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slava [35]
4 years ago
9

How many moles of Al(CN)3 are in 183 g of the compound?

Chemistry
1 answer:
Lyrx [107]4 years ago
4 0
You need to calculate the molar mass for Al(CN)3 using the atomic weights for Al, C, and N given on the periodic table.

1 Al (26.98) + 3 C (3 x 12.01) + 3 N (3 x 14.01) = 105.04 g Al(CN)3 / mole

183 g Al(CN)3 x (1 mole Al(CN)3 / 105.04 g Al(CN)3) = 1.74 moles Al(CN)3
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Answer:

1.2

Explanation:

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to find the moles, you must multiply both sides by volume

so, take 1.5 L * 0.80 M

this equals 1.2 :)

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Keq for the reaction below is 2400. If the initial conditions of the reaction are a 1.0 L flask that contains 0.024 mol NO (g),
podryga [215]

Answer:

The answer to your question is it is not at equilibrium, it will move to the products.

Explanation:

Data

Keq = 2400

Volume = 1 L

moles of NO = 0.024

moles of N₂ = 2

moles of O₂ = 2.6

Process

1.- Determine the concentration of reactants and products

[NO] = 0.024 / 1 = 0.024

[N₂] = 2/1 = 2

[O₂] = 2.6/ 1= 2.6

2.- Balanced chemical reaction

                     N₂ + O₂    ⇒   2NO

3.- Write the equation for the equilibrium of this reaction

                     Keq = [NO]²/[N₂][O₂]

- Substitution

                    Keq = [0.024]² / [2][2.6]

-Simplification

                    Keq = 0.000576 / 5.2

-Result

                    Keq = 1.11 x 10⁻⁴

Conclusion

It is not at equilibrium, it will move to the products because the experimental Keq was lower than the Keq theoretical-

                         1.11 x 10⁻⁴ < 2400

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