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Serhud [2]
3 years ago
9

The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4

Chemistry
1 answer:
Evgesh-ka [11]3 years ago
5 0

In NH3 , let oxidation number of N be x

x + (+1)3 = 0

x = -3

In HNO3 , let oxidation number of N be x

1 + x + (-2)3 = 0

x = +5

In NO2 , let oxidation number of N be x

x + (-2)2 = 0

x = +4
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Which of the following best explains why bonds form between atoms?
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D

Explanation:

I believe the answer is D because atoms are always seeking to fill up their outer electron shell/valence shell and want to gain a full octet.

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_______ is the process in which light energy is transformed into food energy. A.life B.photosynthesis C.abiotic D.biotic
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Oxygen gas was produced in a reaction and collected over water. A 136.1 mL mL sample of gas was collected over water at 25C and
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Answer:

Explanation:

We shall find volume of gas at NTP or at 273 K , 760 mm of Hg .

Pressure of given gas = 1.06 x 760 mm of Hg less vapor pressure of water .

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For it we use gas law formula ,

P₁V₁ / T₁ = P₂V₂ / T₂

781.84 x 136.1 / ( 273 + 25 ) = 760 x V₂ / 273

= 128.26 mL .

= 128.26  x 10⁻³ L .

22.4 L of oxygen will have mass of 32 g

128.26 x 10⁻³ L of oxygen will have mass of 32 x 128.26 x 10⁻³ / 22.4 g

= 183.22 mg .

4 0
3 years ago
Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of
harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g)   (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

n = \frac{m}{M}

Where:    

m: is the mass

M: is the molar mass

  • For <em>hydrogen </em>we have:

n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles

  • And for <em>nitrogen</em>:

n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles

We can see in reaction (1) that <u>3 moles of hydrogen</u> react with <u>1 mol of nitrogen</u>, so the number of hydrogen moles needed to react nitrogen is:

n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles

Since we have <u>3.47 moles of hydrogen</u> and we need <u>7.50 moles</u> to react with all the mass of nitrogen, the <em>limiting reactant</em> is <em>hydrogen</em>.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that <u>3 moles of hydrogen</u> produces <u>2 moles of ammonia</u>, so:

n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

Therefore, hydrogen is the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

   

I hope it helps you!                        

6 0
3 years ago
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