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Luden [163]
3 years ago
7

Predict which of the following changes will cause the volume of the balloon to increase or decrease assuming that the temperatur

e and the gas filling the balloon remain unchanged. Drag the appropriate items to their respective bins.
Standard temperature and pressure (STP) are considered to be 273 K and 1.0 atm.

1. Balloon filled with helium under water at 1.15 atm is released and floats to the surface, which is at STP.
2. Balloon filled with helium at STP floats into the atmosphere where the pressure is 0.5 atm.
3. Balloon filled with helium at STP is submerged under water where the pressure is 1.25 atm.
4. Balloon filled with helium at STP floats into air where the pressure equals 1 atm.

Bins:
a. Volume increases
b. Volume decreases
c. Volume is unchanged
Chemistry
1 answer:
IrinaVladis [17]3 years ago
7 0

Answer:

1. a

2. a

3. b

4. c

Explanation:

The STP is the Standard conditions of Temperature and Pressure, in these conditions, T = 0°C and P = 1 atm. By the ideal gas law,

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

We can observe that pressure and volume are indirect proportional, so when one increase, the other decrease. so:

1. The pressure decreases from 1.15 atm to STP, 1 atm, thus the volume increases.

2. The pressure decreases from STP, 1 atm, to 0.5 atm, thus the volume increases.

3. The pressure increases from STP, 1 atm, to 1.25 atm, thus the volume decreases.

4. The pressure remains constant at 1 atm, thus the volume is unchanged.

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Your lab partner combined chloroform and acetone to create a solution where the mole fraction of chloroform, Xchloroform, is 0.1
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Answer:

Explanation:

[u]Assumptions[/u]

1. There is exactly 1 mole of chloroform

2. The liquids mix together well such that the volume of the solution is the sum of the volumes of the two liquids

Given that the mole fraction of the Chloroform is 0.171

Mole fraction of Chloroform =

Mole of chloroform /(Mole of chloroform + mole of acetone)

According to assumptions, mole of chloroform is equal to 1

Therefore 0.171 =1/(1+mole of acetone)

1 + mole of acetone = 1/0.171

Mole of acetone = 1(/0.171) - 1

Moles of acetone = 4.85mol.

From Stochiometry

Mass of acetone = Mole of acetone * Molar Mass of acetone

Molar mass of acetone = 58.1grams/mol

Mass of acetone = 4.85 *58.1 = 282g =0.282kg

Mass of chloroform = moles of chloroform *Molar mass of Chloroform

Molar mass of chloform = 119.4 grams/mol

Mass of chloroform = 1* 119.4 =119.4g=0.1994kg

Volume of acetone = Mass of acetone / Density of acetone

Volume of acetone = 282/0.791

Volume of acetone = 357mL

Volume of Chloroform = Mass of Chloroform /Density of Chloroform

Volume of Chloroform = 119.4/1.48

= 81mL

Total volume of solution = 357mL+81mL = 438mL = 0.438L

1. Molarity = Moles of solute(chloroform)/mass of solvent (acetone) in kg

Molarity = 1/0.282 = 3.55molal

2. Molarity = moles of solute( chloroform) /Volume of solution

= 1/0.438 =2.28Molar

Therefore the molality and molarity respectively are 3.55 and 2.28.

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