VCE Chemistry Question Thread

[mary1911997]

use the dilution formula.
C1V1 = C2V2

For the first reaction
C1 = 0.25M
V1 = 10mL
C2 = ?
V2 = 50mL

Sub these values into the equation to solve for the "new" concentration.

You don't need to use units in litres as the mL units will cancel out.

what about the other three reactions, what values would C1 V1 and V2 take

[Jakeybaby]

what about the other three reactions, what values would C1 V1 and V2 take

C1 is the concentration of the sodium thiosulfate, which is constant at 0.25M
V1 is the volume of the sodium thiosulfate, which changes i.e. 10, 25, 40, 45
V2 is the volume after dilution, which is also constant at 50mL.

[lzxnl]

what about the other three reactions, what values would C1 V1 and V2 take

Don't go around rote learning formulas. Think of it this way. You're adding water in your dilution (presumably; I haven't read much of the recent posts) and you haven't changed the amount of thiosulfate ions. So the number of thiosulfate ions is the same. Set this quantity before and after the dilution constant.

[HopefulLawStudent]

Sounds good to me 

Never actually seen this, but I assume you're referring to when an acid is added.
If an acid is added, you increase the overall hydronium ion concentration, and by Le Chaterlier's principle, the equilibrium has to shift to the left as it favours the backward reaction (since more H3O+).

After googling, the acidic properties of metal hydrides: https://en.wikipedia.org/wiki/Transition_metal_hydride
It says they may be weak acids. Weak acid means they don't fully ionise in water. I.e. they're equilibrium constant is low, and the backward reaction is preferred, hence equilibrium shifts to the left.

Again, this was just a guess from what I saw on the wikipedia page. Jyce or |zxn| may know the acidic properties of HMe's and be able to explain it better.

Thank youuuuuu!!! It makes so much sense now.

[HopefulLawStudent]

Hey:

Could someone please help me with question 32a)ii). Why does the reaction yield increase? By increasing the pressure, wouldn't that result in a net back reaction and therefore mean that the reaction yield [for SO3] decreases? Why do the answers say that it increases?

[HighTide]

Hey:

Could someone please help me with question 32a)ii). Why does the reaction yield increase? By increasing the pressure, wouldn't that result in a net back reaction and therefore mean that the reaction yield [for SO3] decreases? Why do the answers say that it increases?

Increasing pressure favours the side with fewer particles. The reactants side has 3. The products has 2. The forward reaction will occur to a greater extent thus more SO3 will be produced, hence concentration increases.

[HopefulLawStudent]

I can't count.

Thank you so much!

[sweetiepi]

How do you calculate the amount of energy released when 7.29 grams of Mg is burned?

The equation given is:

[Swagadaktal]

How do you calculate the amount of energy released when 7.29 grams of Mg is burned?

The equation given is:

Calculate mol Mg. So 7.29/24.3 = 0.3 mol
You know for 2 mol of Mg reacting in that equaiton, 1200kJ is released (kjmol-1 of that equation)
so 0.3 mols of 2 mols is 0.3/2 = 0.15  (the fraction of 0.3 mols in relation to that equation if that makes any sense)
Therefore the energy released is 0.15* 1200 kJ = 180 kJ

Pre sure that's how you work it out havent done many of these qs so if anyone picks up any errors lmk

[Jakeybaby]

Calculate mol Mg. So 7.29/24.3 = 0.3 mol
You know for 2 mol of Mg reacting in that equaiton, 1200kJ is released (kjmol-1 of that equation)
so 0.3 mols of 2 mols is 0.3/2 = 0.15  (the fraction of 0.3 mols in relation to that equation if that makes any sense)
Therefore the energy released is 0.15* 1200 kJ = 180 kJ

Pre sure that's how you work it out havent done many of these qs so if anyone picks up any errors lmk

Correct answer, although, my working out was slightly different.

We know that the thermochemical equation stated that 1200kJ of energy is released per 2 mol of Mg is burned, so therefore 600kJ of energy is released when burning 1 mol of Mg.

n(Mg) = 7.29/24.3 (as n=m/M)

Therefore:
Energy Released = (7.29/24.3) * (1200/2)
Energy Released = 180kJ

[Swagadaktal]

Correct answer, although, my working out was slightly different.

We know that the thermochemical equation stated that 1200kJ of energy is released per 2 mol of Mg is burned, so therefore 600kJ of energy is released when burning 1 mol of Mg.

n(Mg) = 7.29/24.3 (as n=m/M)

Therefore:
Energy Released = (7.29/24.3) * (1200/2)
Energy Released = 180kJ

OOH - different ways to answer stuff thanks for sharing <3

[sweetiepi]

Calculate mol Mg. So 7.29/24.3 = 0.3 mol
You know for 2 mol of Mg reacting in that equaiton, 1200kJ is released (kjmol-1 of that equation)
so 0.3 mols of 2 mols is 0.3/2 = 0.15  (the fraction of 0.3 mols in relation to that equation if that makes any sense)
Therefore the energy released is 0.15* 1200 kJ = 180 kJ

Pre sure that's how you work it out havent done many of these qs so if anyone picks up any errors lmk

Correct answer, although, my working out was slightly different.

We know that the thermochemical equation stated that 1200kJ of energy is released per 2 mol of Mg is burned, so therefore 600kJ of energy is released when burning 1 mol of Mg.

n(Mg) = 7.29/24.3 (as n=m/M)

Therefore:
Energy Released = (7.29/24.3) * (1200/2)
Energy Released = 180kJ

Thanks y'all!

[larissaaa_]

How does a catalyst increase the rate of reaction?

[sweetcheeks]

How does a catalyst increase the rate of reaction?

A catalyst increases the rate of reaction by lowering the activation energy required for a reaction to occur. Collision theory tells us that molecules are constantly colliding, however only a fraction of these has energy equal to or greater than the amount of energy required for the reaction to occur (that hump on the enthalpy graph). A catalyst lowers this hump, allowing molecules with less energy to collide successfully and react.

[geminii]

This question has been bugging me for quite some time now...what is the difference between core charge and electronegativity?
I know that core charge is the attraction between the nucleus and the valence electrons while electronegativity is the ability of an atom to attract electrons towards itself, but in terms of their trends in the periodic table, I'm not sure what the difference is.
For example, apparently core charge remains the same down a group but how can that be? Shouldn't it decrease since there are more shells, meaning the valence electron is further from the nucleus, so there is less attraction between the valence electrons and the nucleus?
Another example is how, in the explanations of why electronegativity decreases down a group and increases across a period, the textbook mentions attraction between the valence electrons and the nucleus, and not between the atom and other electrons, like the definition of electronegativity states.
Also I'm not sure of the difference between the reason why electronegativity increases across a period and why core charge increases across a period.
I am so confused by this. Any help is appreciated! Thanks