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stiks02 [169]
3 years ago
10

Match each statement with a layer of the atmosphere. There will be two statements for each layer.

Chemistry
1 answer:
LuckyWell [14K]3 years ago
6 0

<u>The troposphere: </u>

H. This layer can have thunderstorms or clear, sunny skies.

A. The biosphere interacts most with this layer. 

<u>The stratosphere:</u>

B. It is the second layer from Earth's surface. 

G. Winds are strong and steady in this layer. 

<u>The mesosphere:</u>

E. It is heated by the ozone layer beneath it. 

D. This layer is where most meteor showers occur. 

<u>The thermosphere :</u>

F. It contains the ionosphere and exosphere. 

C. It contains layers of single, unmixed gas. 

<u>Explanation:</u>

Depending on the Earth's  temperature the atmosphere can be separated into layers. The troposphere, the stratosphere, the mesosphere and the thermosphere are those layers. The lowest layer is named as Troposphere (0-10 km from the Earth outer surface), it comprises about 75% of the atmosphere's total air and nearly most the water vapor.

Stratosphere (10-30) includes much of the surface ozone. The change in height temperature arises as this ozone absorbs ultraviolet (UV) radiation from the sun. The temperature in Mesosphere (30-50 Km) declines again with height, hitting a minimum of about -90 ° C at the "mesopause." Above this thermosphere (50-400 Km) is settled which is a area where temperatures rise with height once again. The penetration of intense  UV and X-ray radiation from the sun induces this temperature rise.

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A propane stove burned 470 grams propane and produced 625 grams of water (this is the actual yield) C3H8 +5O2=3CO2+4H20. What wa
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81.3%

Explanation:

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The balanced equation for the reaction:

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C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

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