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Lina20 [59]
3 years ago
7

Calculate the number of moles of aluminum, sulfur, and oxygen atoms in 9.00 moles of aluminum sulfate, al2(so4)3. express the nu

mber of moles of al, s, and o atoms numerically, separated by commas.
Chemistry
1 answer:
klasskru [66]3 years ago
5 0
<span>Ans : The chemical formula for aluminum sulfate is Alâ‚‚(SOâ‚„)â‚ 6.00 mol Alâ‚‚(SOâ‚„)â‚ Al: 6.00 mol Alâ‚‚(SOâ‚„)â‚ x (2 mol Al / 1 mol Alâ‚‚(SOâ‚„)â‚ ) = 12.00 mol Al S: 6.00 mol Alâ‚‚(SOâ‚„)â‚ x (3 mol S / 1 mol Alâ‚‚(SOâ‚„)â‚ ) = 18.00 mol S O: 6.00 mol Alâ‚‚(SOâ‚„)â‚ x (12 mol O / 1 mol Alâ‚‚(SOâ‚„)â‚ ) = 72.00 mol O</span>
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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
Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

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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A bacteria culture begins with 15 bacteria which double in amount at the end of every hour. Solve for the number of bacteria tha
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The number of bacteria is given by:
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Where N(t) is the number after n hours have passed and N(o) is the original number which is 15.
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Answer:

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Explanation:

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