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adoni [48]
3 years ago
9

What happens to the average kinetic energy of the molecules of a warm substance when the temperature of the substance is slightl

y decreased? (5 points) Group of answer choices It increases. It decreases. It becomes zero. It remains unchanged.
Chemistry
2 answers:
iris [78.8K]3 years ago
5 0

Answer: Option (b) is the correct answer.

Explanation:

Average kinetic energy is defined as half of mass of gas times the square of root mean square velocity.

Mathematically,    Average K.E = \frac{1}{2}mv^{2}_{rms}

Also,   Kinetic energy = \frac{3}{2}kT

So, when there is increase in temperature then there will be increase in movement of the molecules. As a result, kinetic energy will also increase. Since, kinetic energy is proportional to temperature.

Therefore, with decrease in temperature there will be decrease in average kinetic energy of the molecules.

Thus, we can conclude that the average kinetic energy of the molecules of a warm substance also decreases when the temperature of the substance is slightly decreased.

Ber [7]3 years ago
3 0

Answer:

decreases

Explanation:

the particles have less kinetic energy as the thermal heat is decreased

You might be interested in
It is necessary to know the time of day and the position of the Sun or the stars relative to one's longitude line because ____.
allochka39001 [22]

Answer: the longitude line rotates with Earth, 360 degrees in 24 hours.

Explanation: In the old days, the position of the sun was used to tell which time of the day it is. But more modern findings, has shown that the longitude line matters too.

3 0
3 years ago
. An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend in altitude until the pressure is 0.
Tju [1.3M]

Answer:

The answer to your question is Volume = 11.4 L

Explanation:

Data

Volume 1 = V1 = 6 L

Pressure 1 = P1 = 1 atm

Temperature 1 = T1 = 22°C

Volume 2 = V2 = ?

Pressure 2 = 0.45 atm

Temperature 2 = -21°C

Process

1.- Convert temperature (°C) to °K

T1 = 273 + 22 = 295°K

T2 = 273 + (-21) = 252°K

2.- Use the combined gas law to solve this problem

                   P1V1 / T1 = P2V2 / T2

-Solve for V2

                   V2 = P1V1T2 / T1P2

-Substitution

                   V2 = (6)(1)(252) / (295)(0.45)

- Simplification

                   V2 = 1512 / 132.75

- Result

                   V2 = 11.38 L

3 0
3 years ago
I need help plz with the first question plz
Cloud [144]
Ok so put it in order
7 0
3 years ago
Properties of group 7 (aka. The Halogens)?
ella [17]
Halogens can be characterized as :

1) High Electronegativity 
2) Low melting & boiling points
3) Forms Ionic compounds
4) Non-metallic in nature

Hope this helps!
7 0
3 years ago
A sample of an unknown compound is vaporized at 150.°C . The gas produced has a volume of 960.mL at a pressure of 1.00atm , and
IrinaK [193]

Answer:

34.02 g.

Explanation:

Hello!

In this case, since the gas behaves ideally, we can use the following equation to compute the moles at the specified conditions:

PV=nRT\\\\n=\frac{1.00atm*0.960L}{0.08206\frac{atm*L}{mol*K}*(150+273)K} =0.0277mol\\\\

Now, since the molar mass of a compound is computed by dividing the mass over mass, we obtain the following molar mass:

MM=\frac{0.941g}{0.0277mol} \\\\MM=34.02g/mol

So probably, the gas may be H₂S.

Best regards!

6 0
3 years ago
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