Can somebody answer these chemistry questions?

1.
Let z be the mass in grams of BaO in the mixture.
Then 3.50 - z is the mass of CaO in the mixture.

Supposing both oxides undergo the same sort of reaction in equimolar amounts.

z / (153.3271 g BaO/mol) x ( 233.3909 g BaSO4/mol) = g BaSO4

(3.50 - z) / (56.0778 g CaO/mol) x (136.1416 g CaSO4/mol) = g CaSO4

So the sum of the last two expressions equals 5.7204 g.

Written algebraically:

(z / (153.3271) x ( 233.3909 )) + ((3.50 - z) / (56.0778 ) x (136.1416 )) = 5.7204
Solve for z:
z = 3.07 g BaO
3.50 - 3.07 = 0.43 g CaO

What "numeral"?

2.
2 AgBr + Cl2 → Br2 + 2 AgCl

The weight loss is due to the difference in molar mass between AgBr and AgCl.
(0.1470 g) / ((187.7723 g AgBr/mol) - (143.3214 g AgCl/mol)) = 0.00330702 mol AgBr converted

(0.00330702 mol AgBr) x (187.7723 g AgBr/mol) = 0.620967 g AgBr in the original answer

1.00 g total - 0.620967 g AgBr = 0.379033 g AgCl

(0.379033 g AgCl) / (143.3214 g AgCl/mol) x (1 mol Cl / 1 mol AgCl) x (35.4532 g Cl/mol) / (1.00 g) =
0.0938 = 9.38% Cl by mass