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balu736 [363]
3 years ago
13

Using the following equation for the combustion of octane, calculate the heat associated with the formation of 100.0 g of carbon

dioxide. The molar mass of octane is 114.33 g/mol. The molar mass of carbon dioxide is 44.0095 g/mol.
Chemistry
1 answer:
givi [52]3 years ago
7 0

Answer: Heat associated with the formation of 100.0 g of carbon dioxide is 1563.2 kJ.

Explanation:

Reaction equation will be as follows.

    2C_{8}H_{18} + 25O_{2} \rightarrow 16CO_{2} + 18H_{2}O;  \Delta H^{o}_{rxn} = -11018 kJ

Mass of CO_{2} = 100 g

Hence, moles of CO_{2} present will be calculated as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{100 g}{44.0095 g/mol}

                             = 2.27 mol

Therefore, heat produced by 2.27 mol for the given reaction will be calculated as follows.

         2.27 mol \times \frac{11018 kJ}{16 mol CO_{2}}

             = 1563.2 kJ

Thus, we can conclude that heat associated with the formation of 100.0 g of carbon dioxide is 1563.2 kJ.

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Answer:

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Explanation:

One way to calculate the lattice energy is to use Hess's Law.

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We must generate this reaction rom the equations given.

(1)  Li(s) + ½Cl₂ (g) ⟶ LiCl(s);      ΔHf°     = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g);                          ΔHsub =    161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g)                     BE        =   243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻                   IE₁         =   520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g)                EA₁       =  -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

                                                <u>E/kJ </u> 

(5) Li⁺(g) +e⁻ ⟶ Li(g)                520

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The lattice energy of LiCl is \boxed{\textbf{-862 kJ/mol}}.

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A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati
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Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.  

\text{calculating the initial HI}= \frac{mol}{V}

                                       =\frac{9.3 \times 10 ^ -3}{2}

                                      =0.00465 \ Mol

\text{Similarly}\ \  I_2 \ \  \text{follows} \ \  H_2 = 0 }

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

HI  = 0.00465 - 2x\\\\ I_{2}  \ eq = H_2 \ eq = 0 + x \\\\

It is defined that:

I_2 = 6.29 \times 10^{-4}  \ M \\\\x = I_2 \\\\

HI \ eq= 0.00465 - 2x \\

          =0.00465 -2 \times 6.29 \times 10^{-4} \\\\ =  0.00465 -\frac{25.16 }{10^4}  \\\\   = 0.003392\  M

Now, we calculate the position:  

For the reaction H 2(g) + I 2(g)\rightleftharpoons  2HI(g), you can calculate the value of Kc at 1000 K.  

data expression for Kc

2HI \rightleftharpoons  H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}

         = \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386

calculating the reverse reaction

H_2(g) + I_2(g)\rightleftharpoons  2HI(g)

Kc = \frac{1}{Kc} \\\\

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