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statuscvo [17]
3 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]3 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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the land is heated through radiant energy,

Explanation:

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In an unconformity between two layers of rock, how is the lower layer usually described?
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A is the answer to the question
7 0
3 years ago
Read 2 more answers
0:55:00
Eva8 [605]

Answer:

About 366 joules

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PEg = mgh

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8 0
3 years ago
if you started with 2.34 grams of methane and 8.32 grams of oxeygyn and the combustion reaction went to completion how many gram
Romashka-Z-Leto [24]

Answer:

Mass of CO₂ produced  = 5.72 g

Explanation:

Given data:

Mass of methane = 2.34 g

Mass of oxygen = 8.32 g

Mass of CO₂ produced = ?

Solution:

Chemical equation:

CH₄ + 2O₂    →      CO₂ + 2H₂O

Number of moles of methane:

Number of moles = mass/molar mass

Number of moles =  2.34 g/ 16 g/mol

Number of moles = 0.146 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles =  8.32 g/ 32 g/mol

Number of moles = 0.26 mol

Now we will compare the moles of carbon dioxide with oxygen and methane.

                         CH₄             :              CO₂

                           1                 :              1

                        0.146            :           0.146

                         O₂                :               CO₂

                          2                 :                   1

                     0.26                :              1/2×0.26 = 0.13 mol

Less number of moles of CO₂ are produced by oxygen thus oxygen will react as limiting reactant.

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 0.13 mol ×  44 g/mol

Mass = 5.72 g

4 0
3 years ago
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