Using v1/t1=v2/t2
v1=500
v2=?
t1=75=368k
t2=225=498
500/368=v2/498
1.4x498=v2
v2=697.2ml
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:
b because no death and no emigration
Explanation:
Mass number<span> is the </span>number<span> of protons </span>and<span> neutrons in an atom.
</span>Atomic mass<span> is the average </span>mass<span> of all the isotopes of a certain type.</span>
