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
Answer:
The electrons are lost from the valence shell (outermost electron shell) of the atom.
Explanation:
This is able to be inferred not only because valence electrons being lost first is a trend but also because the atom in question has actually 3 valence electrons (13-2-8 = 3).
Answer:
=zero degrease, 1 atm)? so the volume of an ideal gas is 22.l/mol at STP this, 22l.4Lis probably the most remembered and least useful number in chemistry
I'm pretty sure that significant figures is just the amount of numbers there is. So, in this case I think the answer would be D. 5
