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3 years ago

6

In the free radical chlorination of methylcyclopentane, what is the total number of distinct monochlorinated compounds, includin

g stereoisomers, that would be formed?
A) 4
B) 6
C) 8
D) 10

1 answer:

Morgarella [4.7K]3 years ago

8 0

Answer:

B) 6

Explanation:

Methylcyclopentane consist of 3 types of hydrogen which are 1° (primary), 2° (secondary) and 3° (tertiary) hydrogen atoms. In the chlorination of methylcyclopentane, there are four monochlorinated products possible and two stereoisomers.

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When comparing the two elements K and Ge , the more metallic element is__________ based on periodic trends alone.

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3 years ago

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

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$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

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pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

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$pOH=-\log[OH^-]$

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$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

<u>Initial:</u> 0.1

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

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The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

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$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

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$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

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3 years ago

Suppose you used 0.5 M NaOH to titrate your vinegar sample instead of 0.1 M. What effect does the concentration of base added ha

laiz [17]

Answer: the reliability will be worse

Explanation:

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4 0

3 years ago