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jekas [21]
3 years ago
10

Explain how a motorcar engine produces nitrogen oxides

Chemistry
2 answers:
olya-2409 [2.1K]3 years ago
4 0

Answer: combustion causes a chemical reaction between nitrogen and oxygen in the engine.

Explanation:

Nitrogen oxides are produced in combustion processes, partly from nitrogen compounds in the fuel, but mostly by direct combination of atmospheric oxygen and nitrogen in flames. Nitrogen oxides are produced naturally by lightning, and also, to a small extent, by microbial processes in soils.

stepladder [879]3 years ago
4 0

Answer:

when fuels are burned in vehicle engines high temperatures are reached. at these high temperature nitrogen and oxygen from the air combine to produce nitrogen

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How many moles of co2 are found in a 168.2 gram sample
andrew11 [14]

Answer:

Number of moles = 3.82 mol

Explanation:

Given data:

Number of moles of CO₂ = ?

Mass of CO₂ = 168.2 g

Solution:

Number of moles = mass/molar mass

Molar mass of CO₂ = 44 g/mol

By putting values,

Number of moles = 168.2 g/ 44 g/mol

Number of moles = 3.82 mol

6 0
3 years ago
Explain the scientific word meaning of random
astraxan [27]
Odd, unusual, or unexpected.
5 0
3 years ago
9 How many grams of O₂ are needed to produce 15.5 g Fe₂O3 in the following reaction? Fe(s) + O₂(g) → Fe₂O3 (s)​
Slav-nsk [51]

Answer:

Explanation:

so u can work out the amount of moles in FeO3 by doing mr of fe3o3 is 55.8*3+16*3=215.4

moles= mass/mr so you do 15.5g/215.4=0.0719 moles

then using 1 to 1 ratio so O2 moles is 0.0719

then use the equation mass=mole*mr

so 0.0719*16=1.15g

hope this make sense :)

8 0
3 years ago
A compound is found to have 55.7% hafnium+and+44.3%+chlorine.+what+is+the+empirical+formula?
uranmaximum [27]

The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

<h3>How to calculate empirical formula?</h3>

The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.

The empirical formula of the given compound can be calculated as follows:

  • Hafnium = 55.7% = 55.7g
  • Chlorine = 44.3% = 44.3g

First, we convert mass values to moles by dividing by the molar mass of each element

  • Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
  • Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol

Next, we divide each mole value by the smallest

  • Hafnium = 0.312 ÷ 0.312 = 1
  • Chlorine = 1.25 ÷ 0.312 = 4

Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

Learn more about empirical formula at: brainly.com/question/14044066

#SPJ1

7 0
1 year ago
What two methods are used for determining the age of rocks and fossils?
Alexandra [31]

Explanation:

first one because absolute dating is a thing for rock layers

7 0
3 years ago
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