Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
There are 2 molecules of Carbon dioxide(CO2)
<h3>Further explanation</h3>
Given
Molecules of CO2
Required
The number of molecules
Solution
The coefficient of a molecule shows the number of that molecule, while the subscript after the name of the atom indicates the number of that atom in the molecule
Because there is a coefficient of 2 in front of the CO2 molecule, then the number of CO2 molecules is 2
Hello!
Reading the worksheet shows you how to find the concentration of liquid water in a substance, but when converting a decimal to a percentage, remember that ALL percentages are out of 100, so, we multiply the decimal by 100 to convert it into a percentage.
Liquid A: total amount: 10 millimeters & amount of water: 7 millimeters
7/10 = 0.7 × 100 = 70%
Liquid B: total amount: 100 millimeters & amount of water: 92 millimeters
92/100 = 0.92 × 100 = 92%
Liquid C: total amount: 15 millimeters & amount of water: 13 millimeters
13/15 = 0.867 × 100 = 86.7%
Liquid D: total amount: 28 millimeters & amount of water: 22 millimeters
22/28 = 0.786 × 100 = 78.6%
<u>Final answers</u>:
- Liquid A: 70%
- Liquid B: 92%
- Liquid C: 86.7%
- Liquid D: 78.6%
