To find the mass you need to find the weight of a mol of the molecules by adding up the atomic mass.
N = 14.007 g/mol
H = 1.008 g/mol
S = 32.065 g/mol
O = 16 g/mol
2(14.007) + 8(1.008) + 32.065 + 4(16) = 132.143 g/mol
Now you know how much an entire mol weight you multiply it by how much you actually have
0.00456 * 132.143 = 0.603 g
Answer:
B) 7.7
Explanation:
For the reaction Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq)
Kc = (CO₃²⁻) / (CrO₄²⁻)
and the Ksp given are
Ag₂CO₃ ⇒ 2 Ag⁺(aq) + CO₃²⁻(aq) Ksp₁ = (Ag⁺)²(CO₃²⁻)
Ag₂CrO₄ ⇒ 2 Ag⁺(aq)+ CrO₄²⁻(aq) Ksp₂ = (Ag⁺)²(CrO₄²⁻)
Where (...) indicate concentrations M
Notice if we divide the expressions for Ksp we get:
Ksp₁/Ksp₂ = (CO₃²⁻) / (CrO₄²⁻) = 8.5 x 10⁻¹² / 1.1 x 10⁻¹² = 7.7
which is the desired answer.