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goldfiish [28.3K]
3 years ago
11

Using the data below, calculate the enthalpy for the combustion of C to CO

Chemistry
1 answer:
Leona [35]3 years ago
6 0

Answer:

ΔH3 = -110.5 kJ.

Explanation:

Hello!

In this case, by using the Hess Law, we can manipulate the given equation to obtain the combustion of C to CO as shown below:

C(s) + 1/2O2(g) --> CO(g)

Thus, by letting the first reaction to be unchanged:

C(s) + O2(g)--> CO2 (g) ; ΔH1 = -393.5 kJ

And the second one inverted:

CO2(g) --> CO(g) + 1/2O2(g) ; ΔH2= 283.0kJ

If we add them, we obtain:

C(s) + O2(g) + CO2(g) --> CO(g) + CO2 (g) + 1/2O2(g)

Whereas CO2 can be cancelled out and O2 subtracted:

C(s) + 1/2O2(g)  --> CO(g)

Therefore, the required enthalpy of reaction is:

ΔH3 = -393.5 kJ + 283.0kJ

ΔH3 = -110.5 kJ

Best regards!

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3.75 litres is the volume of the balloon indoors at a temperature of 25°C.

Explanation:

Data given:

initial temperature of the gas in balloon  = -35°C or 238.15 K

initial volume = 3 litres

final temperature = 25 °C  or 298.15 K

final volume =?

pressure remains constant

From the data given when pressure is constant Charles' law is applied.

\frac{V1}{T1} = \frac{V2}{T2}

Rearranging the equation to know the final volume of the gas in balloon

V2 = \frac{V1T2}{T1}

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V2 = 3.75 Litres

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Answer:

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Explanation:

The equation bellow can be explained by the dissociation of each specie:

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Answer:

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Nataly_w [17]

Answer:

The answer to your question is a) N₂     b) 3.04 g of NH₃

Explanation:

Data

mass of H₂ = 2.5 g

mass of N₂ = 2.5 g

molar mass H₂ = 2.02 g

molar mass of N₂ = 28.02 g

molar mass of NH₃ = 17.04 g

Balanced chemical reaction

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A)

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The limiting reactant is N₂ (nitrogen) because the experimental proportion was higher than the theoretical proportion.

B)

             28.02 g of N₂ -------------------- (2 x 17.04) g of NH₃

               2.5 g of N₂    --------------------   x

                            x = (2.5 x 2 x 17.04) / 28.02

                            x = 85.2 / 28.02

                           x = 3.04 g of NH₃

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