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Ba(NO3)2 Molarity = moles of solute/Liters of solution ( 100 ml = 0.100 liters ) 0.10 M Ba(NO3)2 = moles Ba(NO3)2/0.100 liters = 0.01 moles Ba(NO3)2 (261.32 grams/1 mole Ba(NO3)2) = 2.6 grams of Ba(NO3)2 needs to be put into that 100 milliliters of solution.
Cu(NO3)2 -----------Cu2+ + 2(NO3)-
The chemical formula of lead(II) nitrate is Pb(NO3)2. The solution don't change the formula - but in the solution exist the anion (NO3)- and the cation Pb2+.
3Mg + 2Fe(NO3)3 ---> 2Fe + 3Mg(NO3)2
The answer is: 2Al + 3Pb(NO3)2 = 2Al(NO3)3 + 3Pb.
Ba(NO3)2 Molarity = moles of solute/Liters of solution ( 100 ml = 0.100 liters ) 0.10 M Ba(NO3)2 = moles Ba(NO3)2/0.100 liters = 0.01 moles Ba(NO3)2 (261.32 grams/1 mole Ba(NO3)2) = 2.6 grams of Ba(NO3)2 needs to be put into that 100 milliliters of solution.
You get 300 mL of 0.0100 M Sr(NO3)2 solution
Cu(NO3)2 -----------Cu2+ + 2(NO3)-
The chemical formula of lead(II) nitrate is Pb(NO3)2. The solution don't change the formula - but in the solution exist the anion (NO3)- and the cation Pb2+.
The chemical equation is: (K+ + I-)(aq) + (Ag+ + [NO3]-)(aq) --> AgI (s) + (K+ + [NO3]-)(aq) or The chemical equation is: K+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + K+[NO3]- (aq)
3Mg + 2Fe(NO3)3 ---> 2Fe + 3Mg(NO3)2
The precipitates are of AgCl and BaSO4 because they are insoluble in water
The answer is: 2Al + 3Pb(NO3)2 = 2Al(NO3)3 + 3Pb.
0.9 M Al(NO3)3 = 0.9 M *3 M (NO3-) = 2.700 M nitrate ion's (=NO3-)(Mark the bold numbers: 3above !! because they correspond to e. o.)
NO3- stands for nitrate anion. It has to have one negative charge ( .- ) in superscript position.
Pb(NO3)2
Nitrate or NO3 is an inorganic anion. A 50 mg/l Nitrate solution (as NO3) is the equivalent to a 11.3 mg/l Nitrate solution (as N).