1. Which reagent could be used to separate Br- from NO3- when added to an aqueous solution containg both?
a. CuSO4
b. AgNO3
c. NaI
d. Ba(OH)2
All choices are aqueous.
I don't understand what the problem is asking. Is it saying that these anions are combined together in a solution as in BrNO3? But is this not possible because both have the same sign in charge? Do I set up an hypothetical equation with reactants and products?
2. Using the following sequence of reactions, determine the order of oxidizing agents (strongest to weakest).
Q + U+ --> Q(+) + U (1)
U + W+ --> U(+) + W (2)
Q + W+ --> Q(+) + W (3)
a. W+ > U+ > Q+
b. W > U > Q
c. Q > U > W
d. Q+ > U+ > W+
Is it W+ > U+ > Q+ ?
Here is my reasoning (without charges, which were looked at later):
(1) U = oxidizing agent Q =reducing agent
(2) W is oxidizing agent U = reducing
(3) W = oxidizing Q=reducing
(Hopefully, I'm not confusing with W > U > Q!)
3. Ammonia reacts with hypobromite, OBr-, to form nitrogen gas and Br- anion. Find M of OBr- if 2.83 mL of solution reacts with 4.11 mg (.00411g) of NH3 (molar mass = 17.03). Balance equation by half reaction method in acid medium.
NH3 + OBr- --> Br- + N2
NH3 --> N2
OBr- --> Br-
Now
2NH3 --> N2 + 6H+
2H+ + OBr- --> Br- + H2O
Then
2e- + 2H+ + OBr- --> Br- + H2O
2NH3 --> N2 + 6H+ + 6e-
So
3(2e- + 2H+ + OBr- --> Br- + H2O) = 6e- + 6H+ + 3OBr- -->3Br- + 3H2O
2NH3 --> N2 + 6H+ + 6e-
Finally,
3OBr- + 2NH3 --> N2 + 3Br- + 3H2O
M = mol OBr-/.00283 L OBr-
0.00411 g NH3 * (1 mol NH3 / 17.03 g) * (3 mol OBr- / 2 mol NH3) = 0.0003620 mol OBr-
so 0.0003620 mol OBr- / .00283 L OBr- = 0.1279 M??
Please reply. Thanks.