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Usimov [2.4K]
3 years ago
14

Increasing the temperature of gas in a container that cannot expand is a way to _______. decrease the volume of the gas decrease

the volume of the gas increase the pressure of the gas increase the pressure of the gas decrease the pressure of the gas decrease the pressure of the gas increase the volume of the gas
Chemistry
1 answer:
alex41 [277]3 years ago
4 0

Answer:

Increase the pressure of the gas

Explanation:

According to the Pressure law, for a fixed mass of gas, at a constant volume (V), the pressure (P) is directly proportional to the absolute temperature (T).

From the kinetic molecular theory, gases are composed of particles which are in constant motion, colliding with themselves as well as with the walls of their container.

When the temperature of these gas molecules is increased, the molecules acquire more kinetic energy and the rate of collisions increases. Since the container cannot expand, the increase in pressure is due to the increase in collisions between the molecules of the gas as well as with the walls of their container.

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How many mL of a 0.375 M solution can be made from 35 g of calcium phosphate?
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Answer:

300 mL

Explanation:

the unit formula of calcium phosphate is Ca3(PO4)2

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n = m/M = 35 g/(310 g/mol)

c = n/V

V = n/c = [35 g/(310 g/mol)]/0.375 mol/L

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A balloon contains 2.0 L of air at 101.5 kPa . You squeeze the balloon to a volume of 0.25 L.
8_murik_8 [283]
<h3>Answer:</h3>

812 kPa

<h3>Explanation:</h3>
  • According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
  • Mathematically, P\alpha \frac{1}{V}

At varying pressure and volume;

P1V1=P2V2

In this case;

Initial volume, V1 = 2.0 L

Initial pressure, P1 = 101.5 kPa

Final volume, V1 = 0.25 L

We are required to determine the new pressure;

P2=\frac{P1V1}{V2}

Replacing the known variables with the values;

P2=\frac{(101.5)(2.0L)}{0.25L}

           = 812 kPa

Thus, the pressure of air inside the balloon after squeezing is 812 kPa

8 0
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