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AlekseyPX
2 years ago
8

Why do vehicle engines produce nitrogen oxides​

Chemistry
2 answers:
Monica [59]2 years ago
5 0

Answer:

When fuels are burned in vehicle engines, high temperatures are reached. At these high temperatures, nitrogen and oxygen from the air combine to produce nitrogen monoxide. When this nitrogen monoxide is released from vehicle exhaust systems, it combines with oxygen in the air to form nitrogen dioxide.

Explanation:

<h2>please mark me as brainlist please</h2>
oksian1 [2.3K]2 years ago
5 0

Explanation:

Nitrogen and oxygen from the air combine at high temperatures in the engine to produce nitrogen (II) oxide by the reaction :

N2(g) + O2(g) → 2NO(g).

Nitric oxides rise in the atmosphere and are oxidized to nitrogen dioxide, NO2 which dissolves in the precipitating water to form acid rain.

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Whats the molar mass of potassium nitride
Lyrx [107]
<span>The answer is
101.1032 g/mol</span>
6 0
3 years ago
Which element would form an ionic bond with Oxygen?
koban [17]

Nitrogen will form an ionic bond

3 0
3 years ago
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10095 m/sec to mile/ sec. 1 km = 0.6214 mile
Anarel [89]

Answer:

Explanation:

to convert any value in miles per second to meters per second, just multiply the value in miles per second by the conversion factor 1609.344. So, 10095 miles per second times 1609.344 is equal to 1.625 × 107 meters per second.

4 0
3 years ago
The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be
Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 4.90°C/m

m_{solute} = Given mass of solute (naphthalene) = 2.60 g

M_{solute} = Molar mass of solute (naphthalene) = 128.2 g/mol

W_{solvent} = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC

Hence, the freezing point of solution is 5.35°C

3 0
3 years ago
Some oxygen gas takes up a volume of 2.00 liters at 0.99 atm and 273 K. Its volume doubles and its temperature decreases to 137
nataly862011 [7]

Answer:

The answer to your question is : letter B. 0.25 atm

Explanation:

To solve this problem we need to use the combined gas law:

    <u>P₁V₁</u>   =    <u>P₂V₂</u>

      T₁              T₂

Data

P1 = 0.99 atm  V1 = 2 l  T1 = 273K

P2 = ?               V2 = 4 l   T2 = 137K

Now, the clear P2 from the equation and we get

       P2 = P1V1T2 / T1V2

Substitution         P2 = (2 x 0.99 x 137)/(273 x 4)

       P2 = 271.26 / 1092

Result       P2 = 0.248 atm ≈ 0.25 atm

5 0
3 years ago
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