Ask question

Ask question

All categories

3 years ago

8

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.

You might be interested in

To the nearest 0.5 mL, what is<br> the water volume in the<br> graduated cylinder?<br> mL?

AnnZ [28]

Answer:

14cm^3

Explanation:

6 0

3 years ago

Ghella [55]

Answer:

B sediment

Explanation:

The excess mileage leads to additional pollution. Car exhaust contains toxic gases, and car fluids often leak (oil drips, antifreeze leaks, etc...). The exhaust also contains nitrogen oxide, which can lead to nutrient-related problems in the environment.

3 0

3 years ago

Read 2 more answers

What is the number of moles of 27.6 grams of lithium

Leya [2.2K]

Lithium's atomic weight is 6.94 g/mol.

$27.6 g Li (\frac{mol}{6.94 g}) = 3.98 mol Li$

Need anything else?

6 0

3 years ago

If 24k gold is the highest purity (100%), then a 6 Karat piece of gold jewelry would contain what percentage of gold?

krok68 [10]

Answer:

25% gold.

Explanation:

6 is 25 percent of 24, so 6 karat gold would be 25% gold.

4 0

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)

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

$m=2.5403 mol/kg$

4 0

3 years ago