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3 years ago

Calculate the decrease in temperature when 6.00 L at 25.0 °C is compressed to 2.50 L.

Chemistry

1 answer:

N76 [4]3 years ago

4 0

Answer:

-149.0K

Explanation:

Charles's law states that the volume of a gas is directly proportional to the absolute temperature under constant pressure.

The equation is:

V1/T1 = V2/T2

Where V1 is initial volume = 6.00L

T1 is absolute initial temperature = 25.0°C + 273 = 298K

V2 is final volume = 2.50L

T2 is our incognite

Replacing:

6.00L/298K = 2.50L/T2

T2 = 2.50L * 298K / 6.00L

T2 = 124.17K

The temperature is 124.17K - 273.15K =

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kiran lit a candle. she placed a 100cm glass jar over the candle. the candle flame went out after 2 seconds. why did the flame g

Radda [10]

Answer:

The candle took up all the oxygen under the glass

Explanation:

Carbon dioxide molecules are heavier than air. Because of this, they push the oxygen and other molecules in the air out of the way as they sink down over the flame and candle. When oxygen is pushed away from the wick, it can't react with the wax anymore.

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4 years ago

WHAT MASS OF 1,1 DICHLOROEHTANE MUST BE MIXED WITH 100G OF 1,1 DICHLOROTETRAFLUOROEHTANE TO GIVE A SOLUTION WITH VAPOR PRESSURE

vaieri [72.5K]

This is an incomplete question.

The complete question is:

1,1-dichlorotetrafluoroethane, CF3CCL2F, has a vapor pressure of 228 torr. What mass of 1,1-dichloroethane must be mixed with 100.0 g of 1,1-dichlorotetrafluoroethane to give a solution with vapor pressure 157 torr at 25 degrees celsius?

Answer: 46.9 g of 1,1 dichloroethane must be fixed with 100 g of 1,1 dichlorotetrafluoroethane to give a solution with vapor pressure of 157 torr at $25^0C$

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$ = relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

$x_2$ = mole fraction of solute

= $\frac{\text {moles of solute}}{\text {total moles}}$

Given : x g of solute is present in 100 g of solvent

moles of solute (1,1 DICHLOROEHTANE) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{xg}{98.96g/mol}moles$

moles of solvent (1,1 DICHLOROTETRAFLUOROEHTANE ) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{100g}{170.92g/mol}=0.58moles$

Total moles = moles of solute + moles of solvent = $\frac{xg}{98.96g/mol}+0.58$

$x_2$ = mole fraction of solute = $\frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$\frac{228-157}{157}=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$0.45=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$x=46.9g$

Thus 46.9 g of 1,1 dichloroethane must be fixed with 100 g of 1,1 dichlorotetrafluoroethane to give a solution with vapor pressure of 157 torr at $25^0C$

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3 years ago

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Explanation:

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3 years ago