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:
Pressure = 4313.43mmHg
Explanation:
P1 = ?
V1 = 0.335L
V2 = 1700mL =1700*10^-3L = 1.7L
P2 = 850mmhg
From Boyle's law, the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature remains constant.
P = k / v
K = pv. P1V1 = P2V2 = P3V3 =........=PnVn
P1V1 = P2V2
Solve for P1,
P1 = (P2*V2) / V1
P1 = (850 * 1.7) / 0.335
P1 = 4313.43mmHg
The pressure of the gas was 4313.43mmHg
Answer: The answer is D :)
Explanation:
Answer:
7.335 moles of Cl₂ are required to react with 4.89 miles of Al.
Explanation:
Given data:
Moles of Al = 4.89 mol
Number of moles of Cl₂ required = ?
Solution:
Chemical equation:
2Al + 3Cl₂ → 2AlCl₃
Now we will compare the moles of Al and chlorine from balance chemical equation.
Al : Cl₂
2 : 3
4.89 : 3/2×4.89 =7.335 mol
Thus, 7.335 moles of Cl₂ are required to react with 4.89 miles of Al.
