the answer of these question is weight
substitute: <span><span>t<span>1/2</span></span>=<span><span>ln(2)</span>k</span>→k=<span><span>ln(2)</span><span>t<span>1/2</span></span></span></span>
Into the appropriate equation: <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−kt</span></span></span>
<span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−<span><span>ln(2)</span><span>t<span>1/2</span></span></span>t</span></span></span>
<span>[A<span>]t</span>=(250.0 g)∗<span>e<span>−<span><span>ln(2)</span><span>3.823 days</span></span>(7.22 days)</span></span>=67.52 g</span>
Answer:
C) acid-base neutralization
Explanation:
NaOH + CH₃COOH = CH₃COONa + H₂O
Break the solutions apart:
NaOH = Na⁺ + OH⁻
CH₃COOH = CH₃COO⁻ + H⁺
Combine the resulting solution after the reaction:
OH⁻ + H⁺ = H₂O
