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Allisa [31]
2 years ago
14

A container with a volume 2. 0 L is filled with a gas at a pressure of 1. 5 atm. By decreasing the volume of the container to 1.

0 L, what is the resulting pressure? Type in your answer using the correct number of significant figures. Remember to use the formula for Boyle's law: P1V1 = P2V2 atm.
Chemistry
1 answer:
worty [1.4K]2 years ago
5 0

The resulting pressure of the gas after decreasing the initial volume from 2 L to 1 L is 3 atm.

<h3>What is Boyle's Law?</h3>

According to the Boyle's Law at constant temperature, pressure of the gas is inversely proportional to the volume of that gas.

For the given question we use the below equation is:

P₁V₁ = P₂V₂, where

P₁ = initial pressure of gas = 1.5 atm

V₁ = initial volume of gas = 2 L

P₂ = final pressure of gas = ?

V₂ = final volume of gas = 1 L

On putting all these values on the above equation, we get

P₂ = (1.5atm)(2L) / (1L) = 3 atm

Hence required pressure of the gas is 3 atm.

To know more about Boyle's Law, visit the below link:
brainly.com/question/469270

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Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

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Number of moles  = 35.5 g / 24 g/mol

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Number of moles = Mass / molar mass

8 mol  = Mass / 2 g/mol

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Number of moles  = 1.6 g /48  g/mol

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Number of moles = Mass / molar masa

Number of moles  = 54 g / 18 g/mol

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a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO    →    2Fe  + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

                        Fe          :           F₂O₃                

                           2          :             1

                          1.78       :        1/2×1.78 = 0.89 mol

Mass of  F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂   →  2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

                          Al            :         AlI₃    

                          2             :           2

                         0.05         :        0.05

                           I₂            :         AlI₃

                           3            :          2

                         0.1           :           2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:                            

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂   → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

               Pb         :        O₂

                2          :         1

               0.03     :      1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass =  0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂   → 2PbO

   

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