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lyudmila [28]
3 years ago
5

Propane (C3 H8 (g), Delta.Hf = –103.8 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Delta.Hf = –393.5 kJ/mol ) and

water (H2 O, Delta.Hf = –241.82 kJ/mol) according to the equation below. Upper C subscript 3 upper H subscript 8 (g) plus 5 upper O subscript 2 (g) right arrow 3 upper C upper O subscript 2 (g) plus 4 upper H subscript 2 upper O (g). What is the enthalpy of combustion (per mole) of C3 H8 (g)? Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants.
Chemistry
2 answers:
MrRissso [65]3 years ago
6 0

Answer:

a

Explanation:

becasue i said so. boom roasted.

xeze [42]3 years ago
4 0

Answer:

-2,044.0 kJ/mol

Explanation:

<em> I just did the test on edg</em>

<em />

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Suppose that you have 115 mL of a buffer that is 0.460 M in both benzoic acid ( C 6 H 5 COOH ) and its conjugate base ( C 6 H 5
77julia77 [94]

<u>Answer:</u> The maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

<u>Explanation:</u>

We are given:

Concentration of buffer = 0.460 M

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})        .....(1)

We are given:

pK_a = negative logarithm of acid dissociation constant of benzoic acid = 4.2

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[C_6H_5COO^-]=0.460M

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pH=4.2+\log(\frac{0.460}{0.460})\\\\pH=4.2

When the pH of the buffer changes by 1 unit, the buffering capacity is said to be lost.

pH change for loosing buffer capacity = [4.2 - 1] = 3.2

Calculating the ratio of conjugate base and its acid by using equation 1:

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3.2=4.2+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})\\\\\frac{[C_6H_5COO^-]}{[C_6H_5COOH]}=0.1

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}        ......(2)

For benzoic acid and its conjugate base:

Molarity of benzoic acid and its conjugate base = 0.460 M

Volume of solution = 115 mL

Putting values in equation 2, we get:

0.460=\frac{\text{Moles of benzoic acid and its conjugate base}\times 1000}{115mL}\\\\\text{Moles of benzoic acid and its conjugate base}=\frac{0.460\times 115}{1000}=0.053mol

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C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-

Let the moles of acid added to carry out the change is 'x' moles

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\frac{[C_6H_5COO^-]-x}{[C_6H_5COOH]+x}=0.1\\\\\frac{0.053-x}{0.053+x}=0.1\\\\x=0.0434

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