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:
For H-Cl, the direction is towards the chlorine atom
For F-CH3, the direction is towards the flourine atom.
Explanation:
The dipole moment is a vector quantity. This implies that it has both magnitude and direction.
Thus, the direction of the dipole moment always points from the positive atom towards the negative atom.
This explains the fact that it points to chlorine in HCl and points to flourine in F-CH3
Answer:
The correct answer is d. 6H20 + 6CO2.
The reactant in the chemical reaction 6H2O + 6CO2 ---> C6H12O6 + 6O2 is 6H20 + 6CO2. Remember that the reactant is always at the left side of the equation. So the correct answer is 6H20 + 6CO2 since it's in the left of the equation. I hope this answer helped you.
Explanation:
Since Au is a symbol for Gold, and once you split the name into to giving each ion its charge... you'll see that this compound has Au+2 and Cl03- .... so the name would be
Gold(II) Chlorate
Hope this helps!
