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Ilya [14]
3 years ago
9

Carbon dioxide in the atmosphere can be reduced by _.*​

Chemistry
2 answers:
Dima020 [189]3 years ago
7 0

Answer:

Less driving by yourself and more carpooling to reduce the amount of cars releasing Carbon dioxide into the air.

Explanation:

pychu [463]3 years ago
6 0

Answer:

As the use of plants as carbon sinks can be undone by events such as wildfires, the long-term reliability of these approaches has been questioned. Carbon dioxide that has been removed from the atmosphere can also be stored in the Earth's crust by injecting it into the subsurface, or in the form of insoluble carbonate salts (mineral sequestration).

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Explanation:

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timofeeve [1]

Answer:

melting

Explanation:

liquid is a good conductor of electricity

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water droplets tend to have a spherical, round shape, Explain ths fact in terms of the polar nature of water molecules
Doss [256]
What does it happen in smallest scale? 
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<span>"TWO-PHASE INTERFACIAL FORCEs". </span>
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6 0
3 years ago
What can create sediment
bearhunter [10]
Sediment is created by wind and water.

4 0
4 years ago
Read 2 more answers
If the percent yield for the following reaction is 75.0%, and 25.0 g of NO₂ are consumed in the reaction, how many grams of nitr
victus00 [196]

Answer:

17.1195 grams of nitric acid are produced.

Explanation:

3NO_2+H_2O\rightarrow 2HNO_3+NO

Moles of nitrogen dioxide :

\frac{25.0 g}{56 g/mol}=0.5434 mol

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.

Then 0.5434 moles of nitrogen dioxides will give:

\frac{2}{3}\times 0.5434 mol=0.3623 mol of nitric acid.

Mass of 0.3623 moles of nitric acid :

0.3623 mol\times 63 g/mol=22.8260 g

Theoretical yield = 22.8260 g

Experimental yield = ?

\%Yield=\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

75\%=\frac{\text{Experimental yield}}{22.8260 g}

Experimental yield of nitric acid = 17.1195 g

7 0
4 years ago
he standard enthalpies of formation for S (g), F (g), SF4 (g), and SF6 (g) are +278.8, +79.0, -775, and -1209 kJ per mole, respe
JulsSmile [24]

Answer:

+60.54 kJ/mol

Explanation:

The enthalpy of formation of a compound is the sum of the enthalpies of the formation of its constituents and the bond of them. To form a compound, energy must be lost, so, they're negative.

For SF4, the enthalpy is formed by the energy of one S, two F, and 4 S-F bond (Hb):

H = - (278.8 + 4*79.0 + 4*Hb)

-775 = -(594.8 + 4Hb)

594.8 + 4Hb = 775

4Hb = 180.2

Hb = +45.05 kJ/mol

For SF6, the enthalpy is formed by the energy of one S, six F, and 6 S--F bonds (Hb):

H = -(278.8 + 6*79.0 + 6*Hb)

-1209 = -(752.8 + 6Hb)

752.8 + 6Hb = 1209

6Hb = 456.2

Hb = +76.03 kJ/mol

Thus, the energy of S--F bond must be the average of these two:

(45.05 + 76.03)/2 = +60.54 kJ/mol

3 0
3 years ago
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