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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 reaction of aluminum with chlorine gas is shown 2Al + 3Cl2 -> 2AlCl3 based on this equation how many molecules of chlorin
Vanyuwa [196]

45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

The balanced equation for the reaction is given below:

2Al + 3Cl₂ —> 2AlCl₃

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

Therefore,

30 atoms of Al will require = \frac{30 * 3}{2}\\ = 45 molecules of Cl₂.

Thus, 45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

Learn more: brainly.com/question/24918379

3 0
2 years ago
One of the products of a fermentation reaction is
Alex73 [517]
There are two types of fermentation aerobic and anaerobic. Aerobic are those that need oxygen to ferment while anaerobic are those who do not need oxygen. Products of fermentation are usually water, ethanol, lactic acid and carbon dioxide. Other products are acetone and butyric acid.
5 0
3 years ago
In the experiment, 40 mL of 3 M sodium hydroxide is used to extract the benzoic acid. In order to recover the benzoic acid from
nikdorinn [45]

Answer:

20 mL OF 6 M HYDROCHLORIC ACID WILL BE NEEDED

Explanation:

M1 V1 = M2 V2

M1 = Molarity of sodium hydroxide = 3 M

V1 = volume of sodium hydroxide = 40 mL

M2 = Molarity of hydrochloric acid = 6 M

V2 = Volume of hydrochloric acid = unknown

Rearranging the equation, we have:

V2 = M1 V1 / M2

V2 = 3 * 40 mL / 6

V2 = 120 / 6

V2 = 20 mL

To precipitate the benzoic acid by 6 M of hydrochloric acid, 20 mL volume will be needed.

6 0
3 years ago
Calculate the maximum volume in ml of 0.15M HCl that each of the following antacid formulations would be expected to neutralize.
vlada-n [284]

a. 34 mL; b. 110 mL

a. A tablet containing 150 Mg(OH)₂


Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O


<em>Moles of Mg(OH)₂</em> = 150 mg Mg(OH)₂ × [1 mmol Mg(OH)₂/58.32 mg Mg(OH)₂

= 2.572 mmol Mg(OH)₂


<em>Moles of HCl</em> = 2.572 mmol Mg(OH)₂ × [2 mmol HCl/1 mmol Mg(OH)₂]

= 5.144 mmol HCl


Volume of HCl = 5.144 mmol HCl × (1 mmol HCl/0.15 mmol HCl) = 34 mL HCl


b. A tablet containing 850 mg CaCO₃


CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + H₂O


<em>Moles of CaCO₃</em> = 850 mg CaCO₃ × [1 mmol CaCO₃/100.09 mg CaCO₃

= 8.492 mmol CaCO₃


<em>Moles of HCl</em> = 8.492 mmol CaCO₃ × [2 mmol HCl/1 mmol CaCO₃]

= 16.98 mmol HCl


Volume of HCl = 16.98 mmol HCl × (1 mL HCl/0.15 mmol HCl) = 110 mL HCl


5 0
3 years ago
A substance with a high [h ] would likely have which additional characteristics?
Ray Of Light [21]

Answer:

Is [h] really [H]?  What characteristics are mentioned?

Explanation:

A high [H} means high acid concentration.  

6 0
2 years ago
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