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2 years ago

Question 3

Chemistry

1 answer:

harina [27]2 years ago

4 0

Answer: 0.118M

Explanation:

The formula for molarity is: $M=\frac{mol solute}{L solution}$

First, we need to find the number of moles of NaCl.

$2.70gNaCl*\frac{1molNaCl}{58.44gNaCl} =0.05molNaCl$

Next, we must convert millimeters to liters. We can do that by dividing the number of mL by 1000.

$\frac{425mL}{1000} =0.425L$

Now we have our needed data! All we need to do now is plug in our data to the molarity formula.

$M=\frac{0.05mol}{0.425L} =0.118M$

I hope this helps! Pls mark brainliest!! :)

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pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:

pH = pKb + log([HB⁺]/[B])

pKb = -log(1.8×10⁻⁶) = 5.7

Chemical reaction for C₆H₅NH₂ is:

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Change: -x x x

Equilibrium: 0.25-x x x

Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]

Kb = x² / 0.25 - x

x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:

1.8×10⁻⁶ = x² / 0.25

x² = (1.8×10⁻⁶)(0.25)

x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]

On putting all these values on the above equation of pH, we get

pH = 5.7 + log(0.67×10⁻³/0.25)

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To know more about Henderson Hasselbalch equation, visit the below link:
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The percent yield of chloro-ethane in the reaction is 82.98%.

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As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

$\frac{1}{1}\times 9.1549 mol=9.1549 mol$ of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

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The percent yield of chloro-ethane in the reaction is 82.98%.

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