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brilliants [131]
3 years ago
13

A weather balloon is released in Houston when the pressure is 758.0 mmHg and 32.7°C. After the balloon reaches an altitude of 60

00 m it measures a volume of 425 L with a pressure of 364.2 mmHg and a temperature is 3.10°C. What was the initial volume of the balloon? Show all your work and round your answer to the hundredths place.
A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon dioxide. The total pressure of the mixture is 100.0 kPa. If the partial pressure for helium is 83.1 kPa and 0.1000 kPa for carbon dioxide, what is the partial pressure of oxygen? Show all your work and round your answer to the tenths place.
Chemistry
1 answer:
erica [24]3 years ago
4 0

Answer:

See explanation

Explanation:

From the information provided;

P1 = initial pressure = 758.0 mmHg

T1 = Initial temperature = 32.7°C + 273 = 305.7 K

V1 = Initial volume?

P2 = Final pressure = 364.2 mmHg

V2 = Final volume =  425 L

T2= Final Temperature = 3.10°C + 273 = 276.1 K

P1V1/T1 = P2V2/T2

P1V1T2 =P2V2T1

V1 = P2V2T1/P1T2

V1 = 364.2 * 425 * 305.7/758.0 * 276.1

V1 = 226. 18 L

B) Total pressure = 100.0 kPa

Pressure of Helium = 83.1 kPa

Pressure of carbon dioxide = 0.1000 kPa

Pressure of  oxygen ?

From Dalton's law of partial pressures;

PT = P1 + P2 +P3 +.........

Hence;

Poxygen = 100.0 kPa - (83.1 kPa + 0.1000 kPa)

Poxygen = 16.8 kPa

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The answer to your question is below

Explanation:

                                         Trial 1                   Trial 2

mass of Mg                      0.255 g                0.353 g

mass of MgO                   0.418 g                 0.576 g

Chemical reaction

                                2Mg(s)    +  O₂(g)     ⇒   2MgO(s)

Question 1.

Atomic mass of Mg = 24.31 x 2 = 48.62 g

Molecular mass of MgO = 2(24.31 + 16) = 80.62 g

Trial 1

                          48.62 g of Mg ----------------- 80.62 g of MgO

                          0.255 g            ----------------   x

                          x = (0.255 x 80.62)/48.62

                          x = 0.422 g of MgO

Trial 2                48.62 g of Mg ----------------- 80.62 g of MgO

                          0.353 g            ----------------   x

                          x = (0.353 x 80.62)/48.62

                          x = 0.585 g of MgO

Question 2

Trial 1          

Percent yield = 0.418/0.422 x 100 = 99%

Trial 2

Percent yield = 0.576/0.585 x 100 = 98.5%

Question 3

Average = (99 + 98.5)/2

              = 98.75%

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