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Sever21 [200]
3 years ago
8

A 0.100 M solution of bromoacetic acid (BrCH2COOH) is 13.2% ionized. Calculate [H+]? Calculate BrCH2COO^ - ? Calculate BrCH2COOH

. (looking for Molarity)
Chemistry
1 answer:
Natali [406]3 years ago
8 0

<u>Answer:</u> The concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M

<u>Explanation:</u>

We are given:

Molarity of bromoacetic acid = 0.100 M

Percent of ionization = 13.2 %

The chemical equation for the ionization of bromoacetic acid follows:

BrCH_2COOH\rightarrow BrCH_2COO^-+H^+

1 mole of bromoacetic acid produces 1 mole of bromoacetate ion and 1 mole of hydrogen ion

Molarity of hydrogen ion = 13.2 % of 0.100 = \frac{13.2}{100}\times 0.100=0.0132M

Molarity of bromoacetate ion = molarity of hydrogen ion = 0.0132 M

Molarity of bromoacetic acid = Molarity of solution - Molarity of ionized substance

Molarity of bromoacetic acid = 0.100 - 0.0132 = 0.0868 M

Hence, the concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M

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An alkene with the molecular formula C8H16 undergoes ozonolysis to yield a mixture of (CH3)2C=O and (CH3)3CCHO. The alkene is:
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2,4,4-trimethyl-2-pentene yields mixture of (CH_{3})_{2}C=O and (CH_{3})_{3}CHO

Explanation:

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8 0
3 years ago
Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
Katen [24]

<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

K_c=\frac{[HI]^2}{[H_2][I_2]}

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0
3 years ago
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