1. Write the equilibrium expressions for the following reactions: a) 2N2O(g) + O2(g) rightarrow 4NO(g)
b) H2(g) + CO2(g) rightarrow H2O(g) + CO(g)
c) 4HCl(g) + O2(g) rightarrow 2Cl2(g) + 2 H2O(g)
d) 3H2(g) + N2(g) rightarrow 2NH3(g)
e) H2(g) + Cl2(g) rightarrow 2HCl(g)
f) 2SO2(g) + O2(g) rightarrow 2SO3(g)
2. Using the above equilibrium expressions, calculate the equilibrium constant, K, for the following concentrations:
a) [N2O] = 0.0035 M, [O2] = 0.0027 M, [NO] = 5.6 x 10-9 M
b) [H2] = 0.61 M, [CO2] = 1.6 M, [H2O] = 1.1 M, [CO] = 1.4 M
c) [HCl] = 1.2 x 10-3 M, [O2] = 3.8 x 10-4 M, [H2O] = 5.8 x 10-2 M, [Cl2] = 5.8 x 10-2 M
d) [NH3] = 0.62 M, [H2] = 0.14 M, [N2] = 0.45 M
e) [H2] = 0.0045 M, [Cl2] = 0.0045 M, [HCl] = 0.0625 M
f) [SO2] = 2.00 M, [O2] = 1.50 M, [SO3] = 3.00 M

i belive it is water

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

intermolecular forces are the forces between molecules or particles.

there are two kinds of van der waals forces: weak london dispersion forces and stronger dipole-dipole forces.

in molecule of water there are dipole-dipole forces between negatively charged oxygen from one molecule of water and positively charger hydrogen of another molecule of water.

hydrogen bond is an electrostatic attraction between two polar groups that occurs when a hydrogen atom (h), covalently bound to a highly electronegative atom such as flourine (f), oxygen (o) and nitrogen (n) atoms.