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vovikov84 [41]
3 years ago
13

If 0.756L of gas exerts a pressure of 94.6kPa, what would the volume be at standard pressure (101.33kPa)?

Chemistry
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

The answer to your question is Volume = 0.7 l

Explanation:

Data

Volume 1 = V1 = 0.746 l

Pressure 1 = P1 = 94.6 kPa

Volume 2 = V2 = ?

Pressure 2 = 101.33 kPa

Process

To solve this process use Boyle's law.

           P1V1 = P2V2

-Solve for V2

           V2 = P1V1/P2

-Substitution

          V2 = (94.6 x 0.746) / 101.33

-Simplification

           V2 = 70.57 / 101.33

-Result

            V2 = 0.696 l ≈ 0.7 l

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Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T
Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 730.5 mmHg  

V = Volume of the gas = 9.49 L

T = Temperature of the gas = 20^oC=[20+273]K=293K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol

According to the reaction shown below as:-

2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when \frac{2}{3}\times 0.3794 moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 735.5 mmHg  

V = Volume of the gas = 9.99 L

T = Temperature of the gas = 20^oC=[20+273]K=293K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0
3 years ago
Where are galciers found
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Answer: Antarctica or anywhere

Explanation:

I hope this helps im not sure if I’m correct but I’m sure u can find them anywhere cold

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You drop a rock weighing 23.2 g into a graduated cylinder that contains 55 mL. The level
snow_lady [41]

Answer:

<h2>3.31 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 23.2 g

volume = final volume of water - initial volume of water

volume = 62 - 55 = 7 mL

We have

density =  \frac{23.2}{7}  \\  = 3.314285

We have the final answer as

<h3>3.31 g/mL</h3>

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