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Oxides of nitrogen are pollutant gases which are emitted from car exhausts.

In urban traffic, when a car travels one kilometre, it releases 0.23 g of an oxide of nitrogen NxOy, which occupies 120 cm3.

What are the values of x and y?
(Assume 1 mol of gas molecules occupies 24.0 dm3.)

A x = 1, y = 1

B x = 1, y = 2

C x = 2, y = 1

D x = 2, y = 4

Respuesta :

Let's look at the molar weight of the answers: 
NO is 30 g/mol 
NO2 is 46 
N2O is 44 
N2O4 is 124 

We have the grams of the product, so we need the moles in order to calculate the molar weight. We us PV=nRT for this, assuming standard temperature and pressure. 
You were given the liters (.120L) 
Std pressure is 1 atmosphere 
You're looking for n, the number of moles 
Temp is 293.15 kelvin, thats standard 
And r is the gas constant in liters-atm per mol kelvin 

(.120 liters)(1atm)=n(293.15K)(.08206) 
Solving for n is .0049883835 mol 

.23g divided by .0049883 mol is about 46g/mol. You're answer is B I think, NO2

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Answer:

The correct answer is option B.

Explanation:

Mass of [tex]N_xO_y[/tex] ,m= 0.23 g

Molar mass of [tex]N_xO_y=M=14 g\mol\times x+16 g/mol\times y[/tex]

[tex]n=\frac{m}{M}[/tex]

Volume occupied by the [tex]N_xO_y=120 cm^3=0.120 dm^3[/tex]

[tex]1 cm^3 = 0.001 dm^3[/tex]

1 mol of gas molecules occupies 24.0 [tex]dm^3[/tex].

Then n moles of [tex]N_xO_y[/tex] will occupy volume of [tex]0.120 dm^3[/tex].

[tex]n\times 24.0 dm^3/mol=0.120 dm^3[/tex]

n = 0.005 moles

[tex]0.005 mol=\frac{0.23 g}{14g/mol\times x+16 g/mol\times y}[/tex]

[tex]14x+16y=46[/tex]

Putting values from option A:  x = 1, y = 1

14 × 1+16× 1 ≠ 46

Putting values from option B:  x = 1, y = 2

14 × 1+16× 2 = 46  (The correct answer)

Putting values from option C:  x = 2, y = 1

14 × 2+16× 1 ≠ 46

Putting values from option D:  x = 2, y = 4

14 × 2+16× 4 ≠ 46

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