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blsea [12.9K]
2 years ago
14

Silver bromide (AgBr) has a solubility of 7.07 × 10^-7 mol/L

Chemistry
1 answer:
kirill115 [55]2 years ago
7 0

a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)

b.  Ksp AgBr = s²

c. 5 x 10⁻¹³  mol/L

<h3>Further explanation</h3>

Given

solubility AgBr = 7.07 x 10⁻⁷ mol/L

Required

The dissolution reaction

Ksp

The solubility product constant

Solution

a. dissolution reaction of AgBr

AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)

b. Ksp

Ksp AgBr  = [Ag⁺]  [Br⁻]

Ksp AgBr = (s) (s)

Ksp AgBr = s²

c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³  mol/L

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Answer:

=> 2.8554 g/mL

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m)     = 16.59 g

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From our question, we are to determine the density (rho) of the rock.

The formula:

p = \frac{m}{v}

Substitute the values into the formula:

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Therefore, the density (rho) of the rock is 2.8554 g/mL.

5 0
1 year ago
A compound consisting of B, N, and H undergoes elemental analysis. The % composition by mass is found to be 40.28% B, 52.20% N,
swat32

Empirical formula is the simplest ratio of components making up a compound.

The percentage composition of each element has been given

therefore the mass present of each element in 100 g of compound is

                      B                                   N                         H

mass          40.28 g                         52.20 g                 7.53 g

number of moles  

                 40.28 g / 11 g/mol      52.20 g / 14 g/mol    7.53 g / 1 g/mol

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3 years ago
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