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Margarita [4]
3 years ago
8

To test the purity of sodium bicarbonate, you dissolve a 3.50g sample in water and add sulfuric acid. if 1.04g of carbon dioxide

forms, what is the percent purity of the sodium bicarbonate?
Chemistry
1 answer:
Andrews [41]3 years ago
8 0

Answer is: the percent purity of the sodium bicarbonate is 56.83 %.

1. Chemical reaction: 2NaHCO₃ + H₂SO₄ → 2CO₂ + 2H₂O + Na₂SO₄.

2. m(NaHCO₃) = 3.50 g

n(NaHCO₃) = m(NaHCO₃) ÷ M(NaHCO₃).

n(NaHCO₃) = 3.50 g ÷ 84 g/mol.

n(NaHCO₃) = 0.042 mol.

3. From chemical reaction: n(NaHCO₃) : n(CO₂) = 1 : 1.

n(CO₂) = 0.042 mol.

m(CO₂) = 0.042 mol · 44 g/mol.

m(CO₂) = 1.83 g.

4. the percent purity = 1.04 g/1.83 g  ·100%.

the percent purity = 56.8 %.

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Isobutyl propionate is the substance that provides the flavor for rum extract. combustion of a 1.152 g sample of this carbon-hyd
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Answer;

C7H14O2

Solution;

Isobutyl contains , oxygen, carbon and hydrogen (total mass is 1.152 g)

Mass of carbon = 12/44 × 2.726 g

                           = 0.743455 g

Mass of Hydrogen  = 2/18 × 1.116 g

                            =  0.124 g

Mass of oxygen = 1.152 - (0.7435 + 0.124)

                           =  0.2845 g

Moles of carbon ;  0.7435/12 = 0.06196 moles

Moles of hydrogen; 0.124/1 = 0.124 moles

Moles of oxygen;  0.2845/16 = 0.01778 moles

Ratios ; 0.06196/0.01778 ; 0.124/0.01778 : 0.01778/0.01778

         =  3.5 :  7.0 : 1

To make them whole numbers ; we multiply the ratios by 2  to get;

(3.5 :  7.0 : 1 )2 = 7 : 14 : 2

Thus, the empirical formula of Isobutyl propionate  is C7H14O2


4 0
3 years ago
Convert 1.25 atm to bar.
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Answer: 1.27 bar

Explanation:

1 atm = 1.01325 bar

1.25 atm = Z (let Z be the unknown value)

To get the value of Z, cross multiply

Z x 1 atm = 1.25 atm x 1.01325 bar

1 atm•Z = 1.2665625 atm•bar

To get the value of Z, divide both sides by 1 atm

1 atm•Z/1 atm = 1.2665625 atm•bar/1atm

Z = 1.2665625 bar

(Round up Z to the nearest hundredth as 1.27 bar)

Thus, 1.25 atm when coverted gives 1.27 bar

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Explanation:

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