Regard the principle of utilization of two gas.
Make a consistent control of hardware containing gas.
Make a consistent control of weight diminishing valves giving gas.
No smoking zone.
Answer:
ΔH = 125.94kJ
Explanation:
It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:
1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ
2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ
-1/2 (1):
WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ
3/2 (2):
3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ
The sum of last both reactions:
WO3(s) + 3H2(g) → W(s) + 3H2O(g)
ΔH = 842.7kJ -716.76kJ
<h3>ΔH = 125.94kJ </h3>
Answer:
it would be the second choice
Answer:
it will float back up because it's not dense enough to go down
Answer:
There are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.
Explanation:
We can calculate the number of moles (η) of BeO as follows:

Where:
m: is the mass = 250 g
M: is the molar mass = 25.0116 g/mol
Hence, the number of moles is:

Therefore, there are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.
I hope it helps you!