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statuscvo [17]
2 years ago
9

In an isothermal gas chromatography experiment using an ECD detector, 1.69 nmols of nchlorohexane, C6H13Cl, was added as an inte

rnal standard to an unknown amount of nchlorodecane, C10H21Cl. The area of the first peak to elute was 32434 units and the area of the second peak to elute was 2022 units. Calculate the amount of chlorodecane in the unknown.
Chemistry
1 answer:
prohojiy [21]2 years ago
4 0

Answer:

The amount of Chlorodecane in the unknown is 0.105nmols

Explanation:

a) Since the GC is in an isothermal state, Chlorohexane C6H13Cl (1.69 nmols) because of its lower boiling point will elute first and Chlorodecane C12H21Cl will elute second.

The area of the first peak corresponding to Chlorohexane is 32434 units.

The area of the second peak corresponding to chlorodecane is 2022 units.

Since the response factor of the compound is not given in question and considering the response factor is same for both the compounds, the answer will be as follow:

1.69 nmols of Chlorohexane gives 32434 units

How much of chlorodecane gives 2022 units

By cross multiplication;

Moles of Chlorodecane = 2022*1.69/32434

                                         =0.105nmols

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<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.

<u>Explanation:</u>

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}

where,

k = rate constant = ?

t = time taken = 1.52 hrs

[A_o] = Initial concentration of reactant = 100 g

[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}

To calculate the half life period of first order reaction, we use the equation:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life period of first order reaction = ?

k = rate constant = 0.2098hr^{-1}

Putting values in above equation, we get:

t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs

Hence, the half life of the sample of silver-112 is 3.303 hours.

6 0
3 years ago
A reaction was suppose to produce 20 L of gas. Only 12 litters of gas were produced in the lab. What is the percent yield for th
pogonyaev

Answer:

\boxed {\boxed {\sf 60 \%}}

Explanation:

Percent yield is a ratio of the actual yield to the theoretical yield. It is found using this formula:

\% \ yield= \frac {actual \ yield}{theoretical \  yield} *100

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The theoretical yield is 20 liters, because that was the expected yield.

\% \ yield= \frac {12 \ L}{20\ L} *100

\% \ yield= 0.6*100

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For this reaction, the percent yield is 60%.

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That law is known as Boyle's Law, "The volume of a given mass of a gas is inversely related to pressure when the temperature is constant"
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Answer:

TRUE.

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See the image below-⬇⬇-

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