Inorganic Chemistry Answers

A stimulant in coffee and tea is caffeine, a substance of molar mass 194 g/mol. When 0.376 g of caffeine was burned, 0.682 g of carbon dioxide, 0.174 g of water, and 0.110 g of nitrogen were formed. Determine the empirical and molecular formulas of caffeine, and write the equation for its combustion.

A stimulant in coffee and tea is caffeine, a substance of molar mass 194 g/mol. When 0.376 g of caffeine was burned, 0.682 g of carbon dioxide, 0.174 g of water, and 0.110 g of nitrogen were formed. Determine the empirical and molecular formulas of caffeine, and write the equation for its combustion.

Determine the hybridization of the metal centre in the complexes below using Ligand Field Theory.

[RhCl3(OH)3]3-
[Cr(en)2(PMe3)2]

What is the % yield of 7.50g of nitrogen reacted with 1.30g of hydrogen and 5.75 g of ammonia is produced

how to use Molecular orbital theory to determine the hybridization of the metal centre in the complex.

1. [RhCl3(OH)3]3-

2. [Cr(en)2(PMe3)2]

Determine the hybridization of the metal centre in the complexes below using Ligand Field Theory.

1. [RhCl3(OH)3]3-

2. [Cr(en)2(PMe3)2]

Given the following data:

2 H2 (g) + C (s) → CH4 (g) ΔG° = -51 kJ

2 H2 (g) + O2 (g) → 2H2O (ℓ) ΔG° = -474 kJ

C (s) + O2 (g) → CO2 (g) ΔG° = -394 kJ

Calculate ΔG° for CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (ℓ).

Calculate the volume of 10.0 carat diamond (1 carat at = 200.0mg). The density of diamond is 3.51g/ml?

Balance the following reaction using changes in oxidation number,with steps

a MnO2. +. bH+.----->c Mn2+ dMnO-4. + e H2O

Thank you very much, sir.
I will like to point out please that I believe question b) Kcr (i.e. Kc for the reverse reaction)= 1÷ 0.65 = 1.5. Kcf (i.e. Kc for the forward reaction)= 0.65 while Kpf= 0.0002
While,
question c, why can't I just use the expression,
Kc3=√Kc1
Kc3=√0.65= 0.81
Where Kc1= N2(g) + 3H2(g) = 2NH3(g)
And Kc3= 1/2N2(g) + 3/2H2(g) = NH3(g)
, please?
PS: please take the = sign to mean a reversible reaction.
Thank you!!!!!!!