Bazza's 3/4 chemistry questions

[Shenz0r]

yeah that sounds right, thanks

um
i have no idea what the text is doing

7a) write ionic half equations for the:
reduction of MnO₂ to Mn²⁺.

The answer is
MnO₂(s) + 4H⁺(aq) + 2e^- -> Mn²⁺(aq) + 2h_2o (l)

now i have 2 problems with this answer; first, i don't think it's in half equation form, secondly, where do the random hydrogen atoms come from? it hasn't stated it's an acidic solution ?

thanks (if this is wrong, correct answer would be appreciated)

Add water to balance oxygen.
1. MnO₂ (s) --> Mn²⁺ (aq) + 2H₂O (l)
Balance hydrogen by adding H+ ions.
2. MnO₂(s) + 4H⁺(aq) --> Mn²⁺ (aq) + 2H₂O (l)
Balance the charges on both side by adding electrons.
3.2e^- +  MnO₂(s) + 4H⁺(aq) --> Mn²⁺ (aq) + 2H₂O (l)

There is your ionic equation.

[Panicmode]

yeah that sounds right, thanks

um
i have no idea what the text is doing

7a) write ionic half equations for the:
reduction of MnO₂ to Mn²⁺.

The answer is
MnO₂(s) + 4H⁺(aq) + 2e^- -> Mn²⁺(aq) + 2h_2o (l)

now i have 2 problems with this answer; first, i don't think it's in half equation form, secondly, where do the random hydrogen atoms come from? it hasn't stated it's an acidic solution ?

thanks (if this is wrong, correct answer would be appreciated)

Add water to balance oxygen.
1. MnO₂ (s) --> Mn²⁺ (aq) + 2H₂O (l)
Balance hydrogen by adding H+ ions.
2. MnO₂(s) + 4H⁺(aq) --> Mn²⁺ (aq) + 2H₂O (l)
Balance the charges on both side by adding electrons.
3.2e^- +  MnO₂(s) + 4H⁺(aq) --> Mn²⁺ (aq) + 2H₂O (l)

There is your ionic equation.

That's correct but before you balance oxygens/hydrogens you must first balance all other elements.

For instance in the equation Cr2O7 2-(aq)---> Cr 3+ (aq) , one would first balance the chromium atoms before moving on to balancing the oxygens.


It's also worth noting that this only works in acidic conditions. Later on in the course you may be required to balance an equation in alkaline conditions (particularly if you're working with an alkaline fuel cell). To do this is simple,balance as if in acidic conditions, then add as many OH- ions to both sides of the equation as there are H+ ions. The H+ ions and the OH- ions combine to form water. You can then cancel the waters out if necessary.

[WhoTookMyUsername]

THis is something i should definitely know by now, but is there an easy ish way of telling what state a substance is in? or do you just have to remember the state for hundreds of compounds and molecules that may come up in an exam?

[Panicmode]

THis is something i should definitely know by now, but is there an easy ish way of telling what state a substance is in? or do you just have to remember the state for hundreds of compounds and molecules that may come up in an exam?

Generally, it will be given to you in the question or it will be stated someplace in the data booklet. Of course, you are expected to recognise the states of some obvious compounds such as H2O(l) in aqueous equations. The state can generally be inferred from the context of the equation. Also, remember that the heavier compounds become, the more they tend towards the liquid/solid state. This can be evidenced when looking at simple alkane hydrocarbons. Methane exists as a gas at room temperature but as the molecular weight of the compounds increases, so does the boiling temp so that by pentane, the mixture is a liquid at room temperature.

[WhoTookMyUsername]

Well for questions like

Write ionic half equations for the

a) reduction of MnO2 to Mn(2+)

would we be expected to know those states/?

[slammy]

Well for questions like

Write ionic half equations for the

a) reduction of MnO2 to Mn(2+)

would we be expected to know those states/?

The question would probably have information about MnO2 being a solid, but I think throughout the year you'll find that you'll just encounter the same substances and hence become familiar with each
And for Mn2+, ions are always in aqueous form

[WhoTookMyUsername]

1) are ions always in aqueous form !!?!?! Are transition  metal ions always in aqueous form? What are the "rules' similar to this ?

2) For the half equation where concentrated hydrochloric acid forms chlorine gas, why is the Hydrogen in HCl excluded from the half equation? (as compared to say magnesium dioxide -> magnesium (II) where the O2 is still accounted for)>?


thanks

[Panicmode]

1) are ions always in aqueous form !!?!?! Are transition  metal ions always in aqueous form? What are the "rules' similar to this ?

