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dimaraw [331]
3 years ago
11

the volume of gas in a container was originally 3.24L, while at standard pressure 1.00 atm. what will the volume be if the press

ure is increased to 1.20 atm?
Chemistry
1 answer:
Simora [160]3 years ago
8 0
Hello!

To solve this problem, we will use the Boyle's Law, which describes how pressure changes when volume changes and vice-versa. The equation for this law is the following one, and we'll clear for V2:

P1*V1=P2*V2 \\ \\ V2= \frac{P1*V1}{P2}= \frac{3,24 L * 1 atm}{1,20 atm}= 2,7 L

So, the final volume after increasing the pressure would be 2,7 L. That means that volume decreases when the pressure increases

Have a nice day!
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3 years ago
If 30.0 mL of 0.150 M CaCl₂ is added to 38.5 mL of 0.100 M AgNO3, what is the mass of the AgCl precipitate?
Ksenya-84 [330]

If 30.0 mL of 0.150 M CaCl₂ is added to 38.5 mL of 0.100 M AgNO3. The mass of the AgCl precipitate is 0.552 g.

<h3>What is Stoichiometry ?</h3>

Stoichiometry helps us use the balanced chemical equation to measure quantitative relationships and it is to calculate the amounts of products and reactants that are given in a reaction.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

Now we have to write the balanced equation

2AgNO₃ + CaCl₂ →  2AgCl + Ca(NO₃)₂

Molecular Weight of CaCl₂ = 110.98 g/mol

Molecular Weight of AgNO₃ = 170.01 g/mol

Molecular Weight of AgCl = 143.45 g/mol

Here,

Volume of CaCl₂ = 30.0 mL = 0.03L

Volume of AgNO₃ = 38.5 mL = 0.0385 L

Now find the number of moles

Number of moles = Volume × Molarity

                         

Number of moles of CaCl₂ = 0.03 L × 0.150

                                             = 0.00456 mol

Number of moles of AgNO₃ = 0.0385 L × 0.100

                                               = 0.00385 mol

The stoichiometric ratio of AgNO₃ to CaCl₂ is 2:1.

= \frac{0.00385}{2}

= 0.001925 mol                      

According to Stoichiometry Mass of AgCl

= 0.0385  \times \frac{0.1}{1\ \text{mol}} \times \frac{2\ \text{mol} AgCl}{2\ \text{mol} AgNO_3} \times \frac{143.4\ g}{1\ \text{mol}}

= 0.552 g AgCl

Thus from the above conclusion we can say that If 30.0 mL of 0.150 M CaCl₂ is added to 38.5 mL of 0.100 M AgNO3. The mass of the AgCl precipitate is 0.552 g.

Learn more about the Stoichiometry here: brainly.com/question/16060223

#SPJ1

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