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kap26 [50]
3 years ago
9

If the pressure changes from 1 atm to 3 atm, what does the volume of 30 L change to

Chemistry
1 answer:
DiKsa [7]3 years ago
5 0

Answer: 10L

Explanation:

Given that:

Initial pressure P1 = 1 atm

New pressure P2 = 3 atm

Initial volume V2 = 30 L

New volume V2 = ?

Since pressure and volume are involved, apply the formula for Boyle's law

P1V1 = P2V2

1 atm x 30L = 3 atm x V2

30 atm L = 3 atm x V2

V2 = (30 atm L / 3 atm)

V2 = 10L

Thus, volume changed to 10 liters

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What is the difference between polar and non-polar
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2 years ago
When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin
LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine (C_3H_7NO_2) are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is 5.9^oC lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is 7.2^oC lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

<u>Answer:</u> The van't Hoff factor for potassium bromide in X is 1.63

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\Delta T_f=i\times K_f\times m

OR

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

  • <u>When alanine is dissolved in mystery liquid X:</u>

\Delta T_f=5.9^oC

i = Vant Hoff factor = 1 (for non-electrolytes)

K_f = freezing point depression constant

m_{solute} = Given mass of solute (alanine) = 177. g

M_{solute} = Molar mass of solute (alanine) = 89 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m

  • <u>When KBr is dissolved in mystery liquid X:</u>

\Delta T_f=7.2^oC

i = Vant Hoff factor = ?

K_f = freezing point depression constant = 2.37^oC/m

m_{solute} = Given mass of solute (KBr) = 177. g

M_{solute} = Molar mass of solute (KBr) = 119 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63

Hence, the van't Hoff factor for potassium bromide in X is 1.63

7 0
3 years ago
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