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Alexus [3.1K]
1 year ago
5

Help me plsnayone here50 pointsfreee ​

Chemistry
2 answers:
tatyana61 [14]1 year ago
8 0

Answer:

what is the question? Because all i see is that theirs answer choices so i guess the answer is A

Explanation:

scoray [572]1 year ago
3 0

8).

(a) Acid Rain is formed due to precipitation of acids (oxides of nitrogen and sulfur dioxide) mixing with the oxygen and water in the clouds. Then they pour on the ground with the rain water leading huge problems like damaging habitat for animals and etc.

(b) Acid rain is formed due to Sulphur dioxide formed by combustion of fuels and nitrogen oxides from hot engines, they dissolve and cause acid rain.

Equation's:

formation of Sulphur dioxide : S + O₂ = SO₂

formation of sulfuric acid : SO₂ + ½O₂ + H₂O = H₂SO₄

(c) Some harmful effects of acid rain are:

   i) Damages tree and plants.

   ii) Kills fish and animals in river.

   iii) Damages buildings and structures.

Additional: Acid rain can be prevented by addition of catalytic converters in car and using slaked lime in factories to neutralize the acidic oxides.

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In order to become stable, Potassium, K, will _______ and have a resulting charge of _____.
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2 years ago
1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
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