Answer:
5.52atm
Explanation:
Using the pressure law formula:
P1/T1 = P2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the question, the following information were provided;
P1 = 4.72 atm
P2 = ?
T1 = -3.50°C = -3.50 + 273 = 269.5K
T2 = 42°C = 42 + 273 = 315K
Using P1/T1 = P2/T2
4.72/269.5 = P2/315
CROSS MULTIPLY
4.72 × 315 = 269.5 × P2
1,486.8 = 269.5P2
P2 = 1,486.8 ÷ 269.5
P2 = 5.52atm
Answer:
13,200 mL
Explanation:
multiply by 1000 to go from L to mL
CuCl2 + 2NaNO3 ----> Cu(NO3)2 + 2NaCl
using molar masses:-
Theoretical yields:-
63.54 + 2(35.45) g of CuCl2 produces 2(22.98 + 35.45) g of NaCl
134.44 g .................................................... 116.86 g
31.0 g ....................................................31.0 * 116.86 /134.44=26.95g
So percentage yield is 21.2* 100 / 26.95 = 78.7% to nearest tenth
Answer:
a) pH = 4.68 (more effective)
b) pH =4.44.
Explanation:
The pH of buffer solution is obtained by Henderson Hassalbalch's equation.
The equation is:
![pH =pKa +log\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3DpKa%20%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
a) pKa of acetic acid = 4.74
[salt] = [CH₃COONa] = 1.4 M
[acid] = [CH₃COOH] = 1.6 M

This is more effective as there is very less difference in the concentration of salt and acid.
b) pKa of acetic acid = 4.74
[salt] = [CH₃COONa] = 0.1 M
[acid] = [CH₃COOH] = 0.2 M

Answer:
At the top of Group 11 above silver and gold.
Period 4
Explanation: