From the given question we see that the heat of solution is positive. This implies that when you dissolve lead nitrate in water, the result is an endothermic process
Therefore we can say that the crystallization ( which is the opposite of dissolution) would be exothermic process.
When the temperature of the solution would be lowered, the reaction would favor a exothermic process.
The primary reason for dissolution is that when the temperature is reduced all the steps involved in dissolving are overturned.
Dissolution of any molecule involves 3 major steps:
The first 2 steps are endothermic and depend on the temperautre of the solution as temperature is an indication of the energy available. So you can see that the reduction in temperature is actually a reduction in the energy in the solvent molecules, which implies that the first 2 steps are now becoming less feasible.
Eventually crystal nuclei will be formed where the molecules that have reformed have accumulated and newly formed molecules will join these nuclei.
Part a please Question 3 (30 pts) Lead nitrate (Pb(NO3)2) is a toxic compound used in...
an aqueous solution is .907M Pb(NO3)2. what is the molality of the lead(II) nitrate in this solution? the density of the solution is 1.252g/ml
When a 6.07-g sample of solid lead(II) nitrate dissolves in 31.9 g of water in a coffee-cup calorimeter (see above figure) the temperature falls from 22.00 oC to 18.47 oC. Calculate H in kJ/mol Pb(NO3)2 for the solution process. Pb(NO3)2(s) Pb2+(aq) + 2 NO3-(aq) The specific heat of water is 4.18 J/g-K. Hsolution = ?kJ/mol Pb(NO3)2.
How many grams of lead nitrate, Pb(NO3)2, are required to make a 7.06 % w/v aqueous solution in a 50.0 mL volumetric flask? in grams
part 1) An aqueous solution of magnesium nitrate, Mg(NO3)2, contains 2.03 grams of magnesium nitrate and 17.8 grams of water. The percentage by mass of magnesium nitrate in the solution is %. part 2) The mole fraction of lead acetate, Pb(CH3COO)2, in an aqueous solution is 1.80×10-2 . The percent by mass of lead acetate in the solution is %. part 3) An aqueous solution is 40.0 % by mass potassium bromide, KBr, and has a density of 1.37 g/mL. The mole...
12. A solution of 0.00016 M lead (II) nitrate, or Pb(NO3)2, was poured into 450 mL of 0.00023 M sodium sulfate, Na2S04. Would a precipitate of lead(II)sulfate, PbSO4, be expected to form if 250 mL of the lead nitrate solution were added? Write the equilibrium equation of PbSO4 dissociation in water.
A26. What will be observed when 15.0 mL of 0.040 M lead(II) nitrate, Pb(NO3)2, is mixed with 15.0 mL of 0.040 M sodium chloride? (lead chloride Ksp = 1.7 × 10–5). (A) A clear solution with no precipitate will result. (B) Solid PbCl2 will precipitate and excess Pb2+ ions will remain in solution. (C) Solid PbCl2 will precipitate and excess Cl– ions will remain in solution. (D) Solid PbCl2 will precipitate and there will be no excess ions in solution....
A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 10.47 g PbCl2(s) 10.47 g PbCl 2 ( s ) is obtained from 200.0 mL 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) solution. concentration: M
8. A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq), until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 12.12 g PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution. concentration: M
explain how to obtain the answer for part a KClo3 part b Pb(NO3)2 partc Ce2(SO4)3 Part A KCIO3 E 80 Express the mass in grams as an integer. E 60 30 20 0 10 20 30 40 50 60 70 80 90 100 Incorrect: Try Again; 2 attempts remaining By referring to the figure above, determine the minimal mass of each of the following salts required to form a saturated solution in 250 g of water at 30 C To...
Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 15.93 g of PbCl2(s) is obtained from 200.0 mL of the original solution Calculate the molarity of the Pb(NO3)2(aq) solution. Number