In thermodynamics<span>, </span>work<span> performed by a system is the energy transferred by the system to its surroundings. It can be calculated by the expression:
</span>
W = PdV
Integrating,
We will have,
W = P(V2 - V1)
133.7 (1 litre-atm / 101.325 Joule) ( <span>760 Torr / atm ) </span>= 783 (V2 - .0737 )
V2 = 1.35 L
Hope this answers the question. Have a nice day.
When waters salinity increases it’s freezing point will decrease. This is one of the reasons why people salt the roads after a snow storm.
Answer:
Explanation:
A tertiary alcohol is a compound (an alcohol) in which the carbon atom that has the hydroxyl group (-OH) is also bonded (saturated) to three different carbon atoms.
Based on the question, the only <u>tertiary alcohol that can result from C₆H₁₄O that have a 4-carbon chain</u> is
2-hydroxy-2,3-dimethylbutane
H OH H H
| | | |
H - C - C - C - C - H
| | | |
H CH₃ CH₃ H
From the above, we can see that the carbon atom having the hydroxyl group is also bonded to three other carbon atoms. And since we aren't considering stereochemistry, this is the only tertiary alcohol we can have with a 4-carbon chain
Answer:
The value of the equilibrium constant for reaction asked is 
.
Explanation:
![K_{goal}=\frac{[C][O_2]}{[CO_2]}](https://tex.z-dn.net/?f=K_%7Bgoal%7D%3D%5Cfrac%7B%5BC%5D%5BO_2%5D%7D%7B%5BCO_2%5D%7D)
..[1]
![K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BCH_3COOH%5D%5BO_2%5D%5E2%7D%7B%5BCO_2%5D%5E2%5BH_2O%5D%5E2%7D)
..[2]
![K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2%5D%5E2%5BO_2%5D%7D)
..[3]
![K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_3%3D%5Cfrac%7B%5BC%5D%5E2%5BH_2%5D%5E2%5BO_2%5D%7D%7B%5BCH_3COOH%5D%7D)
[1] + [2] + [3]
( on adding the equilibrium constant will get multiplied with each other)
![K=\frac{[C]^2[O_2]^2}{[CO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC%5D%5E2%5BO_2%5D%5E2%7D%7B%5BCO_2%5D%5E2%7D)
On comparing the K and 
:
The value of the equilibrium constant for reaction asked is .
