Answer:
Percentage abundance of 121 Sb is = 57.2 %
Percentage abundance of 123 Sb is = 42.8 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope, 121 Sb :
% = x %
Mass = 120.9038 u
For second isotope, 123 Sb:
% = 100 - x
Mass = 122.9042 u
Given, Average Mass = 121.7601 u
Thus,
Solving for x, we get that:
x = 57.2 %
<u>Thus, percentage abundance of 121 Sb is = 57.2 %
</u>
<u>percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %</u>
2h-7 I had that before unless I’m tripping but I did rechecked
Answer:
mass = 242.372 grams
Explanation:
1- getting the number of moles of HCl:
molarity = number of moles of solute / liters of solution
4 = number of moles of HCl / 5.2
number of moles of HCl = 4 * 5.2 = 20.8 moles
2- getting the number moles of magnesium:
From the balanced equation given, we can note that one mole of magnesium is required to react with two moles of HCl. To get the number of moles required to react with 20.8 moles of HCl. we will simply do cross multiplication as follows:
1 mole of Mg ...............> 2 moles of HCl
?? moles of Mg ...........> 20.8 moles of HCl
Number of moles of Mg = 20.8 / 2 = 10.4 moles
3- getting the mass of Mg:
number of moles = mass / molar mass
Using the periodic table, we can find that the molar mass of magnesium is 24.305 grams.
This means that:
10.4 = mass / 24.305
mass = 24.305 * 10.4
mass = 242.372 grams
Hope this helps :)
Do you know the formula? I can’t find it.
Uno y dos es pequeno 3 y 4 3s primo
