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3 years ago

A sample of krypton gas with a volume of 10.00 L has a temp of 303 K and exerts

Chemistry

1 answer:

marshall27 [118]3 years ago

7 0

Answer: The final volume will be 14.85 L.

Explanation:

To calculate the final volume when temperature increases, we use Charles' Law.

This law states that volume is directly proportional to the temperature of the gas if number of moles and pressure remains constant.

$V\propto T\\\\\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ = Initial volume and temperature

$V_2\text{ and }T_2$ = Final volume and temperature

We are given:

$V_1=10L\\
T_1=303K\\
V_2=?L\\
T_2=450K$

Putting values in above equation, we get:

$\frac{10L}{303K}=\frac{V_2}{450K}$

$V_2=14.85L$

Hence, the final volume of the gas is 14.85L

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2 years ago

Read 2 more answers

Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

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3 years ago

When 0.105 mol propane, C3H8 is burned in an excess of oxygen, how many moles of oxygen are consumed? The products are carbon di

torisob [31]

Answer: The amount of oxygen gas consumed is 0.525 moles

Explanation:

We are given:

Moles of propane burned = 0.150 moles

The chemical equation for the combustion of propane follows:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

By Stoichiometry of the reaction:

1 mole of propane reacts with 5 moles of oxygen gas

So, 0.150 moles of propane will react with = $\frac{5}{1}\times 0.105=0.525mol$ of oxygen gas

Hence, the amount of oxygen gas consumed is 0.525 moles

6 0

3 years ago