March 19, 2024, 07:12:32 AM
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1
Hello,

I am working on ideas to create an educational video about the chemistry of copper and nitrogen oxides. The goal is to used copper ii nitrate to form nitrogen dioxide for production of nitric acid. On my initial searching I found that copper ii nitrate may be heated to around 180 C to liberate the NO2. I was wondering if there was any way to have the gas be released at a lower temperature. Are there any catalysts or methods of reducing the temperature of this reaction? Thanks in advance for any *delete me*
2
Note: polymerization reactions take many steps, they are typically of the (approximate) form

nX :rarrow: Xn

Or, when you have a mix of things reacting it can be something closer to

nX + nY :rarrow: (XY)n

(these are not exact reaction equations, they are just signaling the general idea).

As n can take many values in the same mixture, this is not a simple equilibrium reaction. It is quite difficult to apply correctly simple LCP logic here, not understanding well why and when it works for simpler cases.
3
If add now again 100 ml 0,05 m  H3PO4 again the volume increase to 200 ml

Quote
Molarity changed to 0,075 M

Now do the math adding 0.1 M H3PO4.
4
i'm just looking into le chatlier now, and this doesn't directly address your question but this video

Le Chatelier's Principle
Professor Dave Explains
https://www.youtube.com/watch?v=XmgRRmxS3is

In the first minute of the video he goes into adjusting concentrations of reactants and products and the effect on equilibrium.

That might help in your study of this

Also the first 3min 30sec of this video

Which way will the Equilibrium Shift? (Le Chatelier's Principle)
chemistNATE

https://www.youtube.com/watch?v=BPDkl92NCUs

Explains what happens if you add/remove reactants and products for an example reaction.

I don't know about your reaction specifically or how it translates in a lab setting, but that info above might help in your investigations.

5
That is what I meant. If adjusted the pH with phosphoric acid, then the molarity will change, of course the volume will also increase. pH can only adjusted with an different acid like HCl . To increase pH NaOH can be used, what has no effect of the phosphate beside also volume effect.

Example

Lets have a 0,1 M buffer half dihydrogenphosphate and half phosphoric acid. Means both chemicals are 0,05 M

According HH equation is pH = pKa = 2,1 ( first pka)

Volumen should be 100 ml
If add now again 100 ml 0,05 m  H3PO4 again the volume increase to 200 ml, the concentration of salt is half because dilution.  The concentration of acid is the same, because same concentration is added.

In HH equation pH = pka  + log( c salt/c acid)
pH = 2,1 + log (0,5 * csalt/ c acid)

pH drops to 1,79

Molarity changed to 0,075 M

6
Based on what you said, it seems obvious that the adjusting the pH with H3PO4 will increase the amount of phosphate in the buffer which has already 5mM of ammonium phosphate monobasic, so the molarity of phosphate will increase and it can't stay the same or decrease.
What is missing in this paragraph is a consideration of the volume of H3PO4.  Does that help?
7
Yes,

That's not a correct answer.


Why, what do you mean
You cut out only the Word yes
Should it be NO. For what reason?

I am assuming you answered

So the addition of phosphoric acid should affect the final concentration of phosphate buffer. right?

Neither "yes" nor "no" are correct answer.

Sad that you both attempt to guess instead of doing the math.

I think YOU are wrong, unless you can explain.
8
Yes,

That's not a correct answer.


Why, what do you mean
You cut out only the Word yes
Should it be NO. For what reason?

I am assuming you answered

So the addition of phosphoric acid should affect the final concentration of phosphate buffer. right?

Neither "yes" nor "no" are correct answer.

Sad that you both attempt to guess instead of doing the math.
9
Yes,

That's not a correct answer.


Why, what do you mean
You cut out only the Word yes
Should it be NO. For what reason?


10
It is the question doesn't need the calculation. This is the yes or no question. Why do I need to do the calculation?

Learning opportunity - once you do the calculations you will realize where is the mistake in your thinking.
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