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Rasek [7]
3 years ago
5

A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A

gNO3. If the mass of the AgBr precipitate that forms is 0.7127 g, what is the percent by mass of Br in the original compound?
Chemistry
1 answer:
Pavel [41]3 years ago
4 0

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

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salantis [7]

Answer:

n=17.85 moles

Explanation:

Given mass is, m = 1.5 kg = 1500 g

The molar mass of sodium bicarbonate is, M = 84.007 g/mol

We need to find the no of moles in 1.5 kg of Sodium bicarbonate . We know that, no of moles is equal to given mass divided by molar mass.

n=\dfrac{m}{M}\\\\n=\dfrac{1500}{84.007 }\\\\n=17.85

<h2>So, there are 17.85 moles in 1.5 kg of Sodium bicarbonate.</h2>
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Explanation:

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6 0
3 years ago
What is the molarity when 0.181 moles of MgNO3)2 are dissolved in enough water to make 0.750 L solution?
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Answer:

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Explanation:

Given data:

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Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

by putting values,

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Hope this helps!

Explanation:

5 0
2 years ago
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