QUESTION 1 43.2 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 36.1 mL of 0.415 M NaCl. The equation for the precipitate reaction is: Pb(NO3)2 (aq) + 2 NaCl (aq) --> PbCl2 (s) + 2 NaNO3 (aq) How many moles of PbCl2 are formed? (with correct sig figs)
QUESTION 2 What volume (in mL!!!) of 1.28 M HCl is required to react with 3.33 g of zinc (65.41 g/mol) according to the following reaction? Zn(s) + 2 HCl (aq) → ZnCl2(aq) + H2(g) (with correct sig figs)
QUESTION 3 Assuming volumes are additive, how much water (in mL) should be added to 29.5 mL of a 1.77 M solution of hydrochloric acid (HCl) to produce a 0.100 M solution?(with correct sig figs)
QUESTION 4 What is the concentration of fluoride ions (in M) in a 7.95 M solution of cobalt(III) fluoride: CoF3 (With correct sig figs)
QUESTION 5 What is the concentration (in M) of a calcium nitrate (Ca(NO3)2 164.10 g/mol) solution prepared by dissolving 14.5 g of Ca(NO3)2 (164.10 g/mol) in enough water to produce 393 mL of solution? (Do not drop significant digits.)
QUESTION 1 43.2 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 36.1 mL of 0.415...
44.2 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 31.6 mL of 0.415 M NaCl. The equation for the precipitate reaction is: Pb(NO3)2 (aq) + 2 NaCl (aq) --> PbCl2 (s) + 2 NaNO3 (aq) The concentration of NO3- ion in the reaction solution is _____ M.
If 300 mL of some Pb(NO3)2 solution is mixed with 600 mL of 2.70 x 10−2 M NaCl solution, what is the maximum concentration of the Pb(NO3)2 solution added if no solid PbCl2 forms? (Assume Ksp = 2.00 x 10−5 M at this temperature.) Enter the concentration in M.
If 250 mL of some Pb(NO3)2 solution is mixed with 450 mL of 5.90 x 10−2 M NaCl solution, what is the maximum concentration of the Pb(NO3)2 solution added if no solid PbCl2 forms? (Assume Ksp = 2.00 x 10−5 M at this temperature.) Enter the concentration in M.
If 400 mL of some Pb(NO3)2 solution is mixed with 400 mL of 1.10 x 10-2 M NaCl solution, what is the maximum concentration of the Pb(NO3)2 solution added if no solid PbCl2 forms? (Assume Ksp = 2.00 x 10-5 M at this temperature.) Enter the concentration in M. Submit Answer Tries 0/99
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
How many grams of PbCl2 are formed when 35.0 mL of 0.520 M KCl react with Pb(NO3)2? 2KCl(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbCl2(s) How many grams of PbCl2 are formed when 35.0 mL of 0.520 M KCl react with Pb(NO3)2? 2KCl(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbCl2(s) 10.1 g 25.3 g 5.06 g 15.2 g 2.53 g
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 15.59 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
If 450 ml of some Pb(NO3)2 solution is mixed with 250 ml of 1.10 x 10-2 M NaCl solution, what is the maximum concentration of the Pb(NO3)2 solution added if no solid PbCl2forms? (Assume Ksp = 2.00 x 10-5 M at this temperature.) Enter the concentration in M.
Answer the following questions based on this reaction: Pb(NO3)2 (aq) + 2 NaCl (aq) PbCl2 (s) + 2 NaNO3 (aq) a) If 225 mL of 12.95 M Pb(NO3)2(aq) are reacted with a solution made with 5.05 g of NaCl (aq), how many grams of lead (II) chloride will be precipitated? b) If the actual yield of lead (II) chloride is 1.06 g, what is the percent yield?