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2 years ago

14

You have an object with a mass of 14 g and a volume of 20 mL. Calculate the density and identify which material you have from th

e list.

1 answer:

arsen [322]2 years ago

8 0

Answer:

0.7 g/ml. Whichever material matches that closest is the right one.

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3 years ago

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4 years ago

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Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 l of carbon dioxide at 20.0 Â°c an

kvv77 [185]

Answer:

Percentage by mass of calcium carbonate in the sample is 93.58%.

Explanation:

Assumptions:

calcium carbonate was dissolved completely and the amount of carbon dioxide released was proportional to the amount of calcium carbonate.
the sample did not contain any other compound that released or reacted with carbon dioxide.

$PV = nRT$

$n = \frac{PV}{RT}$

$P = 791 mmHg = \frac{791 mmHg}{760 mmHg} x 1 atm = 1.040789 atm.$

$R = 8.314 \frac{J}{K*mol} = 0.082 \frac{L*atm}{K*mol}$

$T = 20^{0}C + 273^{0} C = 293 K$

$n = \frac{1.040789 atm * 1.14 L}{0.082 \frac{L * atm}{K * mol} * 293 K }$

$n = 0.0493839 mol.$

given the equation of the reaction:

$CaCO_{3} + 2HCL = CaCl_{2} + CO_{2} + H_{2}O$

from assumption:

amount carbon dioxide = amount of calcium carbonate

$n(CaCO_{3} ) = n(CO_{2} )$

reacting mass ( m ) = Molar Mass ( M ) * Amount ( n )

$m(CaCO_{3} ) = n(CaCO_{3} ) * M(CaCO_{3} )$

m = 0.04938397 $mol$ * 100 $\frac{g}{mol}$ = 4.93839 $g$

percentage by mass of CaC $O_{3}$ = $\frac{mass of pure }{mass of impure}*100 = \frac{4.93839g}{5.28g}*100 = 93.5802%$ .

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3 years ago

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3 years ago

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