Your lab partner turns to get a pencil, and accidentally knocks over and breaks a test tube full of sulfuric acid. What should y
2 answers:
You might be interested in
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:
where, 'x' and 'y' are the subscripts of Carbon and hydrogen respectively.
We are given:
Mass of
Mass of
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, 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, 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.
194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.
Explanation:
In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.
It is known that 1 moles of any element has 6.022×10²³ molecules.
Then 1 molecule will have moles.
So
Thus, 1.66 moles are included in BCl₃.
Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.
As it is known as 1 mole contains molecular mass of the compound.
As the molecular mass of BCl₃ will be
Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.
Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.
So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.