I believe the answer is B. Magnesium bromide , if I helped..... You're welcome
Yo sup??
Let the percentage of K-39 be x
then the percentage of K-40 is 100-(x+0.01)
We know that the net weight should be 39.5. Therefore we can say
(39*x+40*(100-(x+0.01))+38*0.01)/100=39.5
(since we are taking it in percent)
39*x+40*(100-(x+0.01))+38*0.01=3950
39x+4000-40x-0.4+0.38=3950
2x=49.98
x=24.99
=25 (approx)
Therefore K-39 is 25% in nature and K-40 is 75% in nature.
Hope this helps.
So let's convert this amount of mL to grams:
Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):
Then we need to convert moles to atoms using Avogadro's number:
![\frac{6.022*10^{23}atoms}{1mole} *[8.135*10^{-2}mol]=4.90*10^{22}atoms](https://tex.z-dn.net/?f=%20%5Cfrac%7B6.022%2A10%5E%7B23%7Datoms%7D%7B1mole%7D%20%2A%5B8.135%2A10%5E%7B-2%7Dmol%5D%3D4.90%2A10%5E%7B22%7Datoms%20)
So now we know that in 1.2 mL of liquid mercury, there are 
present.
It’s because new discoveries are made all the time sometimes what was considered right may be found out to be wrong
Molarity = moles of solute / liters of solution
M = 0.5 / 0.05
M = 10.0 mol/L⁻¹
hope this helps!
