From the stoichiometry of the balanced reaction equation, the correct statement are;
- For every 1 molecule of methane CH4 that reacts, 2 molecules of H2O are produced.
- For every 20 grams of methane (CH4) that reacts, 40 grams of H2O are produced.
- For every 200 moles of methane (CH4) that reacts, 400 moles of H2O are produced.
<h3>What is combustion?</h3>
The term combustion refers to the burning of fossil fuels for the purpose of energy production. The equation for reaction is CH4 + 2O2 ---> CO2 + 2H2O.
Using this equation as shown, the true statements are;
- For every 1 molecule of methane CH4 that reacts, 2 molecules of H2O are produced.
- For every 20 grams of methane (CH4) that reacts, 40 grams of H2O are produced.
- For every 200 moles of methane (CH4) that reacts, 400 moles of H2O are produced.
Learn more about combustion: brainly.com/question/15117038
Answer:
At 430.34 K the reaction will be at equilibrium, at T > 430.34 the
reaction will be spontaneous, and at T < 430.4K the reaction will not
occur spontaneously.
Explanation:
1) Variables:
G = Gibbs energy
H = enthalpy
S = entropy
2) Formula (definition)
G = H + TS
=> ΔG = ΔH - TΔS
3) conditions
ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction
4) Assuming the data given correspond to ΔH and ΔS
ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K
=> T = [ΔH - ΔG] / ΔS
ΔG = 0 => T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K
This is, at 430.34 K the reaction will be at equilibrium, at T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.
Answer:
Q = -33.6kcal .
Explanation:
Hello there!
In this case, according to the equation for the calculation of the total heat of reaction when a fixed mass of a fuel like ethane is burnt, we can write:
Whereas n stands for the moles and the other term for the enthalpy of combustion. Thus, for the required total heat of reaction, we first compute the moles of ethane in 3 g as shown below:
Next, we understand that -337.0kcal is the heat released by the combustion of 1 mole of ethane, therefore, to compute Q, we proceed as follows:
Best regards!
Answer:
Option A. 9.4 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 8 L
Initial temperature (T₁) = 293 K
Final temperature (T₂) = 343 K
Final volume (V₂) =?
V₁ / T₁ = V₂ / T₂
8 / 293 = V₂ / 343
Cross multiply
293 × V₂ = 8 × 343
293 × V₂ = 2744
Divide both side by 293
V₂ = 2744 / 293
V₂ = 9.4 L
Therefore, the final volume of the gas is 9.4 L
They depend on nitrogen-fixing bacteria, which convert atmospheric nitrogen into a usable form.
