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Taya2010 [7]
3 years ago
8

In the laboratory, a quantity of I2 was reacted with excess H2 to give 1.26 moles of HI. It is also known that the percent yield

of this reaction was 84.0%. How many moles of I2 reacted
Chemistry
1 answer:
Anon25 [30]3 years ago
5 0

Answer:

1.008moles of iodine

Explanation:

Hello,

This question requires us to calculate the theoretical yield of I₂ or number of moles that reacted.

Percent yield = (actual yield / estimated yield) × 100

Actual yield = 1.2moles

Estimated yield = ?

Percentage yield = 84%

84 / 100 = 1.2 / x

Cross multiply and solve for x

100x = 84 × 1.2

100x = 100.8

x = 100.8/100

x = 1.008moles

1.008 moles of I₂ reacted in excess of H₂ to give 1.2 moles of HI

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3 years ago
The plunger on a bicycle pump with a 400 mL volume cylinder is
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Answer:

The answer to your question is V2 = 66.7 ml

Explanation:

Data

Volume 1 = V1 = 400 ml

Pressure 1 = P1 = 1 atm

Volume 2 = V2 = ?

Pressure 2 = P2 = 6 atm

Process

1.- To solve this problem use Boyle's law

                     P1V1 = P2V2

-solve for V2

                     V2 = P1V1 / P2

-Substitution

                      V2 = (1)(400) / 6

-Simplification

                      V2 = 400 / 6

-Result

                      V2 = 66.7 ml

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3 years ago
What is the total pressure of air in lungs of an individual with oxygen at 100. mmHg, nitrogen at 573 mmHg, carbon dioxide at 0.
ryzh [129]

Answer: The total pressure of air in lungs of an individual is 760.28 mm Hg

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

p_{total}=p_A+p_B+p_C...

Given : p_{total} =total pressure of gases = ?

p_{O_2} = partial pressure of oxygen = 100 mm Hg

p_{N_2} = partial pressure of nitrogen = 573 mm Hg

p_{CO_2} = partial pressure of Carbon dioxide = 0.053 atm = 40.28 mm Hg(1 atm = 760 mmHg)

p_{H_2O} = partial pressure of water vapor = 47 torr = 47 mm Hg  (1torr=1 mm Hg)

putting in the values we get:

p_{total}=(100+573+40.28+47)mmHg

p_{total}=760.28mmHg

Thus the total pressure of air in lungs of an individual is 760.28 mm Hg

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