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2 years ago

6

Calculate the mass of water vapour present in a room of volume 400 m3 that contains air at 27 °C on a day when the relative humi

dity is 60 per cent. Hint: Relative humidity is the prevailing partial pressure of water vapour expressed as a percentage of the vapour pressure of water vapour at the same temperature (in this case, 35.6 mbar).

1 answer:

Alex73 [517]2 years ago

6 0

Answer:

Mass of water = 6251. 706g or 6.25Kg

Explanation:

Relative humidity = (actual vapor pressure/saturation vapor pressure) * 100%

Actual vapor pressure, Pw = relative humidity * saturation vapor pressure

Pw = 60% * (35.6 *0.001)atm = 0.0216atm

Note: 1mbar = 0.001atm

Using the ideal gas equation: PV=nRT; where P = Pw= 0.02136atm, V= (400

* 1000)dm^3, R= 0.082 atmdm^3/kmol, T= (27+273)K, n = number of moles

Note: 1m^3 = 1000dm^3,R is the molar gas constant.

Making n subject of the formula, n = PV/RT

n= (0.02136 * 400000)/(0.082 * 300) = 347.317 moles

Mass (g) = number of moles (n) * molar mass

molar mass of water=18g

Mass of water = 347.317 * 18 = 6251. 706g or 6.25Kg

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2CH4(g)⟶C2H4(g)+2H2(g)

Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The balanced reaction of $CH_4$ will be,

$2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)$ $\Delta H^o=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890.3kJ$

(2) $C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)$ $\Delta H_2=-136.3kJ$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=-571.6kJ$

(4) $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)$ $\Delta H_4=-3120.8kJ$

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

(1) $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)$ $\Delta H_1=2\times (-890.3kJ)=-1780.6kJ$

(2) $C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)$ $\Delta H_2=-(-136.3kJ)=136.3kJ$

(3) $H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)$ $\Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ$

(4) $2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)$ $\Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ$

The expression for enthalpy of the reaction will be,

$\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)$

$\Delta H=201.9kJ$

Therefore, the enthalpy change for the reaction is, 201.9 kJ

5 0

1 year ago

ANSWER FAST PLEASE!!

Svetach [21]

Answer:D

Explanation:

Hope I helped

3 0

2 years ago

A 3.5 kg iron shovel is left outside through the winter. The shovel, now orange with rust, is rediscovered in the spring. Its ma

Anton [14]

Answer:

a

Explanation:

8 0

1 year ago

Watch the video to determine which of the following relationships are correct according to Boyle’s law.

scoray [572]

Answer:

Part A

Boyle's Law is given mathematically as

P ∝(1/V) or V ∝(1/P)

Options 4 and 5, if they are properly written.

Part B

At constant temperature, and according to the Boyle's law for an ideal gas,

A. What can cause a Volume increase is a corresponding decrease in pressure.

B. What can cause a Volume decrease is a corresponding increase in pressure.

C. The Volume is unchanged if the pressure of the gas is unchanged too.

Part C

The pressure when the gas occupies a volume of 5.0 L = 40 atm

Part D

The pressure when the gas occupies a volume of 4.5 L = 36 atm

Explanation:

Part A

Boyle's Law states that at constant temperature, the pressure of an ideal gas is inversely proportional to the volume occupied by the gas.

So, mathematically, Boyle's Law is given as

P ∝(1/V) or V ∝(1/P)

Part B

Inverse relationship between two quantities means that the higher the value of one of the quantities go, the lower the value of the other quantity goes and vice versa.

So, at constant temperature, and according to the Boyle's law for an ideal gas.

A. What can cause a Volume increase is a corresponding decrease in pressure.

B. What can cause a Volume decrease is a corresponding increase in pressure.

C. The Volume is unchanged if the pressure of the gas is unchanged too.

Part C

A certain gas occupies a volume of 20 L when the applied pressure is 10 atm, find the pressure when the gas occupies a volume of 5.0 L.

According to Boyle's Law for an ideal gas,

P ∝(1/V)

P = (k/V)

where k is the constant of proportionality

PV = k

Therefore,

P₁V₁ = P₂V₂ = k

P₁ = 10 atm

V₁ = 20 L

P₂ = ?

V₂ = 5.0 L

10 × 20 = P₂ × 5

P₂ = 40 atm

Part D

If a certain gas occupies a volume of 18 L when the applied pressure is 9.0 atm , find the pressure when the gas occupies a volume of 4.5 L

P₁V₁ = P₂V₂ = k

P₁ = 9.0 atm

V₁ = 18 L

P₂ = ?

V₂ = 4.5 L

9 × 18 = P₂ × 4.5

P₂ = 36 atm

Hope this Helps!!!

3 0

2 years ago

Which traits does the common garter snake have that might be adaptive for the environment where it lives?

Licemer1 [7]

Answer:

As the garter snake can be found almost in any kind of habitat, what makes them be able to survive in any environment include:

1. They hibernate to increase their chances of survival in unfavorable weather conditions.

2. They can blend with the background of any environment especially grass to escape being eaten.

3. They produce an odor that is usually unpleasant especially when about to be attacked.

Explanation:

The garter snakes are distinguished by the three stripes running the length of their body and can often be found in forests, places that are even close to water bodies, and almost any place, even in holes.

4 0

1 year ago