Answer:
The answer to your question is: 0.25 l
Explanation:
Data
P1 = 1 atm
V1 = 0.5 l
P2 =2 atm
V2 = ?
T = constant
Formula
V1P1 = V2P2
Clear V2 from the formula
V2 = V1P1/P2
Substitution
V2 = (0.5)(1)/2 substitution
= 0.25 l result
Answer:
Hello
The answer is 18,7(b)
Explanation:we don't know the value of o² so at first we should put (x)instead of gr of o²and then write 1molO²/32grO²×2mol SO²/3mol O²×64gr SO²/1molSO²=25 gr SO².and then just find the value of (x).
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
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2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
Answer:
485.76 g of CO₂ can be made by this combustion
Explanation:
Combustion reaction:
2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)
If we only have the amount of butane, we assume the oxygen is the excess reagent.
Ratio is 2:8. Let's make a rule of three:
2 moles of butane can produce 8 moles of dioxide
Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂
We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g