2) For the half equation where concentrated hydrochloric acid forms chlorine gas, why is the Hydrogen in HCl excluded from the half equation? (as compared to say magnesium dioxide -> magnesium (II) where the O2 is still accounted for)>?


thanks

1. No, ions are not always in aqueous form, they can be found in liquid form as well but you won't encounter that in the course. (For unit 3 at least) Later when you do more on the electrochemical series, you'll learn that sometimes liquid solutions of ions must be used so that water doesn't react preferentially. Liquid solutions of ions only exist at very high temperatures.

2. This is because the HCl is an aqueous substance which is to say that it is dissolved in water and so exists as H+ and Cl- ions (for simplification). This means that the Cl- ions can react without the H+ ions. In the Magnesium oxide circumstance, magnesium oxide exists as a solid combined in the one latice. Therefore, for any reaction of the magnesium to occur, the oxygen in the structure must also react.



Also small nitpick, MnO2 is Manganese dioxide (Manganese (IV) oxide) not Magnesium dioxide. Be sure not to make this mistake later, it could cost you marks.

[WhoTookMyUsername]

Thanks panic mode

For  redox half equations what do you do when ite not a 1:1:1:1 ratio?

For example S₂O₄^2- + H₂O -> S₂O₄^2-+2HSO₃^-1

reasoning please

[thushan]

The liquid 'solution' of ions is really just a molten salt.

[WhoTookMyUsername]

Uh if light absorbed raises electrons to a higher energy state, isnt this light quickly re emitted?


I Thought that was the whole idea of heating substances so they release a particular wavelength of light equal to the wavelength of light absorbed?


Like when something absorbs red light what is actually happening? Where does the light go? Why isnt it immediatly re released (ie. why do we not see the red)



2) looking at the IR spectrum of proanone  ( CH₃COCH₃ there is a peak at around 1800. Looking at the table for infrared absorbance bands it seems to indiciate the CO bond is the stretched one here, but the table also says the location for this stretch is in carboxylic acids and esters, of which propanone is neither. What is going on here?

3) how much do we need to know about spectrometry and chromatography? Do we need to memorise the each way each process works and is measured with equipment etc.?

4) in nmrs is the major peak the one with the furthest chemical shift?



What wavelength of loght is used for uv-visible spectroscopy
Thanks

[WhoTookMyUsername]

Edited

[WhoTookMyUsername]

1) if light absorbed raises electrons to a higher energy state, isnt this light quickly re emitted?


I Thought that was the whole idea of heating substances so they release a particular wavelength of light equal to the wavelength of light absorbed?


Like when something absorbs red light what is actually happening? Where does the light go? Why isnt it immediatly re released (ie. why do we not see the red)

Here's a longer, and better, way i've asked the question

I’ve been going through the chem U3 course and I’ve been having a bit of trouble understanding a certain concept. I’ve asked quite a few of my friends and looked around online but haven’t been able to find a firm answer.


Premise 1
So my current understanding is that when you heat a certain substance, let’s say X, it can emit a certain wavelength of light, and this can be in the visible light spectrum. So when you heat it, the electrons absorb energy and can jump up electrons shells, these electron can then jump back to ground state in a variety of fashions releasing a colour typical of substance X.

Premise 2
When we see a certain object as a certain colour, let’s say yellow, it means that red and blue has been absorbed by the object. This can cause electrons to jump up energy shells, and then return to ground state in a variety of fashions emitting colours, these colours could be the originally wavelength that was absorbed (red + blue).

We should then see all objects as “white” because all colours are emitted back to us and we see them all.

Now either one of my premises is wrong or there is a contradiction; or something more complex is going on.

These are the possible solutions i was currently thinking about:
1) Only a small portion of atoms will release the original red and blue, thus this effect is not noticeable to us (some will be other colours, some non visible). The problem i have with this explanation is why we can see the substance X as a certain colour (though maybe this is because it’s more based in the visible spectrum than most other objects???)

2) the light is remitted, but has lost some energy as heat, and does not escape the object, instead simply heats the object up.

Those two are the most plausible in my mind at the moment, but i’m in no way sure what’s going on here.

2) looking at the IR spectrum of proanone  ( CH3COCH3 there is a peak at around 1800. Looking at the table for infrared absorbance bands it seems to indiciate the CO bond is the stretched one here, but the table also says the location for this stretch is in carboxylic acids and esters, of which propanone is neither. What is going on here?

3) how much do we need to know about spectrometry and chromatography? Do we need to memorise the each way each process works and is measured with equipment etc.?

4) in nmrs is the major peak the one with the furthest chemical shift?

5) For the question (attached)

why is the answer given for (b) in the textbook 425 and not 460?


thankyou

Thanks

[WhoTookMyUsername]

major edit

thanks

[WhoTookMyUsername]

Anyone? XD