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AURORKA [14]
2 years ago
9

How many bromine atoms are present in 37.9 g of CH2Br2?

Chemistry
1 answer:
VARVARA [1.3K]2 years ago
4 0

Answer:

The answer to your question is: 6.55 x 10 ²³ atoms of Br

Explanation:

CH2Br2 = 37.9 g

MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g

                   174 g of CH2Br2 ------------------  160 g of Br2

                   37.9 g of CH2Br2   ---------------     x

                x = 37.9 x 160/174 = 34.85 g of Br

                      1 mol of Br -----------------   160 g Br2

                         x              ----------------    174 g Be2

               x = 174 x 1 /160 = 1.088 mol of Br2

                1 mol of Br -----------------  6.023 x 10 ²³ atoms

            1.088 mol of Br -------------    x

                    x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms

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2 years ago
BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w
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Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>

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K_{a}. K_{b} = K_{w}    

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<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>

Reaction involved: BH⁺  +  H₂O  ⇌  B  +  H₃O+

Initial:                     0.1 M                    x         x            

Change:                   -x                      +x       +x

Equilibrium:           0.1 - x                    x         x

<u>The acid dissociation constant: </u>K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}

\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}

\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}

As, x

\Rightarrow 0.1 - x = 0.1

\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }

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Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

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Answer:

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