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Xelga [282]
3 years ago
14

How many grams of oxygen are required to react with 17.0 grams of octane (c8h18) in the combustion of octane in gasoline?

Chemistry
1 answer:
Keith_Richards [23]3 years ago
8 0
The balanced equation for the combustion of octane is as follows
2C₈H₁₈ + 25O₂ ---> 16CO₂  + 18H₂O
stoichiometry of C₈H₁₈  to O₂ is 2:25
number of octane moles reacted - 17.0 g / 114.2 g/mol = 0.149 mol 
according to molar ratio 
if 2 mol of octane reacts with 25 mol of O₂
then 0.149 mol of octane reacts with - 25 /2 x 0.149 mol = 1.86 mol of O₂
 mass of O₂  - 1.86 mol x 32 g/mol = 59.5 g
59.5 g of O₂ is required to react with 

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determine the frequency and wavelength (in nm) of the light emitted when the e- fell from n=4 and n=2
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Frequency = 6.16 ×10¹⁴ Hz

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Explanation:

In case of hydrogen atom energy associated with nth state is,

En =  -13.6/n²

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E₂ = -3.4 ev

Kinetic energy of electron = -E₂ = 3.4 ev

For n = 4

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Kinetic energy of electron = -E₄ = 0.85 ev

Wavelength of radiation emitted:

E = hc/λ = E₄ - E₂

hc/λ = E₄ - E₂

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev   - (-3.4ev )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J

λ = 4.87×10⁻⁷ m

m to nm:

4.87×10⁻⁷ m ×10⁹nm/1 m

4.87×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /4.87×10⁻⁷ m

Frequency = 6.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 6.16 ×10¹⁴ Hz

3 0
3 years ago
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