Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
Answer:
The atoms in the first period have electrons in 1 energy level. The atoms in the second period have electrons in 2 energy levels. The atoms in the third period have electrons in 3 energy levels. The atoms in the fourth period have electrons in 4 energy levels.
Molar mass RbMnO₄ = 204.40 g/mol
1 mole ---------- 204.40 g
7.88 mole ------ ?
mass = 7.88 * 204.40 / 1
mass = 1610.672 g
hope this helps!
Answer:
a. 27g/mol
b. 1.85 x 10^5 moles
Explanation:Please see attachment for explanation
Answer:
false.nowadays humans are destroying forest,extracted more minerals,making pollution. in my view.
