Answer:
Average atomic mass = 3.9 amu
Explanation:
Given data:
Percent abundance of He-2 = 0.420%
Percent abundance of He-3 = 2.75%
Percent abundance of He-4 = 96.83%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (0.420×2)+(2.75×3) +(96.83×4)/100
Average atomic mass = 0.84 + 8.25 +387.32 / 100
Average atomic mass = 396.41 / 100
Average atomic mass = 3.9 amu.
Answer:
The answer to your question is: letter D. 1.33 L
Explanation:
Data
V1 = 50 ml
C1 = 19.3
To solve this problem use the formula C₁V₁ = C₂V₂
C2 = C1V1 / V2
C = concentration
V = volume
a) 1.15 L
C2 = (19.3)(50) / 1150
C2 = 0.84 M
b) No right answer
c) V2= 0.80 L
C2 = (19.3)(50) / 800
C2 = 1.2 M
d) V2 = 1.33 L
C2 = (19.3)(50) / 1330
C2 = 0.72 M
e) V2 = 350 ml
C2 = (19.3)(50) / 350
C2 = 2.75 M
Answer:
Mass of Rb-87 is 86.913 amu.
Explanation:
Given data:
Average mass of rubidium = 85.4678 amu
Mass of Rb-85 = 84.9117
Ratio of 85Rb/87Rb in natural rubidium = 2.591
Mass of Rb = ?
Solution:
The ration of both isotope is 2.591 to 1. Which means that for 2.591 atoms of Rb-85 there is one Rb-87.
For 100% naturally occurring Rb = 2.591 + 1 = 3.591
% abundance of Rb-85 = 2.591/ 3.591 = 0.722
% abundance of Rb-87 = 1 - 0.722= 0.278
84.9117 × 0.722 + X × 0.278 = 85.4678
61.306 + X × 0.278 = 85.4678
X × 0.278 = 85.4678 - 61.306
X × 0.278 = 24.1618
X = 24.1618 / 0.278
X = 86.913 amu
A chromosome is a tightly coiled X of genetic material that has genes on it.
Hope this helps!
shared to form an ionic bond.
I hope this helps!
