Unit 4 Questions MEGATHREAD!

[pi]

This has got nothing to do with the course, but is for my own interest.

In ice, are H3O+ and OH- ions present? And if not, why? And if so, how do fit in the 'lattice'?

Thanks.

[Graphite]

Very very minimal (perhaps till none) as you know in pure water the concentration of H+ is really less (this value is same for OH-) at freezing temperature, the equilibrium would be shifted to the left further reducing this small amount of H+ and OH-

[pi]

Hmmm, but will they be present at all?

[Mao]

This has got nothing to do with the course, but is for my own interest.

In ice, are H3O+ and OH- ions present? And if not, why? And if so, how do fit in the 'lattice'?

Thanks.

Yes, they are present. They are present in the lattice as crystal defects (point defects).

Moderator action: removed real name, sorry for the inconvenience

[homosapiens]

I've got my calorimetry sac coming up this week and am not too sure about errors and their effect on the calibration factor.
If you add too much water when calibrating, will the CF be higher or lower...

[Graphite]

Use the equation to predict, always. So if you add more water the dT value is smaller so now look at the equation as see how this smaller value affects the CF. Hope this helps

[cltf]

I've got my calorimetry sac coming up this week and am not too sure about errors and their effect on the calibration factor.
If you add too much water when calibrating, will the CF be higher or lower...

CF = E/T.c
So if insulation is bad less energy is recorded cause some is lost E decrease so CF decrease
     if more water is present, temp change is less as more energy would be required to raise temp by same amount. Thus T is greater, and CF decreases.

[nbhindi]

Hey guys, I have a SAC on calorimetry tomorrow and I'm not sure how to answer the following question:

CuSO4(aq) + Zn(s) = Cu(s) + ZnSO4  DH= -210.6kJ/mol

Why does the DH value refer to mole of CuSO4 added rather than Zn?

Any help will be greatly appreciated
Thanks

[nacho]

is that an actual question that you have been given?
Because that's not true, it refers to either one

[nbhindi]

Yeah that's what I though. We did a practical about enthalpy and heat of reactions and that was one of the analysis questions. Also, I am having difficulty answering this question:

For the same exothermic reaction, would the strength of the reactant bonds be greater or less than the strength of the product bonds in the reaction? Why?

Thanks

[b^3]

Hey guys, I have a SAC on calorimetry tomorrow and I'm not sure how to answer the following question:

CuSO4(aq) + Zn(s) = Cu(s) + ZnSO4  DH= -210.6kJ/mol

Why does the DH value refer to mole of CuSO4 added rather than Zn?

Any help will be greatly appreciated
Thanks

I don't know if this is right, but it makes sense.
If in your expirement, CuSO4 was the limiting reactant, then you have to use the number of mol of it to work out the change in H, as the mol of Zn wouldn't be refering to the reaction

For the other part, I think this is a checkpoints questions that I have come across.
It's exothermic so energy is being released when bonds are broken, resluting in bonds of a lower energy state. Lower energy state bonds are stronger because more energy is needed to break them, than of bonds who already have a higher energy.

So the strenght of the bonds in the reactants are weaker compared to the strenght of the bonds in the products as the energy of the products is less than the energy of the reactants.

EDIT: Fixed answer.

[nbhindi]

Hey b^3 thanks for answering the second question. However, for the first question I thought something similar but CuSO4 was in excess...wtf??

[nacho]

Hey guys, I have a SAC on calorimetry tomorrow and I'm not sure how to answer the following question:

CuSO4(aq) + Zn(s) = Cu(s) + ZnSO4  DH= -210.6kJ/mol

Why does the DH value refer to mole of CuSO4 added rather than Zn?

Any help will be greatly appreciated
Thanks

I don't know if this is right, but it makes sense.
If in your expirement, CuSO4 was the limiting reactant, then you have to use the number of mol of it to work out the change in H, as the mol of Zn wouldn't be refering to the reaction

but in terms of the equation, just simply the equation, there is no such thing as the limiting reagent
im still in disagreement as it is a 1:1 ratio

[b^3]

Hey guys, I have a SAC on calorimetry tomorrow and I'm not sure how to answer the following question:

CuSO4(aq) + Zn(s) = Cu(s) + ZnSO4  DH= -210.6kJ/mol

Why does the DH value refer to mole of CuSO4 added rather than Zn?

Any help will be greatly appreciated
Thanks

I don't know if this is right, but it makes sense.
If in your expirement, CuSO4 was the limiting reactant, then you have to use the number of mol of it to work out the change in H, as the mol of Zn wouldn't be refering to the reaction

but in terms of the equation, just simply the equation, there is no such thing as the limiting reagent
im still in disagreement as it is a 1:1 ratio

Yes that is correct in terms of the equation. I was referring to the expirement not the equation. The question doesn't specifically refer to either so it was a different interpretation. I was just trying to find a way to make the answer work because it's a stupid question.

[Mao]

Hey guys, I have a SAC on calorimetry tomorrow and I'm not sure how to answer the following question:

CuSO4(aq) + Zn(s) = Cu(s) + ZnSO4  DH= -210.6kJ/mol

Why does the DH value refer to mole of CuSO4 added rather than Zn?

Any help will be greatly appreciated
Thanks

I don't know if this is right, but it makes sense.
If in your expirement, CuSO4 was the limiting reactant, then you have to use the number of mol of it to work out the change in H, as the mol of Zn wouldn't be refering to the reaction

This. I was initially going to post something similar to this.

As a calorimetry exercise, I would imagine it is immersing a piece of Zinc in a CuSO4 solution. Since this is a spontaneous reaction, the reaction will complete until one of the reactant runs out. In this case, the most likely scenario is the copper solution is limiting and therefore the net amount reacted is the number of moles of copper initially present.