Answer:
= 74.4 grams / mole. Ernest Z. The reaction will produce 15.3 g of KCl
Explanation:
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
So we end up with a total of four oxygen atoms for this calcium acetate unit and guys that truly it for this one.
The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.
Isotope mass amu Relative abundance
1 77.9 14.4
2 81.9 14.3
3 85.9 71.3
Express your answer to three significant figures and include the appropriate units.
Answer: 84.2 amu
Explanation:
Mass of isotope 1 = 77.9
% abundance of isotope 1 = 14.4% = 
Mass of isotope 2 = 81.9
% abundance of isotope 2 = 14.3% = 
Mass of isotope 3 = 85.9
% abundance of isotope 2 = 71.3% = 
Formula used for average atomic mass of an element :

![A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2877.9%5Ctimes%200.144%29%2B%2881.9%5Ctimes%200.143%29%2B%2885.9%5Ctimes%200.713%29%5D)

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu