Help on Chem equestion

[Dream of the iris]

Well, I haven't taken a chem class in 2 years and now I'm thrust into this strange and frustrating world once again. So here's an equation that I think I balanced, but wonder if I could get some feedback.

Original equation: NH3+02 yields N2O4 + H20

My numbers were: 4NH3+7O2 yields 2N2O4 + 6H2O.

This would give 4N, 12H and 14O.

Does this sound about right?

Thanks and I'll probably be here more this semester

[PirateJoe]

Looks good to me.

[Dream of the iris]

Cool that's what I thought. Here's another one that may or may not be right.

0.2918g of oxalic acid dihydrate are dissolved in water, then reacted with 38.40 ml of a solution of potassium hydroxide according to the equation:

2KOH + h2C2O4 yields K2C2O4 + 2H2O

What is the molarity of KOH solution?

I figured out the molar mass of the oxalic acid dihydrate to be 90.03g. So

0.2918g X 1 mole H2C2O4/90.03g H2C2O4 = 0.0032molesh2c2o4/0.3840L

I got as a result of that:

0.0083MKOH.

Sound right? I think I was supposed to balance the equation first, bah. Just realized that. Well, did convert the volumes right at least? I know this such a juvenile question, but you know 2 years off is quite a bit a time to forget a bunch of stuff.

[Cytosine]

Unless you entered it into your calculator wrong, looks good to me.

Erm, assuming the oxalic acid reacted completely, etc.

[Dream of the iris]

Thanks guys!

Another question:

Determine the oxidation number (oxidation state) of the P in Na3P3O10.

I said P=25.

Na=+1

O = -2

5(1) + P + 10(-2)

=

5 +p -20

P=-15.

This doesn't sound right.



Also, I'm having difficulty converting cm-3 to ft-3. The problem is:

Convert 57.90 J cm-3 to kcal ft -3. I don't know where to begin on that one.

[JChifty]

Do your own homework you lazy bastard.

[Dream of the iris]

Woah, I'm just asking for a little help here. Just to check the work mostly.

[JChifty]

Ok, fine, it is all correct, now go fail some exams.

[Ekstaz]

Originally Posted by Dream of the iris

Also, I'm having difficulty converting cm-3 to ft-3. The problem is:

Convert 57.90 J cm-3 to kcal ft -3. I don't know where to begin on that one.

Do you mean cm^3 to ft^3?

1 centimeter = 0.032808399 feet which you will need to multiply through 3 times, or do it in one step after cubing both values.
according to my notes 1 calorie is 4.184 Joules, which would make 1 kcal = 4184 Joules. Shouldn't be too bad if you correctly set up your equation so that the unit of the value on top cancels with the units of the bottom value of the ratio youre multiplying by, and so on.

As for oxidation states, I forgot how to do those. I'm sure you could find similar examples in the book/online though if you look, sorry.

chifty, youre pretty feisty for having contributed nothing and flaming doti, a pretty well respected ex-totsean. Seek love and attention elsewhere, you have only my pity.

[Slave of the Beast]

Originally Posted by Dream of the iris

Thanks guys!

Another question:

Determine the oxidation number (oxidation state) of the P in Na3P3O10.

I said P=25.

Na=+1

O = -2

5(1) + P + 10(-2)

=

5 +p -20

P=-15.

This doesn't sound right.

Do you mean Sodium tripolyphosphate - Na5P3O10? If so...

Na=+1
O = -2

The overall charge of these two elements in the compound is therefore:

10(-2) + 5(1) = -15

The 3 P atoms must balance this charge to zero, and so must have a combined charge of +15, therefore:

+15/3 = +5

Oxidation state of P = +5

This value is consistent with the allowed oxidation states of P, namely -3, +3 and +5.

[The Jitterskull]

Oxidations are important if you're going to pursue biochemistry.

Actually the way we do it is:
H gain or O loss = reduction
H loss or O gain = oxidation
(but thats for bio mainly)


... though I'll see you back here when you have to do voltage ;P
PS: Yes I know you're doing charges, if you're not pursuing a science degree don't worry about doing this ever again.

[stiletto]

I've been fiddling around with a couple chem problems the last few nights and they just don't seem to wanna click.

1 0.4505g sample of CaCO3 was dissolved in HCL and the resulting solution was diluted to 250.0ml in a volumetric flask. A 25.00ml Aliquot of the solution required 24.25ml of an EDTA solution for titration to the Eriochrome Black T end point.

A) how many moles of EDTA are contained in the 24.25 ml used for titration?
B) what is the concentration (molarity) of the EDTA solution?
C) if 100.00ml of a water sample required 23.24ml of EDTA of the concentraion found in problem b, what is the hardness of the water in terms of ppm CACO3 (ppm=mg/L)?

I love chem, but some of these titration problems run me ragged. Hope you've got more luck on these then i do.