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% .
Answer:
120 gram sample of a radioactive element, how many grams of that element will be left after 3 half-lives have passed? If you have a 300
Explanation:
Hope this helps!
Answer:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s
Explanation:
You can predict the order of orbital energies by constructing a diagram as shown below.
Follow the arrows to get the orbitals in order of increasing energy.
The order is
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s
2H₂ + O₂ = 2H₂O
n(H₂)=m(H₂)/M(H₂)
n(H₂)=5g/2.0g/mol=2.5 mol
n(O₂)=m(O₂)/M(O₂)
n(O₂)=40g/32.0g/mol=1.25 mol
H₂ : O₂ = 2 : 1
2.5 : 1.25 = 2 : 1
n(H₂O)=n(H₂)=2n(O₂)=2.5 mol
m(H₂O)=n(H₂O)M(H₂O)
m(H₂O)=2.5mol*18.0g/mol=45.0 g
Answer:
m = 0.531 molal
Explanation:
∴ m fructose = 3.35 g
∴ V water = 35.0 mL
∴ ρ H2O = 1 g/mL
- molality = moles solute / Kg solvent
∴ Mw fructose = 180.16 g/mol
⇒ moles fructose = 3.35 g * ( mol / 180.16 g) = 0.0186 mol fructose
⇒ m H2O = 35.0 mL * ( 1 g/mL ) * ( Kg/1000g) = 0.035 Kg H2O
⇒ molality (m) = 0.0186 mol fructose / 0.035 Kg H2O
⇒ m = 0.531 molal
