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Flura [38]
3 years ago
14

A compound contains 0.5 mol Na, 0.5 mol N, and 10 mol H. The empirical formula of the

Chemistry
1 answer:
lesya [120]3 years ago
8 0

Answer:

NaNH₂₀

Explanation:

0.5 mol Na, 0.5 mol N, and 10 mol H

To obtain the empirical formulae, we find the mole ratio between the elements and this is done by dividing all through by the smallest mol (0.5)

Na = 0.5 / 0.5 = 1

N = 0.5 / 0.5 = 1

H = 10 / 0.5 = 20

The mole ratio is used to write the empirical formulae. It is given as;

NaNH₂₀

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In order for you to calculate for the mass of ammonium carbonate, you need to know the molar mass of it and the nitrogen atoms in the compound. Ammonium carbonate has a molar mass of 96.08 grams per mole. There are two nitrogen atoms in ammonium carbonate which is equal to 28.02 grams per mole. Divide the molar mass of nitrogen to the ammonium carbonate, 28.02/96.08 x 100, we get 29.16wt% nitrogen.

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6. Balance the following equations (3 points)
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The position of an element in the activity series determines the strength as a reducing agent or not.

Elements higher in the series are good reducing agents, those lower in the series are poor reducing agents.

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At equilibrium, a 1.0 L reaction vessel contains 2.3 mol of Mg(OH)₂ and 0.170 mol of OH⁻. What is the equilibrium concentration
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The value of X for this chemical reaction is 1.65 x 10⁻⁴

<u>Explanation:</u>

Mg (OH)₂(s) <==> Mg²⁺+(aq) + 2OH₋₁(aq)  

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KC = [Mg²⁺][OH-]²

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On simplifying the equation we get,

X = 1.65 x 10⁻⁴

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