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sergiy2304 [10]
4 years ago
12

Write a balanced equation for the complete oxidation reaction that occurs when ethane burns in air

Chemistry
1 answer:
damaskus [11]4 years ago
6 0
<h3>Balanced equation : 2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)</h3><h3>Further explanation</h3>

Alkanes are saturated hydrocarbons that have single bonds in chains

General formula for alkanes :

\tt \large{\bold{C_nH_{2n+2}}

Hydrocarbon combustion reactions (specifically alkanes)

\large {\box {\bold{C_nH _ (_2_n _ + _ 2_) + \dfrac {3n + 1} {2} O_2 \Rightarrow nCO_2 + (n + 1) H_2O}}}

So that the burning of ethane with air (oxygen):

\tt C_2H_6+\dfrac{7}{2}O_2\rightarrow 2CO_2+3H_2O

2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)

or we can use mathematical equations to solve equilibrium chemical equations by giving the coefficients for each compound involved in the reaction

C₂H₆ (g) + aO₂ (g) ⟶ bCO₂ (g) + cH₂O (ℓ)

C : left 2, right b ⇒ b=2

H: left 6, right 2c⇒ 2c=6⇒ c= 3

O : left 2a, right 2b+c⇒ 2a=2b+c⇒2a=2.2+3⇒2a=7⇒a=7/2

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Answer:

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Explanation:

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7 0
3 years ago
An atom has radius of 227 pm and crystallizes in a body-centered cubic unit cell. What is the volume of the unit cell
sdas [7]

We will see that the volume of the unit cell is 144,070,699.06 pm^3

<h3>How to get the volume of a body-centered cubic unit cell?</h3>

In a body-centered cubic unit cell, the side length of the cube is given as:

S = \frac{4}{\sqrt{3} } *R

Where R is the radius of the atom.

And the volume of a cube is the side length cubed, then we can see that the volume of our cube will be:

V = S^3 = (\frac{4}{\sqrt{3} }*227pm)^3

Solving that we get:

V  = (\frac{4}{\sqrt{3} }*227pm)^3 = 144,070,699.06pm^3

This is the approximated volume of the unit cell.

If you want to learn more about unit cell structures, you can read:

brainly.com/question/13110055

5 0
2 years ago
8. What is the volume of 5.6 moles of carbon dioxide gas at STP?
pshichka [43]

Answer:

125.5 ×10^-3 m^3= 0.1255 m^3

Explanation:

Volume=5.6mol×22.414dm^3

=125.5dm^3

7 0
3 years ago
In this experiment, 0.070 g of caffeine is dissolved in 4.0 ml of water. The caffeine is then extracted from the aqueous solutio
Damm [24]

2.0ml of methylene chloride solution is used each time to extract caffeine from the aqueous solution.  

Consider the concentration of caffeine obtained during each individual extraction from the aqueous solution to be C.  

The total amount of caffeine obtained during each extraction is calculated as

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution) + (Volume of methylene chloride added during each extraction x distribution coefficient of caffeine x concentration of caffeine obtained during each individual extraction from the aqueous solution)  


Substituting these values we get                                                            

The total amount of caffeine obtained during each extraction                

 = (4.0×C )+ (2.0×4.6 × C)                                                                              

= 13.2 C


The amount of caffeine remaining in the aqueous solution is calculated as  

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution)


Substituting these values we get                                                            

The amount of caffeine remaining in the aqueous solution = 4 × C                                                                                            

The fraction of caffeine remaining in aqueous solution is calculated as  

= (The total amount of caffeine obtained during each extraction)/ (The amount of caffeine remaining in the aqueous solution)                    

=4.0 C/13.2 C                                                                                                

= 1/3.3.  

Therefore the fraction of caffeine left in aqueous solution after 3 extractions is =(1/3.3)^3  =0.028

Therefore, the total amount of caffeine extracted                            

=0.070 × (1-(1/3.3)^3)                                                                                      

= 0.068 g


5 0
3 years ago
Read 2 more answers
What mass of water in grams will fill a tank 126cm long, 47 cm wide, and 13 cm high
Shtirlitz [24]

Answer:

Mass =  76986  g

Explanation:

Given data:

Dimensions of tank = 126 cm× 47 cm× 13 cm

Mass of water required to filled = ?

Solution:

First of all we will calculate the volume of tank which is equal to the volume of water required to fill it.

Volume = length ×height ×width

Volume = 126 cm × 13 cm× 47 cm

Volume = 76986 cm³

Mass of water:

Mass = density × volume

density of water is 1 g/cm³

Mass = 1 g/cm³× 76986 cm³

Mass =  76986  g

8 0
3 years ago
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