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
Because the nuclear charge increases across a period and so it has a stronger pull on the outer electrons and will pull in the radius
Answer:
Explanation:
Principal quantum no "n" = 3
Azimuthal quantum no "l"= 1
Magnetic quantum no "m"= +1/2
Over all is 3pz
Answer:
No, i will not use a water pipe consisting of the two metals
Explanation:
Looking at the reduction potential of the both metals, it is clear that an electrochemical cell is set up with iron as the anode and copper as the cathode.
This will make the iron to quickly corrode and eventually destroy the water pipe. It is better to have a set up in which another metal that is higher than iron in the electrochemical series is combined with it.
