4x^3

x = molar solubility = 2.2 x 10^-6 M

• chemistry - ,

2.2 x 10^-6 M looks ok to me. Perhaps they want g/L instead of mols/L.