00:01
In this problem, the submarathon problem, we're asked to identify five compounds that consist of h, n, and o.
00:07
And for part a, we're told all the compounds are electrolytes.
00:19
What? um...
00:44
Although not all of them are strong.
01:01
C and d are ionic and b is covalent.
01:18
Okay, let's see here.
01:42
So one of my compounds is going to be hno3.
01:49
One will be hno2, ionic.
02:03
Um...
02:11
And i'll have to think about this a little bit more.
02:14
Nitrogen occurs in its highest possible oxidation state in a and c.
02:46
And in c, d, and e, it's in its lowest oxidation state.
03:02
The formal charge on both nitrogens, so that one has two, in c is one plus.
03:28
And the formal charge in b is zero.
03:40
Okay, so my greatest charge for oxygen will be a and c.
03:50
My n will have a plus five.
04:08
So a can be a plus five, which would likely be, what is this? this would be plus five.
04:23
So a or c can be plus, can be hno3.
04:40
And my lowest oxidation state, um...
05:02
Let me get the rest of my stuff written down.
05:05
Let's go to c.
05:09
A and e exist in solution.
05:18
So i'm pretty sure that a now is going to be hno3.
05:24
Both give off gases.
05:37
Commercially valuable of a is 16 molar.
05:59
And e is 15 molar.
06:08
D, e, commercial solutions labeled with a misnomer, implies binary gaseous compound of nitrogen, has reacted with water to produce nh4 and oh-.
07:21
Okay, so this one will likely be ammonium hydroxide, is e.
07:34
What did i have for e? e exists in solution.
07:47
D, e says d is 43 .7 % n and 50 .0 % o.
08:06
If d were a gas at stp, it would have a density of 2 .86 grams per milliliter.
08:25
F, a formula unit of c, has one more o than d.
08:44
C and a have one ion in common when a acts as a strong electrolyte.
09:15
And then g, solutions of c are weakly acidic.
09:27
A are strongly acidic.
09:39
B and e are basic.
09:51
0 .726 grams of b requires 21 .98 milliliters of 1 .00 molar hcl to titrate.
10:14
Okay, let's see.
10:18
So i'm going to say right now that i know a is hno3, c is hno2, and we had e, e is nh4oh.
10:52
This is what i think so far.
10:53
So let's hang on here.
10:57
I'll write my lewis structures after i've got them all figured out.
11:06
So my other substance, b, is also basic.
11:12
So it's going to be, d is going to be nh2oh.
11:18
I'm almost positive.
11:22
And i can figure that out in a moment.
11:38
I'm just going through all my steps here now.
11:50
Okay, let's check this one out for d.
12:20
Okay, so a and e.
12:22
A we know, but this one is e.
12:27
So where is, let me get back up here, c, d.
12:48
Let's do this one.
12:50
So i have 43 .7 divided by 14 .01 equals 3 .12 moles of n...