Are these answers correct?1. A gas occupies 30 L at 760 mm Hg. What is its volume at 350 mm Hg?

P1*V1 = P2*V2

760 mm Hg *30 L = 350 mm Hg * V2

V2 = 22800 mm Hg*L/350 mm Hg

V2 = about 65.14 L


2. A gas occupies 500 mL at 1 atm. What is its pressure if the volume becomes 1500 mL?

P1*V1 = P2*V2

1 atm*500 mL = P2*1500 mL

500 atm*mL/1500 mL = P2

P2 = 1/3 atm


3. If a gas occupies 250 mL at 25°C, what will its volume be at 57°C?

25°C + 273 = 298°K; 57°C + 273 = 330°K

V1/T1 = V2/T2

250 mL/298°K = V2/330°K

0.84 mL/k*330°K = V2

V2 = about 276.85 mL

4. What was the original temperature of a 2 L sample of gas if it initially had the volume 5 L at 27°C?

27°C + 273 = 298°K

V1/T1 = V2/T2

5 L/298°K = 2 L/T2

2 L/(0.017 L/K)= T2

T2 = 119.2°K = -153.8°C

5. At what pressure would 100 mL of a gas be found if it began at 500 mL at 2.5 atm?

P1*V1 = P2*V2

2.5 atm*500 mL = P2*100 mL

125 atm*mL/100 mL = P2

P2 = 1.25 atm

6. What is the volume of a gas at 275 mm Hg if it occupied 750 mL at 1000 mm Hg?

P1*V1 = P2*V2

1000 mm Hg*750 mL = 275 mm Hg*V2

750000 mm Hg*mL/275 mm Hg = V2

V2 = about 2727. 27 mL

7. A gas occupies 60 L at 250°K and 1 atm. What is its volume at 250°C and 1.5 atm?

250°C + 273 = 523°K

P1*V1/T1 = P2*V2/T2

1 atm*60 L/250°K = 1.5 atm*V2/523°K

0.24 (atm*L/K)*523°K = 1.5 atm*V2

125.52 (atm*L)/1.5 atm = V2

V2 = 83.68 L

8. What is the final temperature in Celsius of a gas occupying 3.25 L at 750 mm Hg if it occupied 2250 mL at 37°C and 600 mm Hg?

37°C + 273 = 310°K; 2250 mL = 2.25 L

P1*V1/T1 = P2*V2/T2

600 mm Hg*2.25 L/310°K = 750 mm Hg*3.25 L/T2

2437.35 mm Hg*L/ (4.35 mm Hg*L/K) = T2

T2 = 559.69°K
T2 = about 286.69°C

Answers

Answer 1
Answer: I saw a slight error in number 8. 750*3.25=2437.5 not 2437.35. That threw your final answer off by a little more than half a °C. Other than that, I didn't see anything wrong.

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Which is denser, isopropyll alcohol( rubbing alchohol) or water?

Answers

Isopropyll is less dense than water.

If you were to get a gallon of water and a gallon of Isopropyll alcohol and weigh it, the alcohol would be lighted. Another way is to pour water and the alcohol in a cup and you will see that the alcohol is above the water(floating) because its lighter and less dense. Isopropyll alcohol is used for a lot of things, including cleaning certain components of computers:D

So based on this, water is more dense

Answer:

water is denser

Explanation:

if you were to fill 2 55 gallon barrels of each product the water would weigh more because water is more dense

would you select water or gasoline to try dissolving the following : table sugar, motor oil, rubber from tire marks, adhesive residue from a packing tape

Answers

To know the answer, we determine the polarities of the solvent we have. Water is polar and gasoline is nonpolar. We remember that like dissolves like. Therefore, the answers are:

table sugar = water
motor oil = gasoline
rubber from tire marks = gasoline
adhesive residue from a packing tape = gasoline

Answer:

table sugar: water

motor oil: gasoline

rubber from tire marks: gasoline

adhesive residue from a packing tape: gasoline

Explanation:

Like dissolves like so a polar compound will dissolve in a polar solvent and a non polar compound will dissolve in a non-polar solvent. The intermolecular forces of attraction are responsible for the dissolution of substances in different solvents.  

