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DochEvi [55]
3 years ago
9

A sample of chlorine gas starting at 686 mm Hg is placed under a pressure of 991 mm Hg and reduced to a volume of 507.6 mL. What

was the initial volume of the chlorine gas container if the process was performed at constant temperature
Chemistry
1 answer:
Leviafan [203]3 years ago
3 0

Answer:

The initial volume of the chlorine gas V1=733.28mL

Explanation:

Given:

P1= 686mmHg

P2= 991mmHg

V2= 5076mL

V1=?

According to Boyle's law which states that at a constant temperature, the pressure on a gas increases as it's volume decreases.

It can be expressed as : P1V1 = P2V2

Where P1 is the initial pressure

P2= final pressure

V1= initial volume

V2 = final volume

V1= (P2V2)/P1

V1= (991mmHg*507.6mL)/686mmHg

V1=503031.6/686

V1=733.28mL

Therefore, The initial volume of the chlorine gas V1=733.28mL

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Decane (C10H22) is used in diesel. The combustion for decane follows the equation: 2 C10H22 + 31 O2 à 20 CO2 + 22 H2O. Calculate
creativ13 [48]

The mass of water produced is 792 grams by the combustion of 568 grams of decane.

Given:

Combustion of 568 grams of decane with 2979 grams of oxygen.

2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O

To find:

The mass of water produced by combustion of 568 grams of decane.

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Mass of decane = 568 g

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= \frac{568 g}{142 g/mol}=4 mol

Mass of oxygen gas = 2976 g

Moles of oxygen gas:

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2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O

According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:

=\frac{31}{2}\times 4mol=62\text{ mol of}O_2

But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.

So, this means that oxygen gas is present in an excessive amount. Which simply means:

  • Oxygen gas is an excessive reagent.
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According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:

=\frac{22}{2}\times 4mol=44\text{mol of }H_2O

Mass of 44 moles of water ;

=44mol\times 18g/mol=792g

792 grams of water is produced by the combustion of 568 grams of decane.

Learn more about limiting reagent and excessive reagent here:

brainly.com/question/14225536?referrer=searchResults

brainly.com/question/7144022?referrer=searchResults

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Identify the precipitate(s) formed when solutions of na2so4(aq), ba(no3)2(aq), and nh4clo4(aq) are mixed.
Lostsunrise [7]

Answer : BaSO_{4} will be the precipitate which will be formed.


Explanation : When all the three solutions namely; NaSO_{4}  + Ba(NO_{3})_{2}  + NH_{4} ClO_{4} are mixed together a white precipitate of BaSO_{4} is formed as a product in the solution along with the soluble by product of Ammonium nitrate which is NH_{4} NO_{3} 

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