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3 years ago

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how much heat is required to raise the temperature of a 5.45-g sample of iron (specific heat = 0.450 j/g°c) from 25.0°c to 79.8°

c?

1 answer:

Mamont248 [21]3 years ago

5 0

Answer:

134.397 Joules

Explanation:

Using the formula:

E = C × m × Δθ (where E is Energy, C is specific heat capacity and Δθ is change in temperature)

So E = 0.45×5.45×(79.8-25)

So E = 134.397 Joules

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A gas occupies 525 mL at a pressure of 45.0 kPa. What would the volume of the gas be at a pressure of 65.0 kPa

choli [55]

Answer:

The volume of the gas at a pressure of 65.0 kPa would be 363 mL

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Boyle's Law is a gas law that relates the pressure and volume of a certain amount of gas, without temperature variation, that is, at constant temperature.

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. In other words, the product P · V remains constant at the same temperature:

P*V=k

Being P1 and V1 the pressure and volume in state 1 and P2 and V2 the pressure and volume in state 2 are fulfilled:

P1*V1=P2*V2

In this case:

P1= 45 kPa= 45,000 Pa (being 1 kPa=1,000 Pa)
V1= 525 mL= 0.525 L (being 1 L=1,000 mL)
P2= 65 kPa= 65,000 Pa
V2= ?

Replacing:

45,000 Pa* 0.525 L= 65,000 Pa*V2

Solving:

$V2=\frac{45,000 Pa* 0.525 L}{65,000 Pa}$

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<u><em>The volume of the gas at a pressure of 65.0 kPa would be 363 mL</em></u>

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3 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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4 years ago