a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp AgBr = s²
c. 5 x 10⁻¹³ mol/L
<h3>Further explanation</h3>
Given
solubility AgBr = 7.07 x 10⁻⁷ mol/L
Required
The dissolution reaction
Ksp
The solubility product constant
Solution
a. dissolution reaction of AgBr
AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp
Ksp AgBr = [Ag⁺] [Br⁻]
Ksp AgBr = (s) (s)
Ksp AgBr = s²
c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³ mol/L
We know that in 1 mole there is Avogadro’s number of atoms(no matter of what kind of mole) 6.02 x 10^23.
In order to find the number of moles we just apply the rule of three.
1 mole.............. 6.02 x 10^23.
x mole..............<span>6.02 x 10^25
x = </span>6.02 x 10^25/6.02 x 10^23. = 10^2 = 100 moles
The neutralization reaction is expressed as:
2KOH + H2SO4 = K2SO4 + 2H2O
We are given the volume and the concentration of the KOH that is used in the reaction. We use this and the relations from the reaction to determine the concentration of the acid.
0.150 mol KOH / L ( .025 L ) ( 1 mol H2SO4 / 2 mol KOH ) = 0.00188 mol H2SO4
Concentration of acid = 0.00188 mol / .015 = 0.125 M of sulfuric acid
Rate of Effusion of Unknown gas = rX = 0.5
Rate of Effusion of Hydrogen = rH = 1
M.mass of Hydrogen = mH = 2 g/mol
M.mass of Unknown gas = mX = ?
Formula used:
rH / rX = √mX / mH
taking square on both sides,
r²H / r²X = mX / mH
Solving for mX,
mX = (r²H ÷ r²X) × mH
Putting values,
mX = [1² ÷ (0.5)²] × 2 g/mol
mX = (1 ÷ 0.25) × 2 g/mol
mX = 4 × 2 g/mol
mX = 8 g/mol
Result:
Molecular Mass of Unknown gas is 8 g/mol.