The number of solid precipitate that will be formed is 37.08 g
calculation
write the equation for reaction
=Hg(NO3)2 +Na2SO4 = HgSO4(s) +2NaNO3(aq)
find the moles of each reactant
moles ofHg(NO3)2=126.27/324.6= 0.389 moles
moles of Na2SO4=17.796/142=0.125 moles
NaSO4 is the limiting reagent and by use of mole ratio of NaSO4:HgSO4 which is 1:1 therefore the moles of H2SO4 is also= 0.125 moles
mass HgSO4=moles x molar mass
=0.125 x296.65= 37.08g
Answer:
Well in one glucose they are 6.022*1023 individual glucose molecule.