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Lubov Fominskaja [6]
3 years ago
15

Describe the pattern of temperature changes within the layers of the atmosphere. Why do you think temperature changes follow thi

s unique pattern
Chemistry
1 answer:
ser-zykov [4K]3 years ago
6 0

Answer:

The atmosphere refers to the gaseous envelope of earth, comprised of variable gases with definite proportions. The layers of the earth's atmosphere are as follows-

  • Troposphere- This layer starts from the ground and extends up to a height of about 10 km. Here, the temperature decreases with the increasing altitude. All the weather phenomenon takes place in this layer.
  • Stratosphere- It starts from 10 km and extends up to a height of about 50 km. Here the temperature increases as the altitude increase. This is because of the presence of the ozone layer that receives the harmful UV radiation emitted from the sun.
  • Mesosphere- This layer extends from a height of about 50 km to about 80 km above the earth's surface. Here, again the temperature decreases with the increasing altitude.
  • Thermosphere- This layer starts from a height of about 80 km and extends up to about 500 km above the ground surface. In this region again the temperature increases with the increasing elevation.
  • Exosphere- This layer ranges from about 500 km to 10,000 km above the earth's surface. Here, the temperature gradually increases with the increasing height.

This variation in temperature occurs because of the certain reason. In the troposphere and the mesosphere, the temperature decreases with height because the pressure and height are inversely proportional to each other. The stratosphere experiences increasing temperature because of the presence of the ozone layer that is responsible for holding the greenhouse gases and the harmful UV radiation. The thermosphere and the exosphere experience high temperatures because of the receiving of the direct sunlight. Due to these above reason, there occurs this temperature change in a unique pattern.

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Answer:

a. 1.23 V

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Explanation:

Required:

a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?

b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

If E°cell must be at least 1.10 V (E°cell > 1.10 V),

E°red, cat - E°red, an > 1.10 V

E°red, cat - 0.13V > 1.10 V

E°red, cat > 1.23 V

The minimum standard reduction potential is 1.23 V while there is no maximum standard reduction potential.

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A sample of argon has a volume of 1.2 L at STP. If the temperature is increased to 21 c and the pressure is lowered to 0.80 atm,
Nuetrik [128]

Answer:

The new volume is 1.62 L

Explanation:

Boyle's law says:

"The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure." It is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T}=k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. So this law indicates that the quotient between pressure and temperature is constant.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T}=k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law.

\frac{P*V}{T}=k

Having an initial state 1 and a final state 2 it is possible to say that:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

Standard temperature and pressure (STP) indicate pressure conditions P = 1 atm and temperature T = 0 ° C = 273 ° K. Then:

  • P1= 1 atm
  • V1= 1.2 L
  • T1= 273 °K
  • P2= 0.80 atm
  • V2= ?
  • T2= 21°C= 294 °K

Replacing:

\frac{1 atm* 1.2 L}{273K} =\frac{0.8 atm*V2}{294K}

Solving:

V2=\frac{1 atm*1.2 L}{273 K} *\frac{294 K}{0.8 atm}

V2= 1.62 L

<u><em>The new volume is 1.62 L</em></u>

<u><em></em></u>

8 0
3 years ago
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