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Eduardwww [97]
3 years ago
8

A gas that has a volume of 28 liters, a temperature of 45C, And an unknown pressure has its volume increased to 34 liters and it

's temperature decreased to 35C. If I measure the power after the change to be 2.0 ATM , what was the original pressure of the gas? Don't forget to use the right units in your answer.​
Chemistry
1 answer:
patriot [66]3 years ago
3 0

Answer:

P1 = 2.5ATM

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15K

P1 = ?

P2 = 2ATM

applying combined gas equation,

P1V1 / T1 = P2V2 / T2

P1*V1*T2 = P2*V2*T1

Solving for P1

P1 = P2*V2*T1 / V1*T2

P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

P1 = 21634.2 / 8628.2

P1 = 2.5ATM

The initial pressure was 2.5ATM

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a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding empirical formula is
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Answer: The empirical formula is CH_2.

Explanation:

Mass of C = 1.71 g

Mass of H = 0.287 g

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{1.71g}{12g/mole}=0.142moles

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

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For H =\frac{0.287}{0.142}=2

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4 0
3 years ago
For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided
SVETLANKA909090 [29]

Answer:

\frac{1}{24}

Explanation:

Given:

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To find: fraction of the whole athletic field reserved for each fifth class

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