I suspect that the pressure of this change is constant therefore
The equation is used from the combined gas law. (When pressure is constant both P's will cancel out P/P = 1)
V/T = V/T
Initial Change
Initially we have 2L at 20 degress what temperature will be at 1L.
2/20 = 1/T
0.1 = 1/T
0.1T = 1
T = 1/0.1
T = 10 degress celsius.
Hope this helps if you won't be able to understand what is the combined gas law just tell me :).
Answer:
The answer to your question is P2 = 170.9 torr
Explanation:
Data
Volume 1 = 12.1 l Volume 2 = 21.1 l
Temperature 1 = 241 °K Temperature 2 = 298°K
Pressure 1 = 546 torr Pressure 2 = ?
Process
To solve this problem use the combined gas law.
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = T1V1T2 / T1V2
-Substitution
P2 = (241 x 12.1 x 298) / (241 x 21.1)
-Simplification
P2 = 868997.8 / 5085.1
-Result
P2 = 170.9 torr
Answer: Option (A) is the correct answer.
Explanation:
Force acting on a dam is as follows.
F = .......... (1)
Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.
F' =
F' = ........... (2)
Now, dividing equation (1) by equation (2) as follows.
=
Cancelling the common terms we get the following.
=
4F = F'
Thus, we can conclude that if doubled the depth of the dam the hydrostatic force will be 4F.
Answer:
Explanation:
The ΔH for 6 moles is +84 J. So you just divide +84 J by 6 moles to get +14 J/mol (ΔΗreaction).