Question
In: Chemistry
For the reactions shown below, we added 2.25 mL of 0.0540 M HCl to a test...
For the reactions shown below, we added 2.25 mL of 0.0540 M HCl to a test tube containing one of the two cations (Ag+ or Pb2+) and recovered 0.0174 g of precipitate.
AgNO3(aq) + HCl(aq) →
AgCl(s) + HNO3(aq)
Pb(NO3)2(aq) + 2 HCl(aq) →
PbCl2(s) + 2 HNO3(aq)
How much precipitate in moles would be recovered theoretically if the ion was Ag+? (Enter an unrounded value. Use at least one more digit than given.) (in mol)
How much precipitate in moles would be recovered theoretically if the ion was Pb2+? (Enter an unrounded value. Use at least one more digit than given.) (in mol)
How much precipitate in grams would be recovered theoretically if the ion was Ag+? (in g)
How much precipitate in grams would be recovered theoretically if the ion was Pb2+? (in g)
Based on the precipitate amount recovered, which of the two ions was in the unknown? Ag+ OR Pb2+
Solutions
queen_honey_blossom answered 2 years agoRelated Solutions
For the reactions shown below, we added 2.25 mL of 0.0460 M HCl to a test...
For the reactions shown below, we added 2.25 mL of 0.0460
M HCl to a test tube containing one of the two cations
(Ag+ or Pb2+) and recovered 0.0148 g of
precipitate.
AgNO3(aq) + HCl(aq) →
AgCl(s) + HNO3(aq)
Pb(NO3)2(aq) + 2 HCl(aq) →
PbCl2(s) + 2 HNO3(aq)
How much precipitate in moles would be recovered theoretically
if the ion was Ag+? (Enter an unrounded value. Use at
least one more digit than given.)
How much precipitate in moles would...
For the reactions shown below, we added 2.25 mL of 0.0450 M HCl to a test...
For the reactions shown below, we added 2.25 mL of 0.0450
M HCl to a test tube containing one of the two cations
(Ag+ or Pb2+) and recovered 0.0141 g of
precipitate.
AgNO3(aq) + HCl(aq) →
AgCl(s) + HNO3(aq)
Pb(NO3)2(aq) + 2 HCl(aq) →
PbCl2(s) + 2 HNO3(aq)
How much precipitate in moles would be recovered theoretically
if the ion was Ag+? (Enter an unrounded value. Use at
least one more digit than given.)
mol
How much precipitate in moles...
For the reactions shown below, we added 3.75 mL of 0.0420 M Na2S to a test...
For the reactions shown below, we added 3.75 mL of 0.0420
M Na2S to a test tube containing one of the two
cations (Cu2+ or Cd2+) and recovered 0.0228 g
of precipitate.
Cu(NO3)2(aq) +
Na2S(aq) → CuS(s) + 2
NaNO3(aq)
Cd(NO3)2(aq) +
Na2S(aq) → CdS(s) + 2
NaNO3(aq)
How much precipitate in moles would be recovered theoretically
if the ion was Cu2+? (Enter an unrounded value. Use at
least one more digit than given.)
mol
How much precipitate in moles...
For the reactions shown below, we added 3.75 mL of 0.0450 M Na2S to a test...
For the reactions shown below, we added 3.75 mL of 0.0450
M Na2S to a test tube containing one of the two
cations (Cu2+ or Cd2+) and recovered 0.0161 g
of precipitate.
Cu(NO3)2(aq) +
Na2S(aq) → CuS(s) + 2
NaNO3(aq)
Cd(NO3)2(aq) +
Na2S(aq) → CdS(s) + 2
NaNO3(aq)
How much precipitate in moles would be recovered theoretically
if the ion was Cu2+? (Enter an unrounded value. Use at
least one more digit than given.)
mol
How much precipitate in moles...
In a calorimeter, we added 50 mL of 1M HCl to 50 mL of 1.1 M...
In a calorimeter, we added 50 mL of 1M HCl to 50 mL of 1.1 M NH3
which produces NH4Cl. If the temperature increased 5.0 degrees
Celsius, what is the molar enthalpy? We weren't given a heat
capacity for the calorimeter but it was made out of a glass beaker
and the heat capacity of glass is 0.840 J/goC found on
google.
11 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that...
11 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution
that is 0.010 M HC2H3O2 and 0.08 M
C2H3O2-. What
is the pH of the resulting solution?
When 50.0 mL of 0.45 M HCl was added to 50.0 mL of 0.65 M NaOH,...
When 50.0 mL of 0.45 M HCl was added to 50.0 mL of 0.65 M NaOH,
the temperature change was 5.8 *C. Calculate the heat change for
the reaction (Qsoln) and the heat of neutralization
(ChangeHneut), and the state whether the reaction is
exothermic or endothermic.
When 50.0 mL of 0.45 M HCl was added to 50.0 mL of 0.65 M NaOH,...
When 50.0 mL of 0.45 M HCl was added to 50.0 mL of 0.65 M NaOH,
the temperature change was 5.8 *C. Calculate the heat change for
the reaction (Qsoln) and the heat of neutralization
(ChangeHneut), and the state whether the reaction is
exothermic or endothermic.
1.How many mL of 0.500 M HCl should be added to 155 mL of 0.0500 M...
1.How many mL of 0.500 M HCl should be added to 155 mL of 0.0500
M benzylamine to form a buffer with pH = 8.52? C7H7NH2 pKb = 4.35
C7H7NH2 (aq) + H+ (aq) C7H7NH3 + (aq) This structure of
benzylamine has a carbon at the junction of each line with 1
hydrogen on each 3-bonded ring C and 2 hydrogens on the C in the
side chain attached to the ring carbon. Note that the N atom has...
A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.300 M HCl....
A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.300
M HCl. The solution was then treated with an excess of aqueous
chromium(III) nitrate, resulting in formation of 2.36 g of
precipitate. Determine the concentration of the solution.
ADVERTISEMENT
ADVERTISEMENT
Latest Questions
- Consider the following network of a project consisting of 7 activities with a scheduled project duration...
- In 1941, President Roosevelt spoke about the “four freedoms.” For many in the country, however, these...
- Your friend is looking for investors in a risky business venture. To convince you to participate,...
- Suppose you observe the following situation: State of Economy Probability of State Return if State Occurs...
- Merger Analysis TransWorld Communications Inc., a large telecommunications company, is evaluating the possible acquisition of Georgia...
- Journal Entries and Trial Balance Elite Realty acts as an agent in buying, selling, renting, and...
- In racket Assume that the elements of a list are indexed starting with 1. Write a...
ADVERTISEMENT