How many moles are there in 45.0 grams of sulfuric acid,<br> H2SO4?

1) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data

earnstyle [38]

Answer:

1) The absolute zero temperature is -272.74 °C

2) The absolute zero temperature is -269.91°C

Explanation:

1) The specific volume of air at the given temperatures are;

At -85°C, v = 1/(1.877) = 0.533 dm³/g, the pressure =

At 0°C, v = 1/1.294 ≈ 0.773 dm³/g

At 100°C, v = 1/0.946 ≈ 1.057 dm³/g

We therefore have;

v₁ = 0.533 T₁ = 188.15 K

v₂ = 0.773 T₂ = 273.15 K

v₃ = 1.057 T₃ = 373.15 K

The slope of the graph formed by the above data is therefore given as follows;

m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K

The equation is therefor;

v - 0.533 = 0.00284×(T - 188.15)

v = 0.00284×T - 0.534 + 0.533 = 0.00284×T - 0.001167

v = 0.00284×T - 0.001167

Therefore, when the temperature, at absolute 0, we have;

v = 0

Which gives;

0 = 0.00284×T - 0.001167

0.00284×T = 0.001167

T = 0.001167/0.00284 ≈ 0.411 K

Which is 0.411 -273.15 ≈ -272.74 °C

The absolute zero temperature is -272.74 °C

2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm

The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)

The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;

0°C = 0 + 273.15 K = 273.15 K

Therefore, the equation of the graph can be presented as follows;

v - 20.00 = 0.0741 × (T - 273.15)

Which gives;

v = 0.0741·T - 0.0741 ×(273.15) + 20

v = 0.0741·T - 20.2404125 + 20

v = 0.0741·T - 0.240415

Therefore at absolute 0, v = 0, we have;

0 = 0.0741·T - 0.240415

0.0741·T = 0.240415

T = 0.240415/0.0741 = 3.2445 K

The temperature in degrees is therefore;

3.2445 K - 273.15 ≈ -269.91°C

The absolute zero temperature is therefore, -269.91°C

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Option B: high pressure and low temperature

A gas is more soluble under high pressure and low temperature conditions.

On increasing temperature of a gas, its kinetic energy increases. The increase in kinetic energy increases the motion of particles of gas this causes most of the gaseous particles to escape from the gas phase. Thus, less particles are available to dissolve in liquid and solubility decreases.

The effect of pressure on solubility of gas can be explained with the help of Henry's law. According to the law, at constant temperature, solubility of gas and partial pressure of gas are related to each other as follows:

$p=k_{h}c$

Here, p is the partial pressure of the gas, $k_{h}$ is Henry's law constant, and

c is the concentrate of the gas.

According to above relation, concentration of gas decreases on decreasing partial pressure. Thus, on increasing pressure, concentration of gas increases this increases the solubility of gas in liquid.

Therefore, solubility of gas is greatest at high pressure and low temperature.

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How many moles are there in 45.0 grams of sulfuric acid, H2SO4?