For example, polar solute exhibit hydrogen bonding and dipole interaction with polar solvents (water). Non-polar solute exhibit London dispersion forces with non-polar solvents(gasoline).

What is a spelling bee​

Answers

Answer: a contest in which you are eliminated if you fail to spell a word correctly.

a contest where you attempt to spell a series of words that you are given by the judges

How are the atoms in an object affected when the temperature of the object increase?

Answers

As the temperature is increased, the atoms displace their neighboring atoms and this chain goes continuously due to which energy is transferred from one atom to their neighbor atom like wise vibration and energy increases.

What is the volume occupied by 0.108 mol of helium gas at a pressure of 0.909 atm and a temperature of 306 K ? Express your answer using three significant figures. Would the volume be different if the gas was argon (under the same conditions)? The volume would be the lower for argon gas. The volume would be the same for argon gas. The volume would be the greater for argon gas.

Answers

Answer: 2.93 L

Explanation:

AI-generated answer

To find the volume occupied by 0.108 mol of helium gas at a pressure of 0.909 atm and a temperature of 306 K, we can use the ideal gas law equation:

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 in Kelvin.

Rearranging the equation to solve for V:

V = (nRT) / P

Plugging in the given values:

n = 0.108 mol

R = 0.0821 L·atm/(mol·K) (the ideal gas constant)

T = 306 K

P = 0.909 atm

V = (0.108 mol * 0.0821 L·atm/(mol·K) * 306 K) / 0.909 atm

Calculating this expression, we find that the volume occupied by 0.108 mol of helium gas at a pressure of 0.909 atm and a temperature of 306 K is approximately 2.93 L.

Now, let's consider the second part of the question: Would the volume be different if the gas was argon (under the same conditions)?

The volume would be the same for argon gas.

According to the ideal gas law, at the same temperature, pressure, and number of moles, the volume occupied by a gas is the same regardless of the gas's identity. Therefore, if we replaced helium gas with argon gas while keeping the same conditions of pressure, temperature, and number of moles, the volume occupied by argon gas would be the same, approximately 2.93 L.

A list of changes is shown.Temperature decreases
Intermolecular bonds become weaker
Freezing
Intermolecular bonds become stronger
Evaporation
Temperature increases
Vibration of molecules slows down
Molecules vibrate faster
Which of the above changes would be accompanied by an increase of the kinetic energy of particles in a liquid?
A, B, C, and H
A, C, D, and G
B, E, F, and H
B, C, F, and G

Answers

With an increase in the kinetic energy of the solid molecule the intermolecular bonds become weaker, temperature increases, melting and molecules vibrating faster are changes.

What is the effect of the increased kinetic energy?

Kinetic energy is the form of the energy that is present in an entity in motion. When the kinetic energy is increased then the temperature of the particles increases and leads to evaporation.

The increased kinetic energy weakens the intermolecular bonds of the liquid particles and the molecules of the particles start vibrating fast.

Therefore, option b. intermolecular bonds become weak, option e. evaporation, option f. temperature increases and option h. molecules vibrate faster are the changes.

Learn more about kinetic energy here:

brainly.com/question/25726474

Final answer:

The changes that increase the kinetic energy of liquid particles are B, E, F, and H which correspond to evaporation, temperature increases, and molecules vibrating faster.

Explanation:

The changes that would be accompanied by an increase of the kinetic energy of particles in a liquid are evaporation, temperature increases, and molecules vibrate faster. During evaporation, molecules gain sufficient energy to overcome intermolecular forces and enter the gaseous phase. When the temperature increases, the average kinetic energy of the molecules increases, causing them to vibrate faster. Consequently, the correct answer is B, E, F, and H.

The changes that would be accompanied by an increase of the kinetic energy of particles in a liquid are B, E, F, and G.

When temperature increases, the average kinetic energy of the particles in a liquid also increases. This is represented by changes B and F. Evaporation (change E) also leads to an increase in the kinetic energy of the particles as they gain enough energy to escape into the vapor phase. Lastly, intermolecular bonds becoming weaker (change G) allows for more movement of the particles, increasing their kinetic energy.