Answer:
6.31g/mol
Explanation:
Using the ideal gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
Mole (n) = mass (m)/molar mass (Mm)
* Mm = m/n
Also, density (p) = mass (m) ÷ volume (V)
PV = nRT
Since n = M/Mm
PV = M/Mm. RT
PV × Mm = m × RT
Divide both sides by V
P × Mm = m/V × RT
Since p = m/V
P × Mm = p × RT
Mm = p × RT/P
Mm = 0.249 × 0.0821 × 293/0.95
Mm = 5.989 ÷ 0.95
Mm = 6.31g/mol
Chemical properties include reactivity with other elements of substances, toxicity, flammability, and chemical stability
just to name a few
Answer:- The natural abundance of
is 0.478 or 47.8% and
is 0.522 or 52.2% .
Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:
Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)
We have been given with atomic masses for
and
as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.
Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of
as n then the abundance of
would be 1-n .
Let's plug in the values in the formula:

151.964=150.919860n+152.921243-152.921243n
on keeping similar terms on same side:


negative sign is on both sides so it is canceled:



The abundance of
is 0.478 which is 47.8%.
The abundance of
is = 
= 0.522 which is 52.2%
Hence, the natural abundance of
is 0.478 or 47.8% and
is 0.522 or 52.2% .