To calculate the mean, you add up all of the data values, and then divide that sum by the *number* of values.
For instance, if you wanted to find the mean score at a home run derby, and you’re given the following numbers for home runs scored by each player:
5, 4, 6, 5, 3, 1
You could calculate the mean by adding all of the score up
5 + 4 + 6 + 5 + 3 + 1 = 24
And dividing by the number of hitters (in this case, 6)
24 / 6 = 4
So the *mean score* of the home run derby would be 4.
Answer:
CH4
Explanation:
The number of moles of carbon and hydrogen has been given as follows:
C = 0.300 mol
H = 1.20 mol
Next, we divide each mole value by the smallest (0.300)
C = 0.300 ÷ 0.300 = 1
H = 1.20 ÷ 0.300 = 4
The empirical ratio of Carbon and Hydrogen is 1:4, hence, the empirical formula is CH4
It is often a sign of kidney problems. :)
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