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Soloha48 [4]
3 years ago
15

Chemistry questions below! (●'◡'●)

Chemistry
1 answer:
bearhunter [10]3 years ago
4 0

(1) Islands are always surrounded by bodies of water, in this case oceans. These islands tend to have more 'stable climates' and 'smaller temperature fluctuations' due to the specific heat of water, which is quite high, 4180 J/kg C. The higher the specific heat, the more the energy needed to raise the temperature of the substance. Land requires less energy to raise it's temperature than water due to it's lower specific heat. Thus, islands tend to have more stable temperatures than regions surrounded by land.

(2) Based on the conditions provided, the thermochemical equation for the reaction should be:  

<u>2NaHCO₃ (s) ---> Na₂CO₃ (s) + H₂O (g) + O₂ (g)</u>

This is a decomposition reaction, as it explains that the baking soda breaks down into the products. All we had left to do is balance the equation, adding 2 as the coefficient of the reactant.

(3) Let's rewrite the given equations:

First chemical reaction: H₂(g) + F₂(g)  ---> 2HF(g), ΔH₁ = –537 kJ.

Second chemical reaction: C(s) + 2F₂(g) ---> CF₄(g), ΔH₂ = –680 kJ

Third chemical reaction: 2C(s) + 2H₂(g) ---> C₂H₄(g), ΔH₃ = +52.3 kJ.

Fourth chemical reaction: C₂H₄(g) + 6F₂(g) ---> 2CF₄(g) + 4HF(g), ΔH₄ = ?

Now the 'fourth chemical reaction' is the one in which we need to determine the enthalpy:

ΔH₄ = 2(ΔH₁) + 2(ΔH₂) - ΔH₃,

ΔH₄ = 2(–537) + 2(–680) - 52.3 =<u> -2486.3 kJ</u>

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Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel
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The question is incomplete, here is the complete question:

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following. The mole fraction of gas B?

A mixture of gases contains 10.25 g of N₂, 2.05 g of H₂, and 7.63 g of NH₃.

<u>Answer:</u> The mole fraction of gas B (hydrogen gas) is 0.557

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For nitrogen gas:</u>

Given mass of nitrogen gas = 10.25 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen gas}=\frac{10.25g}{28g/mol}=0.366mol

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 2.05 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrogen gas}=\frac{2.05g}{2g/mol}=1.025mol

  • <u>For ammonia gas:</u>

Given mass of ammonia gas = 7.63 g

Molar mass of ammonia gas = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia gas}=\frac{7.63g}{17g/mol}=0.449mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B+n_C}

Moles of gas B (hydrogen gas) = 1.025 moles

Total moles = [0.366 + 1.025 + 0.449] = 1.84 moles

Putting values in above equation, we get:

\chi_{(H_2)}=\frac{1.025}{1.84}=0.557

Hence, the mole fraction of gas B (hydrogen gas) is 0.557

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