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3 years ago

Hils

Chemistry

1 answer:

shtirl [24]3 years ago

5 0

<h2>Answer:</h2>

d is your answer hope this helps

Explanation:

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Given the values of ΔGfo given below in kJ/mol, calculate the value of ΔGo in kJ for the combustion of 1 mole of butane to form

xz_007 [3.2K]

Answer:

Δ $_rG=-2753kJ$

Explanation:

Hello,

Butane's combustion is:

$C_4H_{10}+\frac{13}{2} O_2-->4CO_2+5H_2O$

Now the change in the Gibbs free energy for this reaction is computed via (don't forget that oxygen's Gibbs free energy of formation is 0 since it is an element):

Δ $_rG=$ 4Δ $_fG_{CO_2}+$ 5Δ $_fG_{H_2O}$ -Δ $_fG_{C_4H_{10}}$

Δ $_rG=4(-395kJ/mol)+5(-238kJ/mol)-(-17kJ/mol)$

Δ $_rG=-2753kJ/mol$

Now, since the calculation is done for 1 mol of butane, the change in the Gibbs free energy for this fuel is:

Δ $_rG=-2753kJ$

Best regards.

4 0

3 years ago

How long does it take for 87.5 % of each isotope(Krypton-73, Krypton-74, Krypton-76, and Krypton-81) to decay?

fomenos

1st half-life: 100%/2 = 50%
2nd half-life: 50%/2 = 25% + 50% (from first half-life)
3rd half-life: 25%/2 = 12.5% + 75% (from first and second half-lives

I looked it up...

3 0

3 years ago

Which is the noble gas notation for chlorine?

Nikitich [7]

The answer is [Ne] 3s^2 3p^5 because chlorine is the fifth element in the 3rd row of elements in in p orbital

3 0

4 years ago

Read 2 more answers

What is the molarity of a 10 L solution containing 5.0 miles if solute

Bingel [31]

Hey there !

Solution :

Moles of solution = 5.0

Volume of solution = 10 L

Therefore :

M = n / V

M = 5.0 / 10

= 0.5 M

3 0

3 years ago

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 97. g of octane is mi

zhannawk [14.2K]

Answer: 61 grams

Explanation:

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

$\text{Number of moles of octane}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{97g}{114g/mol}=0.85moles$

$\text{Number of moles of oxygen}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{150g}{32g/mol}=4.69moles$

The chemical equation for the combustion of octane in oxygen follows the equation:

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

By stoichiometry of the reaction;

25 moles of oxygen react with 2 moles of octane

4.69 moles of oxygen react with= $\frac{2}{25}\times 4.69=0.37$ moles of octane

Thus, oxygen is the limiting reagent as it limits the formation of product and octane is the excess reagent.

25 moles of oxygen produce 18 moles of water

4.69 moles of oxygen produce= $\frac{18}{25}\times 4.69=3.38$ moles of water.

Mass of water produced= $moles\times {\text{Molar mass}}=3.38\times 18g/mol=61g$

The maximum mass of water that could be produced by the chemical reaction is 61 grams.

6 0

3 years ago