Prelim Chem Thread (1 Viewer)

[Snowflek]

Oh lmao... Thank you for explaining.. Im retarded lol

 

[collin599]

how do i study for a chem prac exam?

 

[pikachu975]

how do i study for a chem prac exam?

Probably go over the practicals you've done so far in class and remember the methods, the general idea of what the results should be, and go over variables/discussions.

 

[1008]

how do i study for a chem prac exam?

For practicals, you have to make sure that you have your theory linked to its practical applications. They would've told you at least what topic the exam would be from, so make sure that you're able to justify all the chemical reactions in that section/chapter with ease, i.e. linking any observations you find during your experiment with the principles. For example, if there's an experiment about the galvanic cell, you should be able to explain the redox reactions, i.e. in which directions do the electrons flow and why (hint: see standard potentials). You should be able to test all the sub-topics in the section, and be able to link any real-world phenomena to what you've learnt. I think the Conquering Chem book is really good for this. Dot point is also good. But again this depends on how much time you've left.

Hope this helps and good luck

 

[dragon658]

Can someone please tell me why when you're writing the equation for 'lithium ion from a lithium atom' it is Li -> Li(superscript +) + e-
And then for Bromine its Br + e- -> Br(superscript -)

Yeah so why is it that there's an electron on the left side for bromine and an electron on the right side for lithium?
Sorry if this is a stupid question its just I really don't know haha. Is it something to do with lithium being a metal and bromine being a non-metal?

 

[BlueGas]

Can someone please tell me why when you're writing the equation for 'lithium ion from a lithium atom' it is Li -> Li(superscript +) + e-
And then for Bromine its Br + e- -> Br(superscript -)

Yeah so why is it that there's an electron on the left side for bromine and an electron on the right side for lithium?
Sorry if this is a stupid question its just I really don't know haha. Is it something to do with lithium being a metal and bromine being a non-metal?

Have you learnt oxidation reduction reactions?

 

[dragon658]

I don't think so... but i think I figured it out myself. If it's an element that has an excess of electrons (lithium) then its got the electron on the right side and then if it's got a deficiency of electrons (bromine -1) then the electron is on the left in the equation.... also... what happens to ion formation equations where you make ammonium bromide. I ended up with 2AL +3Br2 --> 2Al6Br3 (but like that 6 is gone in the answer.. I know its because I do end up having 6 Brs anyway but is that someone you just do.. take the 6 away since it doesn't belong in the middle obviously?) And the only reason I know how to do the equations is because I remember crossing the charge over for equations.. so yeah.

 

[dragon658]

also.. are you just meant to know that oxygen exists as 02 and nitrogen as N2 etc.. what does 2 even refer to? Does it mean it's just a molecule.

 

[30june2016]

also.. are you just meant to know that oxygen exists as 02 and nitrogen as N2 etc.. what does 2 even refer to? Does it mean it's just a molecule.

They exist as diatomic molecules consisting of 2 atoms of the same element

H2, N2, F2, O2, I2, Cl2, Br2

(can remember them by the mnemonic: Have No Fear Of Ice C(l)old B(r)eer)

 

[dragon658]

Can someone please explain this... the question is - Consider .374g of aluminium chloride. It contains how many moles of a) aluminium chloride, which I worked out to be 2.81 x 10^-03. B) aluminium C) Chlorine.

I'm not quite sure how to get b and c. For b my thought process was to get the mass divided by the molar mass of aluminium by itself but that doesn't seem to be right.

 

[30june2016]

Can someone please explain this... the question is - Consider .374g of aluminium chloride. It contains how many moles of a) aluminium chloride, which I worked out to be 2.81 x 10^-03. B) aluminium C) Chlorine.

I'm not quite sure how to get b and c. For b my thought process was to get the mass divided by the molar mass of aluminium by itself but that doesn't seem to be right.

for AlCl₃, the mole ratio is 1:3
therefore you divide the number of mols for AlCl₃ as found in (A) and multiply by 1/4 to get number of mols Al
(or multiply by 3/4 to get number of mols Cl)

 

[dragon658]

Thanks heaps^. Another question... how many decimal places are you supposed to round off to when doing these mass and molecule and mole calculations? For atomic weights as well do I just take Aluminium as 27 or actually 26.98? And when I put the calculations into my calculator should I use the entire scientific answer instead of the rounded one?

 

[jathu123]

Thanks heaps^. Another question... how many decimal places are you supposed to round off to when doing these mass and molecule and mole calculations? For atomic weights as well do I just take Aluminium as 27 or actually 26.98? And when I put the calculations into my calculator should I use the entire scientific answer instead of the rounded one?

If any data is given in the question, always round up your answers to the lowest significant figures given in the question. For example "if 0.20 g of something reacts with 0.7255 g of something else" then your final answer should be in 2 sig figs, cause of 0.20 which has the least sig figs.

However, if there is no data given in the question, I heard that your final answer should be 4 sig figs (which is what is given in the periodic table). But not entirely sure about this part