Hg(No3)2 +NaSO4 --->2NaNO3 + HgSO4(s)
calculate the moles of each reactant
moles=mass/molar mass
moles of Hg(NO3)2= 51.429g/ 324.6 g/mol(molar mass of Hg(NO3)2)=0.158 moles
moles Na2SO4 16.642g/142g/mol= 0.117 moles of Na2SO4
Na2SO4 is the limiting reagent in the equation and by use mole ratio Na2So4 to HgSO4 is 1:1 therefore the moles of HgSO4 =0.117 moles
mass of HgSO4=moles x molar mass of HgSo4= 0.117 g x 303.6g/mol= 35.5212 grams
If element X has 59 protons then element X has 59 electrons.
Side note: as long as an element stays an atom the number of protons and electrons will always have the same value.
Answer : The average atomic mass of an element X is, 63.546 amu
Solution : Given,
Mass of isotope X-63 = 62.9296 amu
% abundance of isotope X-63 = 69.15% = 0.6915
Mass of isotope X-64 = 64.9278 amu
% abundance of isotope X-64 = 30.85% = 0.3085
Formula used for average atomic mass of an element X :
![\text{ Average atomic mass of an element X}=\sum[(62.9296\times0.6915)+(64.9278\times 0.3085)]](https://tex.z-dn.net/?f=%5Ctext%7B%20Average%20atomic%20mass%20of%20an%20element%20X%7D%3D%5Csum%5B%2862.9296%5Ctimes0.6915%29%2B%2864.9278%5Ctimes%200.3085%29%5D)
Therefore, the average atomic mass of an element X is, 63.546 amu
