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Mashutka [201]
4 years ago
7

A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit

hin a cylinder. If the original pressure was 1.00 atm, the final pressure be _ atm. Answer in 3 sig figs.
Chemistry
1 answer:
Evgesh-ka [11]4 years ago
5 0

Answer:

The answer to your question is   P2 = 0.78 atm

Explanation:

Data

Temperature 1 = T1 = 263°K                 Temperature 2 = T2 = 298°K

Volume 1 = V1 = 24 L                             Volume 2 = V2 = 35 L

Pressure 1 = P1 = 1                                  Pressure 2 = P2 = ?

Process

1.- To solve this problem use the Combined gas law

                          P1V1/T1 = P2V2/T2

-Solve for P2

                           P2 = P1V1T2 / T1V2

-Substitution

                          P2 = (1)(24)(298) / (263)(35)

-Simplification

                          P2 = 7152 / 9205

-Result

                          P2 = 0.777

   or                    P2 = 0.78 atm

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24 grams of CH4 was added to the above reaction. Calculate the theoretical yield of CO2. A. 66 grams B. 33 grams c. 132 grams. D
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Answer:

Option A. 66 g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the mass of CH₄ that reacted and the mass of CO₂ produced from the balanced equation. This is illustrated below:

Molar mass of CH₄ = 12 + (4×1)

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Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of CO₂ = 12 + (16×2)

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Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Finally, we shall determine the theoretical yield of CO₂. this can be obtained as follow:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Therefore, 24 g of CH₄ will react to produce = (24 × 44) /16 = 66 g of CO₂.

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3 years ago
When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H
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Answer: The empirical formula is CH_2 and molecular formula is C_4H_8

Explanation:

We are given:

Mass of CO_2 = 18.95 g

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Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, =\frac{12}{44}\times 18.59=5.07g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, =\frac{2}{18}\times 7.759=0.862g of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.422}{0.422}=1

For H =\frac{0.862}{0.422}=2

The ratio of C : H = 1: 2

Hence the empirical formula is CH_2.

The empirical weight of CH_2 = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4

The molecular formula will be=4\times CH_2=C_4H_8

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