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Marat540 [252]
3 years ago
9

The equations given in the problem introduction can be added together to give the following reaction: overall: N2O4→2NO2 However

, one of them must be reversed. Which one?
Chemistry
1 answer:
Murljashka [212]3 years ago
4 0

Answer:

Reverse the 2 NO_2 \longrightarrow 2 NO + O_2  reaction

Explanation:

Reactions:

2 NO_2 \longrightarrow 2 NO + O_2

N_2O_4 \longrightarrow 2 NO + O_2

Overall:

N_2O_4 \longrightarrow 2 NO_2

As can be seen, in the overall reaction we have N_2O_4 in the reactants like in the second reaction and NO_2 in the products. The NO_2 is in the first reaction but as a reactant so we need to reverse that reaction:

2 NO + O_2 \longrightarrow 2 NO_2

N_2O_4 \longrightarrow 2 NO + O_2

Combining:

N_2O_4 + 2 NO + O_2\longrightarrow 2 NO + O_2 + 2 NO_2

N_2O_4 \longrightarrow 2 NO_2

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D. Acids I hope this helped ur welcome
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A gas diffuses 1/6 times faster than hydrogen gas (H2). What is the molar mass of the gas? 59.95 g/mol 66.54 g/mol 68.68 g/mol 7
taurus [48]

Solution us here,

let the rate of diffusion of H2(r1) be x and⁷ the rate of diffusion of gas(r2) be x+x/6=7x/6.

Molar mass of H2 (M1)=2 g

Molar mass of gas(M2)= ?

Now, from the Ghram's law of diffusion of gas,

\frac{r1}{r2}  =   \sqrt{ \frac{m2}{m1}  }   \\ or \:  \frac{x}{ \frac{7x}{6} }  =  \sqrt{ \frac{m2}{2} }  \\ or \:  \frac{6}{7}  =  \sqrt{ \frac{m2}{2} }  \\ squaring \: in \: both \: sides \\  \frac{36}{49}  = \frac{m2}{2}  \\ or m2 = 1 .469

here, it looks like I have done wrong.

But all of the answers in option are wrong.

Because, rate of diffusion of any gas is inversely proportional to the square root of its malor mass.

And in the question, gas has higher rate of diffusion than hrdrogen. So it should have molar mass less than hydrogen.

8 0
4 years ago
Read 2 more answers
How to solve #2, not sure if my work is correct
erica [24]
Go through your work carefully i see a mistake in number 2
5 0
3 years ago
No spam or links! Thanks!
Monica [59]

Answer:

364 K or 91°C

Explanation:

Applying,

V₁/T₁ = V₂/T₂................ Equation 1

Where V₁ = Initial Volume, V₂ = Final volume, T₁ = initial Temperature, T₂ = final Temperature.

make T₂ the subject of the equation,

T₂ = V₂T₁/V₁................. Equation 2

From the question,

Given: V₁ = 375 mL, V₂ = 500 mL, T₁ = 0.0°C = (273+0) K = 273 K

Substitute these values into equation 2

T₂ = (500×273)/375

T₂ = 364 K

T₂ = (364-273) °C = 91 °C

5 0
3 years ago
The student adds 6 m nh3 to be cuso4 solution until the cu2+ ion is essentially all converted to cu(nh3)42+ ion. the voltage of
Gnoma [55]

Here we have to get the residual concentration of Cu²⁺ in the reaction.

The residual concentration of  [Cu²⁺] is 2.137×10⁸ m

The reaction between ammonia (NH₃) and CuSO₄ in ionic form is - Cu²⁺ + 4 NH₃ = [Cu(NH₃)₄]²⁺.

Thus 1 mole of Cu²⁺ will react with 4 moles of NH₃.

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Thus the Nernst equation of the reaction is E_{Cell} = E° + \frac{0.059}{n}log\frac{Cu^{2+} }{Cu}.

Here the number of moles is 3, as 6 moles of ammonia will react with 1.5 moles of cu²⁺. and the standard reduction potential (E°) of Cu²⁺/Cu system is (+) 0.674V and the cell potential is 0.92V

On plugging the values in the equation, we get,

0.92 = 0.674 + \frac{0.059}{2}log[Cu²⁺]

Or, 0.029 log [Cu²⁺]= 0.246

Or, log [Cu²⁺] = 8.33

Or, [Cu²⁺] = 2.137×10⁸

3 0
3 years ago
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