Answer: 3) 39.96 amu
Explanation:
Mass of isotope Ar- 36 = 35.97 amu
% abundance of isotope Ar- 36= 0.337% = 
Mass of isotope Ar- 38 = 37.96 amu
% abundance of isotope 2 = 0.063 % = 
Mass of isotope Ar- 40 = 39.96 amu
% abundance of isotope 2 = 99.600 % = 
Formula used for average atomic mass of an element :
![A=\sum[(35.97\times 3.37\times 10^{-3})+(37.96\times 6.3\times 10^{-4})+(39.96\times 0.996)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2835.97%5Ctimes%203.37%5Ctimes%2010%5E%7B-3%7D%29%2B%2837.96%5Ctimes%206.3%5Ctimes%2010%5E%7B-4%7D%29%2B%2839.96%5Ctimes%200.996%29%5D)
Therefore, the average atomic mass of argon is 39.96 amu
Answer:
3). 1.30 × 10^(24) molecules
Explanation:
From avogadro's law which state that equal volume of all gases at the same temperature and pressure contain the same number of molecules.
We can relate it to this question as;
V₁/n₁ = V₂/n₂
Where;
V₁ is initial volume
n₁ is initial number of molecules
V₂ is final volume
n₂ is final number of molecules
Thus at STP, we have V₁ = V₂ and as such Plugging in the relevant values gives;
5/(1.30 x 10^(24)) = 5/n₂
n₂ = 1.30 x 10^(24) molecules
They both hurt and they both burn lol
they will each create a small burn on your skin if they touch you sometimes visible sometimes not. but both should burn about the same just one is water one is fire
Answer: The volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L
Explanation:
According to ideal gas equation:
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = ?
n = number of moles = 0.684
R = gas constant = 
T =temperature =
(at STP)
Thus the volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L
