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.
Boiling point is the temperature at which a liquid boils and turns to a gas.
Hope this helps!! (:
The oxidation number of H is -1.
Sum of the oxidation numbers in each element =
charge of the complex
CaH₂ has 1 Ca atom and 2H atoms. The charge of
the complex is zero. Let’s say Oxidation number of H is "a".
Then,
<span> (+2)
+ 2 x a = 0 </span>
<span> +2 + 2a = 0</span>
2a = -2
a = -1
Hence, the oxidation number of Hydrogen atom in CaH₂ is -1
Answer:
2 Na + 1 Cl2 -> 2 NaCl
Explanation:
The answer is really simple, because if you have 1 nonmetal element that has a subscript of 2, you need to multiply the product and the first reactant by 2 to balance it.
