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3 years ago

An ionic compound is composed of the following elements:

Chemistry

2 answers:

Amiraneli [1.4K]3 years ago

6 0

Answer: N3 H12 P O3

Explanation:

From the question :

N = 31.57% H = 9.10% P = 23.27%

O= 36.06%

Divide each of the element by their respective relative atomic masses.

N = 31.57 / 14 = 2.26

H = 9.10/ 1 = 9.10

P = 23.27 / 31= 0.750

O =36.06 / 16 = 2.25

Divide each answer by the lowest of them all, we then have:

N = 2.26/ 0.750 = Approx = 3

H = 9.10 / 0.750 = Approx = 12

P = 0.750/ 0.750= 1

O = 2.25 / 0.750 = Approx = 3

The empiral formula is

N3 H12 P O3

Marysya12 [62]3 years ago

6 0

Answer:

The empirical formula for the compound is (NH₄)₃PO₃

Explanation:

Here we have

Nitrogen Molar mass = 14.01 g/mol, Percentage = 31.57%

Hydrogen Molar mass = 1.00784 g/mol, Percentage = 9.10%

Phosphorus Molar mass = 30.973762 g/mol, Percentage = 23.27%

Oxygen Molar mass = 15.999 g/mol, Percentage = 36.06%

Therefore the number of moles of each element per 100 g of the ionic compound is as follows;

Number of moles, n = Mass/(Molar Mass)

Mass of nitrogen, N in 100 g of ionic compound = 31.57 g n = 2.253 moles

Mass of Hydrogen, H in 100 g of ionic compound = 9.10 g, n = 9.029 moles

Mass of Phosphorus, P in 100 g of ionic compound = 23.27 g, n = 0.7513 moles

Mass of Oxygen, P in 100 g of ionic compound = 36.06 g, n = 2.2539 moles

Dividing by the smallest mole ratio, of 0.7513 (that of phosphorus) we have;

N = 3 moles

H = 12.0184 ≈ 12 moles

P = 1 moles

O = 3 moles

Therefore, we have the empirical formula as follows;

N₃H₁₂PO₃ or (NH₄)₃PO₃.

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Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

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3 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.