Answer:
41 g
Explanation:
We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.
pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]
pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]
log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]
log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40
[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M
We can find the mass of NaC₆H₅COO using the following expression.
M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L
mass NaC₆H₅COO = 41 g
4.1 h = 14760 s
<span>t 1/2 = ln 2 / k </span>
<span>k = rate reaction = 4.97 x 10^-5 </span>
<span>ln 0.045 / 0.36 = - 4.97 x 10^-5 t </span>
<span>2.08 = 4.97 x 10^-5 t </span>
<span>t = 41839.9 s = 11 h 37 min 19 s</span>
Answer:
The answer to your question is P2 = 2676.6 kPa
Explanation:
Data
Volume 1 = V1 = 12.8 L Volume 2 = V2 = 855 ml
Temperature 1 = T1 = -108°C Temperature 2 = 22°C
Pressure 1 = P1 = 100 kPa Pressure 2 = P2 = ?
Process
- To solve this problem use the Combined gas law.
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
- Convert temperature to °K
T1 = -108 + 273 = 165°K
T2 = 22 + 273 = 295°K
- Convert volume 2 to liters
1000 ml -------------------- 1 l
855 ml -------------------- x
x = (855 x 1) / 1000
x = 0.855 l
-Substitution
P2 = (12.8 x 100 x 295) / (165 x 0.855)
-Simplification
P2 = 377600 / 141.075
-Result
P2 = 2676.6 kPa
