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icang [17]
3 years ago
10

In an experiment, a student gently heated a hydrated copper compound to remove the water of hydration. the following data was re

corded:
1. mass of the crucible, cover and contents before heating 23.4g.

2. mass of empty crucible and cover.

3. mass of crucible, cover, and contents after heating to constant mass 20.94g
Chemistry
1 answer:
NeX [460]3 years ago
3 0
The unknown of this problem is the experimental percent of water in the compound in order to remove the water of hydrogen, given the following:

Mass of crucible, cover and contents before heating                               23.54 g
Mass of empty crucible and cover                                                            18.82 g
Mass of crucible, cover, and contents after heating to constant mass     20.94 g

In order to get the answer, determine the following:

Mass of hydrated salt used                          = 23.54 g – 18.82 g = 4.72 g
Mass of dehydrated salt after heating          = 20.94 g – 18.82 g = 2.12 g
Mass of water liberated from salt                 = 4.72 g – 2.12 g = 2.60 g

Then solve the percent of water in the hydrated salt by:

% water = (mass of water / mass of hydrated salt) x 100
% water = 2.60 g / 4.72 g x 100
% water = 55.08 % in the compound 
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The mass of Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g

<h3>What mass of Ba(IO3)2 can be dissolved in 500 ml of water at 25 degrees celcius?</h3>

The Ksp of Ba(IO3)2 = 1.57 × 10^-9

Molar mass of Ba(IO3)2 = 487 g/mol?

Dissociation of Ba(IO3)2 produces 3 moles of ions as follows:

Ba(IO_{3})_{2} \leftrightharpoons Ba^{2+} + 2\:IO_{3}^{-}

Ksp = [Ba^{2+}]*[IO_{3}^{-}]^{2}

[Ba(IO_{3})_{2}] =  \sqrt[3]{ksp} =\sqrt[3]{1.57 \times  {10}^{ - 9} } \\  [Ba(IO_{3})_{2}] = 1.16 \times  {10}^{-3} moldm^{-3}

moles of Ba(IO3)2 = 1.16 × 10^-3 × 0.5 = 0.58 × 10^-3 moles

mass of Ba(IO3)2 = 0.58 × 10^-3 moles × 487 = 2.82 g

Therefore, mass Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g.

Learn more about mass and moles at: brainly.com/question/15374113

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Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

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Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

M=d\times V=1.07 g/mL\times 1000 mL=1,070 g

1) The value of %(m/M):

\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%

2) The value of %(m/V):

\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%

Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}

Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}

n = Equivalent mass

n = \frac{\text{molar mass of ion}}{\text{charge on an ion}}

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Moles of calcium ion = 1 mol (1 CaCl_2 mole has 1 mole of calcium ion)

n=\frac{40 g/mol}{2}=20

=\frac{1 mol}{20 g/mol\times 1L}=0.050 N

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 CaCl_2 mole has 2 mole of chlorine ion)

n=\frac{35.5 g/mol}{1}=35.5

=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N

Moles of calcium chloride = 1.00 mol

Mass of solvent =  Mass of solution - mass of solute

= 1,070 g - 111 g = 959  g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

\frac{1 mol}{0.959 kg}=1.043 mol/kg

Moles of calcium chloride = n_1=1mol

Mass of solvent = 959 g

Moles of water = n_2=\frac{959 g}{18 g/mol}=53.28 mol

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842

7) Mole fraction of water =

\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

m'=d\times v=1.07 g/ml\times 100 g= 107 g

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

11.1\%=\frac{m}{100 mL}\times 100

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

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