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olga_2 [115]
3 years ago
11

Which of the following substances has increased markedly due to the use of fossil fuels and contributes to the greenhouse effect

? Which of the following substances has increased markedly due to the use of fossil fuels and contributes to the greenhouse effect? CO2 SO2 NO2 O3
Chemistry
1 answer:
Alenkasestr [34]3 years ago
7 0

Answer:

The correct option is: CO₂

Explanation:

Greenhouse effect refers to the natural phenomenon by which the Sun's radiation increases the temperature of the Earth's surface.

The incoming Sun's radiation is radiated by the greenhouse gases present in the Earth's atmosphere. A fraction of the radiated energy is then absorbed by the Earth's surface, thus resulting in increase in the temperature.

<u>Thus the green house effect depends upon the amount of green house gases present in the atmosphere.</u>

The green house gases are the r<u>adiatively active gases.</u> The main four green house gases are: water vapor (H₂O(g)), carbon dioxide (CO₂), methane (CH₄) and ozone (O₃).

<u>Human activities such as burning of fossil gases and deforestation increases the concentration of </u><u>CO₂ </u><u>significantly.</u><u> Thereby contributing to the greenhouse effect.</u>

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The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa
Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of MgCl_2 (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of MgCl_2 (solute) = 3.37 g

First we have to calculate the mass of solute.

\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2

\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g

Now we have to calculate the mass of solution.

\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

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Answer:

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