Answer:
Part C: P2 = 0.30 atm
Part D: V1 = 16.22 L.
Explanation:
Part C:
Initial pressure (P1) = 2.67 atm
Initial volume (V1) = 5.54 mL
Final pressure (P2) =.?
Final volume (V2) = 49 mL
The final pressure (P2) can be obtained as follow:
P1V1 = P2V2
2.67 x 5.54 = P2 x 49
Divide both side by 49
P2 = (2.67 x 5.54)/49
P2 = 0.30 atm
Therefore, the final pressure (P2) is 0.30 atm
Part D:
Initial pressure (P1) = 348 Torr
Initial volume (V1) =?
Final pressure (P2) = 684 Torr
Final volume (V2) = 8.25 L
The initial volume (V1) can be obtained as follow:
P1V1 = P2V2
348 x V1 = 684 x 8.25
Divide both side by 348
V1 = (684 x 8.25)/348
V1 = 16.22 L
Therefore, the initial volume (V1) is 16.22 L
POH of a 0.0072 M=-lg(0.0072) = 2.1426675
Answer:
pH = 4.34
Explanation:
pH= -1/2(logKa) -1/2(log C)
= -1/2( log 5.98*10^-8) -1/2(log 0.0353)
=-1/2(-7.22)-1/2(-1.45)
=3.61+0.725= 4.34
Answer:
The answer is "3.57 and 0.07".
Explanation:
Using the slop formula:
Given:
length path 
from calibration it is found that
Answer:
pH = 2.0
Explanation:
To find the pH of a solution, take the -log[H+]. In this case, the -log(9.4 x 10^-3) equals 2.02687 which makes 2.0 when accounting for significant figures.
