Answer:
0.158 moles
Explanation:
We are given;
9.50 x 10^22 molecules of CO
We are required to determine the number of moles;
We need to know;
1 mole of a compound = 6.022 × 10^23 molecules
Therefore;
9.50 x 10^22 molecules of CO will be equivalent to;
= 9.50 x 10^22 molecules ÷ 6.022 × 10^23 molecules/mole
= 0.158 moles
Therefore, the number of moles are 0.158 moles
Answer:
Explanation:
Hello.
In this case, given the heat of fusion of THF to be 8.5 kJ/mol and freezing at -108.5 °C, for the required mass of 5.9 g, we can compute the entropy as:
Whereas n accounts for the moles which are computed below:
Thus, the entropy turns out:
Best regards.
Answer:
Yes. Weight is the product of mass times gravitational acceleration. So all you have to do is vary the gravitational field and you vary weight.
Explanation:
Answer:
2 M
Explanation:
mole weight of CaBr2 = 40 + 2 * 79.9 = 199.8 gm
20 gm is then 20/199.8 =.1 mole
.1 mole / .50 liter = 2 M
Answer:
The answer to your question is: ΔHrxm = -23.9 kJ
Explanation:
Data:
2Fe(s)+3/2O2(g)→Fe2O3(s), ΔH = -824.2 kJ (1)
CO(g)+1/2O2(g)→CO2(g) ΔH = -282.7 kJ (2)
Reaction:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
We invert (1) and change the sign of ΔH
Fe2O3(s) → 2Fe(s)+3/2O2(g) ΔH = 824.2 kJ
We multiply (2) by 3
3( CO(g)+1/2O2(g)→CO2(g) ΔH = -282.7 kJ) (2)
3CO(g)+3/2O2(g)→3CO2(g) ΔH = -848.1 kJ
We add (1) and (2)
Fe2O3(s) → 2Fe(s)+3/2O2(g) ΔH = 824.2 kJ
3CO(g)+3/2O2(g)→3CO2(g) ΔH = -848.1 kJ
Fe2O3(s) + 3CO(g)+3/2O2(g) → 2Fe(s)+3/2O2 + 3CO2(g)
Simplify
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) and ΔHrxm = -23.9 kJ
