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lubasha [3.4K]
2 years ago
12

A bicycle pump contains 250 mL of air at a pressure of 102 kPa. If the pump is

Chemistry
1 answer:
CaHeK987 [17]2 years ago
3 0

Considering the Boyle's law, the new volume becomes 1,148,344.444 mL or 1148.34 L.

<h3>Boyle's law</h3>

Boyle's law relates pressure and volume, stating that the volume occupied by a given mass of gas at constant temperature is inversely proportional to pressure. This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

P×V=k

Now it is possible to assume that you have a certain volume of gas V1 which is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and the following is true:

P1×V1=P2×V2

<h3>New volume</h3>

In this case, you know:

  • P1= 102 kPa= 10335.1 atm (being 1 kPa= 101.325 atm)
  • V1= 250 mL
  • P2= 2.25 atm
  • V2= ?

Replacing in the Boyle's law:

10335.1 atm ×250 mL= 2.25 atm×V2

Solving:

(10335.1 atm ×250 mL)÷ 2.25 atm= V2

<u><em>1,148,344.444 mL= 1148.34 L=  V2</em></u>

Finally, the new volume becomes 1,148,344.444 mL or 1148.34 L.

Learn more about Boyle's law:

brainly.com/question/4147359

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Read 2 more answers
Use the following balanced reaction to solve:
Naily [24]

Answer:  60.7 g of PH_3 will be formed.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}    

\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles

The balanced chemical reaction is

P_4(s)+6H_2(g)\rightarrow 4PH_3(g)

H_2 is the limiting reagent as it limits the formation of product and P_4 is the excess reagent.

According to stoichiometry :

6 moles of H_2 produce = 4 moles of PH_3

Thus 2.68 moles of H_2 will produce=\frac{4}{6}\times 2.68=1.79moles  of PH_3

Mass of PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g

Thus 60.7 g of PH_3 will be formed by reactiong 60 L of hydrogen gas with an excess of P_4

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