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GenaCL600 [577]
3 years ago
14

Q.11. Calculate the mass of NaBr needed to prepare a 50.00-mL aqueous solution that would yield the same conductivity as the NaC

l (aq) solution. Show your calculation to your instructor before making the solution and measuring its conductivity.
Chemistry
1 answer:
klemol [59]3 years ago
6 0

Answer:

The mass of NaBr needed is 0.22969 g.

Explanation:

1 mole of NaBr contains 22.4 dm^3 of NaBr

Therefore, 0.05 dm^3 (50/1000 = 0.05 dm^3) of NaBr = 0.05/22.4 = 0.00223 mol of NaBr

MW of NaBr = 23 + 80 = 103 g/mol

Mass of NaBr = number of moles × MW = 0.00223 × 103 = 0.22969 g

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Answer:

The value of dissociation constant of the monoprotic acid is 1.099\times 10^{-3}.

Explanation:

The pH of the solution = 2.46

pH=-\log[H^+]

2.46=-\log[H^+]

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HA\rightleftharpoons H^++A^-

Initially

0.0144         0      0

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(0.0144-x)       x       x

The expression if an dissociation constant is given by :

K_a=\frac{[A^-][H^+]}{[HA]}

K_a=\frac{x\times x}{(0.0144-x)}

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K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}

K_a=1.099\times 10^{-3}

The value of dissociation constant of the monoprotic acid is 1.099\times 10^{-3}.

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Answer:

See explanation

Explanation:

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Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.

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Answer:

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