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Evgen [1.6K]
3 years ago
14

A sample of 1.000 g of a compound containing carbon and hydrogen reacts with oxygen at elevated temperature to yield 0.692 g H₂O

and 3.381 g CO₂. (a) Calculate the masses of C and H in the sample. (b) Does the compound contain any other elements
Chemistry
1 answer:
ollegr [7]3 years ago
3 0

Answer :

(a) 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) There is no other element present in the compound.

Explanation :

(a) Now we have to determine the masses of C and H in the sample.

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Mass of CO_2=3.381g

Mass of H_2O=0.692g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.381 g of carbon dioxide, \frac{12}{44}\times 3.381=0.922g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.692 g of water, \frac{2}{18}\times 0.692=0.0769g of hydrogen will be contained.

Thus, 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) Now we have to determine the compound contain any other elements or not.

Mass carbon + Mass of hydrogen = 0.922 g + 0.0769 g = 0.999 g ≈ 1 g

This means that there is no other element present in the compound.

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A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate
inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams  and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0
3 years ago
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