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leonid [27]
3 years ago
11

Gasoline contains only hydrogen and carbon atoms, yet nitrogen oxide and nitrogen dioxide are produced when gasoline burns. What

is the source of the nitro- gen and oxygen atoms?
Chemistry
2 answers:
White raven [17]3 years ago
7 0

Answer:

The answer to the question is

The source of the nitro- gen and oxygen atoms is from the atmosphere

Explanation:

When gasoline burns oxygen from the atmosphere supports the combustion reaction.

For example the internal combustion engine of a motor vehicle pulls in air from the surrounding atmosphere to make use of the oxygen in aiding the combustion of the gasoline to drive the pistons that ultimately moves the vehicle. The air from which the oxygen is absorbed also contains more amount of Nitrogen by percentage than oxygen. The nitrogen therefore enters into the combustion reaction that eventually produces nitrogen oxide and nitrogen dioxide

gregori [183]3 years ago
7 0

Explanation:

When fuels are burned either in vehicle engines or other processes, high temperatures are reached. At these high temperatures, nitrogen and oxygen from the air(which oxygen and nitrogen have significant percentages) combine to produce nitrogen monoxide. When this nitrogen monoxide is released from these systems, it combines with oxygen in the air to form nitrogen dioxide.

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After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which sid
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Answer:

d. 8 moles of H2O on the product side

Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow  Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow  Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\

Then, we add the half reactions:

2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

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3 years ago
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I hope it helps

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Is it necessary to know the exact volume of the unknown acid or base to be titrated?
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3 years ago
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