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Genrish500 [490]
3 years ago
10

1.0 moles of an ideal gas are held in a container under 2.0 atm of pressure and 310 K. What must be the volume of this container

?
Chemistry
2 answers:
attashe74 [19]3 years ago
8 0

Answer: 12.7 L

Explanation:

P=2.0atm

V= ???

n=1.0 mol

R= 0.0821

T= 310K

PV=nRT--> (2)V=(1)(0.0821)(310)

                     V=12.7 L

horrorfan [7]3 years ago
3 0

Answer:

12.7 L (3 s.f.)

Explanation:

Since it is an ideal gas, we can apply the ideal gas law:

\boxed{pV=nRT}

☆ where p= pressure

V= volume

n= number of moles

R= ideal gas constant

T= temperature

Substitute all the given information into the formula:

2V= 1(0.08206)(310)

2V= 25.4386

V= 25.4386 ÷2

V= 12.7 L (3 s.f.)

Further explanation:

The ideal gas constant, R, has different values depending on what units are being used. Two examples are listed below:

• \: 0.08206 \: L \: atm ^{ - 1}  \: mol \: K

• \: 8.314 \: J \:  {mol}^{ - 1} \: {K}^{ - 1}

In the above question, we use the value 0.08206 because the pressure was given in atm and the temperature was given in Kelvin. Thus, the unit used for volume is L.

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Answer:

The answer to your question is: 6.8 g of water

Explanation:

Data

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1.4 moles of Ca(OH)2

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MW                   2(36.5)               74                       36 g               111 g

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                            x = (1.4 x 74) / 1  = 103.6 g

Grams of water

                        73 g of HCl ------------------   36g of H2O

                        94.9 g        -------------------    x

                     x = (94.9 x 36) / 73 = 46.8 g of water

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