Hydrogen sulfide is bubbled into a solution that is 0.10 M in both Pb 2 + and Fe 2 + and 0.30 M in HCl. After the solution has come to equilibrium it is saturated with H 2 S ( H 2 S [ H 2 S ] = 0.10 M ). What concentrations of Pb 2 + and Fe 2 + remain in the solution? For a saturated solution of H 2 S we can use the equilibrium: H 2 S ( a q ) + 2 H 2 O ( l ) ⇌ 2 H 3 O + ( a q ) + S 2 − ( a q ) K = 1.0 × 10 − 26 (Hint: The [H 3 O + ] changes as metal sulfides precipitate.)
Hydrogen sulfide is bubbled into a solution that is 0.10 M in both Pb 2 + and Fe 2 + and 0.30 M in HCl. After the solution has come to equilibrium it is saturated with H 2 S ( H 2 S [ H 2 S ] = 0.10 M ). What concentrations of Pb 2 + and Fe 2 + remain in the solution? For a saturated solution of H 2 S we can use the equilibrium: H 2 S ( a q ) + 2 H 2 O ( l ) ⇌ 2 H 3 O + ( a q ) + S 2 − ( a q ) K = 1.0 × 10 − 26 (Hint: The [H 3 O + ] changes as metal sulfides precipitate.)
Hydrogen sulfide is bubbled into a solution that is 0.10 M in both Pb2+ and Fe2+ and 0.30 M in HCl. After the solution has come to equilibrium it is saturated with H2S (
H
2
S
[
H
2
S
]
=
0.10
M). What concentrations of Pb2+ and Fe2+ remain in the solution? For a saturated solution of H2S we can use the equilibrium:
H
2
S
(
a
q
)
+
2
H
2
O
(
l
)
⇌
2
H
3
O
+
(
a
q
)
+
S
2
−
(
a
q
)
K
=
1.0
×
10
−
26
(Hint: The [H3O+] changes as metal sulfides precipitate.)
Question 19
0/2 pts 3
Details
You have a mixture of sodium chloride (NaCl) and potassium chloride (KCl) dissolved in water
and want to separate out the Cl- ions by precipitating them out using silver ions (Ag+). The
chemical equation for the net ionic reaction of NaCl and KCl with silver nitrate, AgNO3, is shown
below.
Ag+(aq) + Cl(aq) → AgCl(s)
The total mass of the NaCl/KCl mixture is 1.299 g. Adding 50.42 mL of 0.381 M solution
precipitates out all of the Cl-.
What are the masses of NaCl and KCl in the mixture?
Atomic masses:
g: Mass of NaCl
g: Mass of KCL
Ag = 107.868 g
mol-
1
Cl = 35.453 g mol-
1
K = 39.098 g mol-
N = 14.007 g mol−1
Na = 22.99 g mol−1
0 = 15.999 g mol
1
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Part 1. Draw monomer units of the following products and draw their reaction mechanism (with arrow pushing)
Polyester fiber
Using a) pthalic anhydride + anhydrous sodium acetate + ethylene glycol
B)pthalic anhydride + anhydrous sodium acetate + glycerol
Identify the missing starting materials/ reagents/ products in the following reactions. Show the stereochemistry clearly in the structures, if any.
If there is a major product, draw the structures of the major product with stereochemistry clearly indicated where applicable. Show only the diastereomers (you do not have to draw the pairs of enantiomers).
If you believe that multiple products are formed in approximately equal amounts (hence neither is the major product), draw the structures of the products, and show the detailed mechanism of these reactions to justify the formation of the multiple products.
If you believe no product is formed, explain why briefly. (6 mark for each, except f and g, which are 10 mark each)
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell