A chemical with a bitter taste is most likely a(n) _____
1 answer:
You might be interested in
Answer:
C ≈ 1.44 × 10⁻³ mol/m³
Explanation:
The given information are;
The liquid volume the pond can hold = 104 m³
The volume of inflow into the pond = 5 m³/h
The volume of outflow into the pond = 5 m³/h
The concentration of the chemical in the inflow water = 0.01 mol/m³
The concentration of the chemical discharged directly into the water = 0.1 mol/h
The concentration, , of chemical that enters the water through inflow per hour is given as follows;
= 0.01 mol/m³ × 5 m³/h = 0.05 mol/h
The concentration, , of chemical that enters the water through direct discharge per hour is given as follows;
= 0.1 mol/h
The total concentration that enters the pond per hour is given as follows;
+
= 0.1 mol/h + 0.05 mol/h = 0.15 mol/h
Whereby the water in the pond properly mixes with the pond, we have;
The concentration of chemicals (C) in the outflow water = 0.15 mol/(104 m³) ≈ 0.00144 mol/m³
C ≈ 1.44 × 10⁻³ mol/m³.
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.