Answer:
The correct answer is Density
Explanation:
Hope this helps you
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
Answer:
1. 72.9 atm
2. 0.43937 K
Explanation:
1. Gray- lussacs law is p1/t1=p2/t2 so we use this formula to figure it out by filling in the variables and solving
p1=45.0 atm
t1=323 K
p2= ?
t2=523 K
Now we fill in this in the formula and solve - 45.0 atm/ 323 K = p2/ 523 K
and now we solve for p2 by multiplying 535k by each side to give us p2
2. Using the same formula we get 10.0atm/? = 12.0 atm/ 273.15 k and we divide both sides by 10.0 atm
Answer: m-%(Ca) = 40.08 / 110.98
Explanation: molar mass of CaCl2 is 40.08+ 2·35.45 = 110.98
Think you have one mole substance. It contains 40.08 g Ca