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3 years ago

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Una muestra de 258.4 g de etanol (C2H5OH) se quemó en una bomba calorimétrica

1 answer:

Ilia_Sergeevich [38]3 years ago

4 0

Answer:

Explanation:

Unclear question.

I infer you want a clear rendering, which reads;

A 258.4 g sample of ethanol (C2H5OH) was burned in a calorimetric pump using a Dewar glass. As a consequence, the water temperature rose to 4.20 ° C.

If the heat capacity of the water and the surrounding glass was 10.4 kJ / ° C, calculate the heat of combustion of one mole of ethanol.

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3 years ago

Calculate the concentration of A bottle of wine contains 12.9% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm e

Delicious77 [7]

Answer:

The mass percentage of the solution is 10.46%.

The molality of the solution is 2.5403 mol/kg.

Explanation:

A bottle of wine contains 12.9% ethanol by volume.

This means that in 100 mL of solution 12.9 L of alcohol is present.

Volume of alcohol = v = 12.9 L

Mass of the ethanol = m

Density of the ethanol ,d= $0.789 g/cm^3=0.789 g/mL$

$1 cm^3=1 mL$

$m=d\times v=0.798 g/ml\times 12.9 mL = 10.1781 g$

Mass of water = M

Volume of water ,V= 100 mL - 12.9 mL = 87.1 mL

Density of water = D=1.00 g/mL

$M=D\times V=1.00 g/ml\times 87.1 mL =87.1 g$

Mass percent

$(w/w)\%=\frac{m}{m+M}\times 100$

$\frac{10.1781 g}{10.1781 g+87.1 g}\times 100=10.46\%$

Molality :

$m=\frac{m}{\text{molar mass of ethanol}\times M(kg)}$

M = 87.1 g = 0.0871 kg (1 kg =1000 g)

$=\frac{10.1781 g}{46 g/mol\times 0.0871 kg}$

$m=2.5403 mol/kg$

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3 years ago