<u>Answer:</u> The entropy change of the process is 
<u>Explanation:</u>
To calculate the entropy change for different phase at same temperature, we use the equation:
where,
= Entropy change
n = moles of acetone = 6.3 moles
= enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol (Conversion factor: 1 kJ = 1000 J)
T = temperature of the system = ![-94.7^oC=[273-94.7]=178.3K](https://tex.z-dn.net/?f=-94.7%5EoC%3D%5B273-94.7%5D%3D178.3K)
Putting values in above equation, we get:
Hence, the entropy change of the process is
Answer:
406.45mL
Explanation:
The following data were obtained from the question:
V1 = 350mL
P1 = 720mmHg
P2 = 630mmHg
V2 =?
The new volume can be obtain as follows:
P1V1 = P2V2
720 x 350 = 620 x v2
Divide both side by 620
V2 = (720 x 350) /620
V2 = 406.45mL
The new volume of the gas is 406.45mL
Answer:
the answer of this question is true
Carbon monoxide (CO) is a colorless, odorless, and tasteless gas that is slightly less dense than air. ... Carbon monoxide consists of one carbon atom and one oxygen atom, connected by a triple bond that consists of two covalent bonds as well as one dative covalent bond.
I think it's probably not right.
