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tensa zangetsu [6.8K]
2 years ago
10

What is the major defining property of gasses?.

Chemistry
1 answer:
jonny [76]2 years ago
5 0

Answer:

Gases have no definite shape or volume. They are fluid, allowing particles/molecules to move freely.

The behavior of a gas is that the volume changes directly with temperature. With a constant volume, the pressure will be directly proportional to the amount of gas.

Explanation:

These are some of the properties I can think of

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What is mass in g of 22 moles of copper?
galina1969 [7]
The molar mass of copper is 63.55 g/mol
6 0
3 years ago
Cylinder of air at 1.5 atm of pressure is kept at room temperature while a piston compresses the air from 40 l down to 10 ml. wh
djyliett [7]

The new pressure, P₂ is 6000 atm.

<h3>Calculation:</h3>

Given,

P₁ = 1.5 atm

V₁ = 40 L = 40,000 mL

V₂ = 10 mL

To calculate,

P₂ =?

Boyle's law is applied here.

According to Boyle's law, at constant temperature, a gas's volume changes inversely with applied pressure.

                                        PV = constant

Therefore,

P₁V₁ = P₂V₂

Put the above values in the equation,

1.5 × 40,000 = P₂ × 10

P₂  = 1.5 × 4000

P₂  = 6000 atm

Therefore, the new pressure, P₂ is 6000 atm.

Learn more about Boyle's law here:

brainly.com/question/23715689

#SPJ4

8 0
2 years ago
What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
Vikentia [17]

Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

Temperature = 91^{o}C = (91 + 273) K = 364 K

As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

4 0
3 years ago
Which of these outdoor activities would you most likely participate in during the winter?
Elza [17]

Answer:

D.

Explanation:

Translation:

A. make a barbecue

B. organize a picnic

C. swimming in the pool

D. to go skiing

Barbecue is usually a summer activity, so we eliminate that.

It would be too cold for a picnic!

It would be too cold to go swimming, and that's more of a summer thing!

Skiing is a snow activity, and since it is winter, it is likely to know.

4 0
3 years ago
If a balloon containing 3000 L of gas at 39 C and 99 k Pa rises to an altitude where the pressure is 45.5 kPa and the temperatur
mihalych1998 [28]

Answer:

The new volume of the balloon will be 6046.28 L

Explanation:

Initial pressure (P1) = 99 kpa

initial volume (V1) = 3000 L

Initial temperature = 39 C = 39 + 273 = 312 K

Final pressure (P2) = 45.5 kpa

Final temperature = 16 C = 16 +273 = 289K

Final volume = ????

To calculate the final volume using the general gas equation

                  P1 V1 / T1  = P2 V2 / T2

make V2 the subject of the formular

                   V2 = 99000 ×3000× 289 / 45500×312

                   V2 = 85833000 /14196

                     V2 = 6046.28 litres

5 0
3 years ago
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