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yulyashka [42]
3 years ago
10

The products of a combustion reaction do NOT include ____. (1 point) water carbon dioxide carbon monoxide hydrogen The products

of a combustion reaction do NOT include Hydrogen.
Chemistry
2 answers:
nalin [4]3 years ago
7 0

the answer is water

Hope this help1

Allushta [10]3 years ago
6 0

Answer: Option (d) is the correct answer.

Explanation:

In combustion reaction, a compound reacts with oxygen and results in the formation of carbondioxide and water.

For example, when propane reacts with oxygen it results in the formation of carbondioxide and water.

The reaction is as follows.

      C_{3}H_{8} + O_{2} \rightarrow CO_{2} + H_{2}O

Also, when there is incomplete combustion of carbon then it results in the formation of carbon monoxide. This reaction is as follows.

         C + \frac{1}{2}O_{2} \rightarrow CO

Thus, we can conclude that the products of a combustion reaction do NOT include hydrogen.

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Lady bird [3.3K]

Answer:

4 Ethanol has a boiling point of 89 C at standard pressure.

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3 years ago
Before a bond breaks in a chemical reaction, what happens
Assoli18 [71]
They have to form a chemical bond in order to brake them down first
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3 years ago
A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its
Over [174]
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0
3 years ago
The missing components in the table to the right
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4 0
3 years ago
"calculate the ratio of the velocity of hydrogen molecules to the velocity of carbon dioxide molecules at the same temperature"
Ne4ueva [31]

Answer: 1:4.69

Explanation:

The ratio can be expressed as:

Ua/Ub= √(Mb/Ma)

Where Ua/Ub is the ratio of velocity of hydrogen to carbon dioxide and Ma is the molecular mass of hydrogen gas= 2

Mb is the molecular mass of CO2 = 44

Therefore

Ua/Ub= √(44/2)

Ua/Ub = 4.69

Therefore the ratio of velocity of hydrogen gas to carbon dioxide = 1:4.69

which implies hydogen is about 4.69 times faster than carbon dioxide.

8 0
3 years ago
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