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.
Answer: 10.99
Explanation: because you take the Cao 13.9 and take CO2 which is 10.99 and it makes 24.8 . Which is CaCO3.
Answer:
a. -0.63 V
b. No
Explanation:
Step 1: Given data
- Standard reduction potential of the anode (E°red): -1.33 V
- Minimum standard cell potential (E°cell): 0.70 V
Step 2: Calculate the required standard reduction potential of the cathode
The galvanic cell must provide at least 0.70V of electrical power, that is:
E°cell > 0.70 V [1]
We can calculate the standard reduction potential of the cathode (E°cat) using the following expression.
E°cell = E°cat - E°an [2]
If we combine [1] and [2], we get,
E°cat - E°an > 0.70 V
E°cat > 0.70 V + E°an
E°cat > 0.70 V + (-1.33 V)
E°cat > -0.63 V
The minimum E°cat is -0.63 V and there is no maximum E°cat.
Solution :
a). ![$SO_3$](https://tex.z-dn.net/?f=%24SO_3%24)
This compound is known as sulfur trioxide.
The molecular shape of sulfur trioxide is trigonal planer.
And the bond angle is 120°.
b). ![$N_2O$](https://tex.z-dn.net/?f=%24N_2O%24)
This compound is known as Nitrous oxide. Here, nitrogen is in the center. There is no lone pair around the nitrogen atom and it forms two sigma bonds with the other two atoms.
It is linear in shape.
The bond angle between them is 180°.
c). ![$CH_2Cl_{2}$](https://tex.z-dn.net/?f=%24CH_2Cl_%7B2%7D%24)
This compound is known as the Dichloromethane.
The molecular shape of the compound is tetrahedral.
The bond angles is 120°