They will stay the same through the laws of conservation of mass even when they have changed state so they will have the reactant of Ice will be the same amount as the product of water
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: m = 24.31 g/mol · 1.13 mol
Explanation: 2 mol HCl use 1 mol Mg.
Magnesium is used 0.5 · 2.26 mol = 1.13 mol
M(Mg) = 24.31 g/ mol
The CH4 molecule has the lowest molecular weight, so it has the lowest boiling point.
Hope I helped :)
