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
all cells come from pre existing cells
Answer:
The answer to your question is n = 5, l = 2, m can be -2, -1, 0, 1 or 2
Explanation:
Data
orbital = 5d
values of n, l, m
Process
1.- Determine the value of n
n is the coefficient of the orbital, in this problem n = 5
2.- Determine the value of l
l takes values depending in the sublevel of energy,
if the sublevel is s then l = 0
p l = 1
d l = 2
f l = 3
For this problem l = 2
3.- Determine the value of m
when l = 2, m takes values of -2, - 1, 0, 1 or 2
Answer:
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