Question

Question #6a7f2

Answer 1

All the substances mentioned here are elements not compounds

Explanation:

see the video
reaction with C,S andP
http://www.dailymotion.com/video/x20zrx8_reaction-of-nitric-acid-with-non-metals_fun
reaction with Cu
https://practicum.melscience.com/experiments/reaction-of-copper-with-nitric-acid.html

Answer 2

`"HNO"_3 + 3"H"^+ + 3"e"^{-} -> "NO" + 2"H"_2"O"`

Answer 3

Here's what I got.

Explanation:

As you know, nitric acid, `"HNO"_3`, is a strong oxidizing agent.

Now, when nitric acid reacts with non-metals, it oxidizes them to their highest oxidation state.

The difference between concentrated nitric acid and dilute nitric acid lies with the reduction of the acid.

As a rule of thumb, you'll often see mentions of how concentrated nitric acid is reduced to nitrogen dioxide, `"NO"_2`, and dilute nitric acid is reduced to nitric oxide, `"NO"`.

The reduction of the acid depends on its concentration, so the more dilute the acid is, the greater its reduction. Depending on concentration and temperature, very dilute nitric acid will get reduced to nitrous oxide, `"N"_2"O"`, nitrogen gas, `"N"_2`, and even ammonia,`"NH"_3`.

For your purposes, assume that the dilute nitric acid is concentrated enough to be reduced to nitric oxide, `"NO"`.

So, when dilute nitric acid reacts with elemental carbon, it will oxidize carbon to its highest oxidation state, which is `color(blue)(+4)` in carbon dioxide, `"CO"_2`.

The balanced chemical equation for this reaction is

`3"C"_text((s]) + 4"HNO"_text(3(aq]) -> 3"CO"_text(2(g]) + 4"NO"_text((g]) + 2"H"_2"O"_text((l])`

Notice that nitrogen is reduced from an oxidation state of `color(blue)(+5)` in nitric acid to an oxidation state of `color(blue)(+2)` in nitric oxide.

Sulfur and phosphorus will be oxidized to their highest oxoacids, i.e. acids in which they're in their highest oxidation state. Remember, you're working with dilute nitric acid, so look for it to be reduced to nitric oxide, `"NO"`.

`"S"_text((s]) + 2"HNO"_text(3(aq]) -> "H"_2"SO"_text(4(aq]) + 2"NO"_text((g])`

Here sulfur goes from an oxidation state of `color(blue)(0)` to an oxidation state of `color(blue)(+6)` in sulfuric acid.

Now, I'm not really sure about dilute nitric acid oxidizing phosphorus to phosphoric acid. From what I remember, you needed concentrated nitric acid to do that, but I may very well be mistaken, so I'll skip this reaction.

I know that nitric acid can oxidize white phosphorus to phosphoric acid, but I'm not sure if the acid is concentrated or dilute

`"P"_text(4(s]) + 10"HNO"_text(3(aq]) + "H"_2"O"_text((l]) -> 4"H"_3"PO"_text(4(aq]) + 5"NO"_text((g]) + 5"NO"_text(2(g])`

Finally, dilute nitric acid will react with copper and be reduced to nitric oxide in the process

`3"Cu"_text((s]) + 8"HNO"_text(3(aq]) -> 3"Cu"("NO"_3)_text(2(aq]) + 4"NO"_text(2(g]) + 2"H"_2"O"_text((l])`