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
The equation is already balanced
2AgNO3 + MgCl2 => 2AgCl + Mg(NO3)2
So the ideal gas law is pv=nrt
The letter n stands for the number of moles. divide both side by rt to isolate.
pv/rt=n
Answer:
Explanation:
soluble and insoluble salts are prepared by processes like neutralization reactions, simple displacement reactions and double displacement reactions.
A hydrate is a chemical that has water molecules loosely bonded to it. The water molecules are not ... You will be using the hydrate CuSO4 . ?H2O. Sample Calculation-. An empty crucible has a mass of 12.770 grams.