Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Answer:
82.0 mL
Explanation:
Step 1: Given data
- Concentration of concentrated acid (C₁): 12.2 M
- Volume of concentrated acid (V₁): ?
- Concentration of dilute acid (C₂): 1.00 M
- Volume of dilute acid (V₂): 1.00 L
Step 2: Calculate the required volume of the concentrated acid
We want to prepare a dilute solution from a concentrated one. We can calculate the volume of the concentrated acid using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 1.00 M × 1.00 L / 12.2 M = 0.0820 L = 82.0 mL
The correct answer to the problem is 0.193 which is three significant figures.
<h3>What are significant figures?</h3>
The term significant figures has to do with the figures that have a mathematical meaning. We know that the result has to correspond to the highest number of significant figures.
Hence, If we multiply 0.34 x 0.568 the result ought to be recorded as 0.193 which is three significant figures.
Learn more about significant figures:brainly.com/question/14804345
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Answer:
ΔG° = -5.4 kJ/mol
ΔG = 873.2 J/mol = 0.873 kJ /mol
Explanation:
Step 1: Data given
ΔG (NO2) = 51.84 kJ/mol
ΔG (N2O4) = 98.28 kJ/mol
Step 2:
ΔG = ΔG° + RT ln Q
⇒with Q = the reaction quatient
⇒with T = the temperature = 298 K
⇒with R = 8.314 J / mol*K
⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2
)
⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol
⇒ ΔG° = -5.4 kJ/mol
Part B
ΔG = ΔG° =RT ln Q
⇒with G° = -5.4 kj/mol = -5400 j/mol
⇒
with R = 8.314 J/K*mol
⇒with T = 298 K
⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577
ΔG = -5400 + 8.314 * 298 * ln(12.577)
ΔG = -5400 + 8.314 * 298 * 2.532
ΔG = 873.2 J/mol = 0.873 kJ/mol
