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
I believe the word that you are looking for is "subjective"
-to be influenced by personal opinion or feeling-
PLEASE RATE AS THE BRAINLIEST ANSWER! THANK YOU! :)
The Law of Conservation of Mass states that the mass of reactants entering a reaction must be equal to the mass of the products exiting it. In this case, we only have 2 reactants, Fe and S, and we only have 1 product, FeS. Therefore we expect the total mass of the Fe and S reactants to equal the mass of FeS. This gives us 112 g + 64 g = 176 g of FeS, which is choice D.
Hmm, friction maybe? I guess it depends on how fast she stopped?
Mass of Ni = 0.015 mol Ni × (58.693 g Ni/1 mol Ni) = 0.88 g Ni
