How many atoms are in 1.75 mol of ecules oh NH^3

1 answer:

5 0

1 mole ----------- 6.02x10²³ atoms
1.75 moles ------- ?

atoms = 1.75 * 6.02x10²³ / 1

= 1.053x10²⁴ atoms

hope this helps!

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g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$