Answer:
The three isomers having the molecular formula 
are drawn in the figure below.
Explanation:
Answer:
The percent isotopic abundance of C- 12 is 98.93 %
The percent isotopic abundance of C- 13 is 1.07 %
Explanation:
we know there are two naturally occurring isotopes of carbon, C-12 (12u) and C-13 (13.003355)
First of all we will set the fraction for both isotopes
X for the isotopes having mass 13.003355
1-x for isotopes having mass 12
The average atomic mass of carbon is 12.0107
we will use the following equation,
13.003355x + 12 (1-x) = 12.0107
13.003355x + 12 - 12x = 12.0107
13.003355x- 12x = 12.0107 -12
1.003355x = 0.0107
x= 0.0107/1.003355
x= 0.0107
0.0107 × 100 = 1.07 %
1.07 % is abundance of C-13 because we solve the fraction x.
now we will calculate the abundance of C-12.
(1-x)
1-0.0107 =0.9893
0.9893 × 100= 98.93 %
98.93 % for C-12.
Answer:
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
Explanation:
Data Given
M1 = 6.00 M
M2 = 2.5 M
V1 = 250 mL
V2 = ?
Solution:
As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.
Now
first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution
For this Purpose we use the following formula
M1V1=M2V2
Put values from given data in the formula
6 x V1 = 2.5 x 250
Rearrange the equation
V1 = 2.5 x 250 /6
V1 = 104 mL
So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M aqueous solution of ammonium hydroxide
But we have to prepare 250 mL of the solution.
so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.
in this question you have to tell about the amount of water that is 146 mL
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
Answer:
1,500 mm
Explanation:
if 1 meter = 1000 mm, 0.5 meters is 500 mm, so 1.50 meters is 1,500 mm
Answer:
A. 85.6 g
= 0.0856 kg.
B. 0.00027 mol/g
= 0.27 mol/kg.
C. 8.39 %
Explanation:
Given:
Molar concentration = 0.25 M
Molar weight of sucrose = 342.296 g/mol
Density of solution = 1.02 g/mL
Mass of water = 934.4 g.
Density in g/l = 1.020 g/ml * 1000ml/1 l
= 1020 g/l
Mass of solution in 1 l of solution = 1020 g
Mass of solution = mass of solvent + mass of solute
Mass of sucrose = 1020 - 934.4
= 85.6 g of sucrose in 1 l of solution.
A.
Density of sucrose = mass/volume
= molar mass/molar concentration
= 342.296 * 0.25
= 85.6 g/l
Number of moles = mass/molar mass
= 85.6/342.296
= 0.25 mol
B.
Molality = number of moles of solute/mass of solvent
= 0.25/934.4
= 0.00027 mol/g
C.
% mass of sucrose = mass of sucrose/total mass of solution * 100
= 85.6/1020 * 100
= 8.39 %
