Answer :
B
!!!!!!!!!!!!!!!
<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>
<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>
<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>
<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>
<span>73.0<span>Tf</span>=2561 °C</span>
<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>
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: base times height
Explanation:
Answer:
Explanation:
After the glorious appearance of Full Moon, the lunar shape starts to wane, meaning it gets smaller. It's visible later at night and into the early morning, and we see a steadily shrinking shape of the lunar surface that's being lit up.
