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Eva8 [605]
3 years ago
7

This is the process that cells reproduce and replace old or damaged cells.

Chemistry
2 answers:
r-ruslan [8.4K]3 years ago
6 0
A. Mitosis

Didnndxbdbdn
Temka [501]3 years ago
3 0
The answer is A. mitosis
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A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure
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Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

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The workdone = 130.1 J

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The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

∴ the final volume of the gas is V_2 = 1311.5 mL

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