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ollegr [7]
2 years ago
15

2 moles of a gas have a pressure of 2 atm at a temperature of 300 K. What is the volume of the gas?

Chemistry
1 answer:
qwelly [4]2 years ago
4 0
  • No of moles=2mol=n
  • Pressure=2atm=P
  • Temperature=300K=T
  • Volume=V

\\ \tt\hookrightarrow PV=nRT

\\ \tt\hookrightarrow V=\dfrac{nRT}{P}

\\ \tt\hookrightarrow V=\dfrac{2(8.314)(300)}{2}

\\ \tt\hookrightarrow V=8.314(300)

\\ \tt\hookrightarrow V=2494.2mL

\\ \tt\hookrightarrow V=24.94dm^3

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Irina18 [472]

Answer:

X = 4

Explanation:

Start 2800

End 175

175 = 2800x(0.5)^X

175/2800 = 0.5^X

0.0625 = 0.5^X

log(0.5 x) = log(0.0625)

x · log(0.5) = log(0.0625)

-0.301x = -1.204

x = -1.204/-0.301

x = 4

8 0
3 years ago
Forming a hypothesis is accomplished through___ reasoning
mixer [17]
Scientific would be the word to fill in the blank
5 0
4 years ago
How tungsten conducts electricity
Elis [28]

Answer:

Yes

Explanation:

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8 0
3 years ago
How many moles of Co2 are produced when 0.2 moles of sodium carbonate reacts with excess HCl
Pie
To determine the number of moles of carbon dioxide that is produced, we need to know the reaction of the process. For the reaction of HCl and sodium carbonate, the balanced chemical equation would be expressed as:

2HCl + Na2CO3 = 2NaCl + H2O + CO2

From the initial amount given of sodium carbonate and the relation of the substances from the balanced reaction, we calculate the moles of carbon dioxide as follows:

0.2 moles Na2Co3 ( 1 mol CO2 / 1 mol Na2Co3 ) = 0.2 moles CO2

Therefore, the amount in moles of carbon dioxide that is produced from 0.2 moles sodium carbonate would be 0.2 moles as well.
5 0
3 years ago
What volume (in L) of oxygen will be required to produce 77.4 L of water vapor in the reaction below?
Inga [223]

Answer:

90.3 L

Explanation:

Given data:

Volume of water produced = 77.4 L

Volume of oxygen required = ?

Solution:

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2C₂H₆ + 7O₂  →  4CO₂ + 6H₂O

It is known that,

1 mole = 22.414 L

There are 7 moles of oxygen = 7×22.414 = 156.9 L

There are 6 moles of water = 6×22.414 = 134.5 L

Now we will compare:

                               H₂O           :              O₂    

                               134.5         :              156.9

                                 77.4         :             156.9/134.5×77.4 =90.3 L

So for the production of 77.4 L water 90.3 L oxygen is required.

8 0
3 years ago
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