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kogti [31]
2 years ago
13

HELPP PLEASE!!!WILL PROB GIVE BRAINLY!!! could someone please show a visual representation of 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

involved in a reaction
Chemistry
1 answer:
jeka942 years ago
3 0

Question:

2C

8

H

18

+ 25O

2

 

→

16CO

2

+ 18H

2

O

The combustion of octane, C

8

H

18

, proceeds according to the reaction above. If 346 mol of octane combusts, what volume of carbon dioxide is produced at 37.0 degrees Celsius and 0.995 atm?

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Calculate the mass of H2OH2O produced by metabolism of 2.4 kgkg of fat, assuming the fat consists entirely of tristearin (C57H11
otez555 [7]

Answer:

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

Explanation:

2C_{57}H_{110}O_6+163O_2\rightarrow 114CO_2+110H_2O

Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g

1 kg = 1000 g

Molar mass of fat = M

M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]

Moles of fat = \frac{2400 g}{890 g/mol}=2.6966 mol

According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give  ;

\frac{110}{2}\times 2.6966 mol=148.31 mol of water

Mass of 148.31 moles of water ;

148.31 mol × 18 g/mol = 2,669.58 g

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

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2 years ago
Which pair of atoms is held together by a covalent bond?
Burka [1]
Carbon and oxygen to form carbon
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2 years ago
Assume that 1.0 mol of C4H10 is completely burned in excess oxygen to form carbon dioxide and water. How many moles of CO2 would
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You need to first write a chemical equation and balance it
 C₄H₁₀ + O₂ → CO₂ + H₂O
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
1.0 moles               X moles
1.0 mol C₄H₁₀ (\frac{8 mol CO₂}{2 mol C₄H₁₀}) = 4 moles of CO₂
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How do you determine how many neutrons an element has?
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Answer:

To find the number of neutrons, subtract the number of protons from the mass number. number of neutrons

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4 0
2 years ago
A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed
AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) <em>Balanced equation including N_2 from air</em>  

The balanced equation <em>ignoring</em> N_2 from air is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2  

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2  

<em>Including</em> N_2 from air, the balanced equation is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2  

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>  

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2  

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2  

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2  

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) <em>Minimum mass of air</em>  

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2  

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2  

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2  

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air  

(d) <em>Air:fuel mass ratio for 100 % combustion</em>  

Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air  

Air:fuel = 20 760 kg/1700 kg = 12.2 :1  

6 0
2 years ago
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