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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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If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of
Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.1\ g}{186.04\ g/mol}

Moles\ of\ ferrocene= 0.0005375\ mol

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

\rho=\frac{Mass}{Volume}

or,  

Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.044\ g}{78.49\ g/mol}

Moles\ of\ acetyl\ chloride= 0.0005606\ mol

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms  0.0005375 mole of monoacetylferrocene

Moles of product formed =  0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %</u>

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