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melomori [17]
3 years ago
8

NEED HELP ASAP!!!

Chemistry
1 answer:
Alexxx [7]3 years ago
6 0

At room temperature oxygen, O_2 is present in the gas phase.

A gas is defined as a substance which is characterized by the rapidly movement of molecules which are widely separated. The shape and volume of gas is not definite that is they take on the shape of the container in which it is placed. The gas molecules have minimum inter-nuclear attractions and thus have high kinetic energy.

Thus, the sentences that describes oxygen at room temperature are:

  • It has no definite volume and takes the shape of its container.
  • Its particles move fast enough to overcome the attraction between them.
  • It has more energy than it would be at a cooler temperature as the kinetic energy is directly proportional to the temperature that is kinetic energy increases on increasing the temperature.

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Answer:

The answer is c

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The average lung capacity of a human is 6.0L.
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Answer:

(a) 0.25 mol

(b) 0.11 mol

(c) 8.77 mol

Explanation:

(a)

We use the equation given by ideal gas which follows:

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where,

P = pressure of the gas = 1.00 atm

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Putting values in above equation, we get:

1.00 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1.00\times 6.0}{0.0821\times 298}=0.25mol

(b)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 0.296 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 200 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

0.296 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 200K\\\\n=\frac{0.296\times 6.0}{0.0821\times 200}=0.11mol

(c)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 30 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 250 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

30 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 250K\\\\n=\frac{30\times 6.0}{0.0821\times 250}=8.77mol

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Doing some algebra and converting temperature to Kevin by adding 273, you should obtain the same result.

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