VCE Chemistry Question Thread

[keltingmeith]

Oh okay, so its like theres 1 carbon atom present in the given substance, to every 4 atoms of hydrogen? Why wouldnt we use weight though? Is it because the weights of elements vary, but the Molar mass also varies? Oh no im in a loop once again...

Think of it like this:

1 carbon atom : 4 hydrogen atoms

But, an atom and a molecule are almost interchangable, so:

1 carbon molecule : 4 hydrogen molecules

And we just said that the mole and molecule are basically the same thing, so:

1 mole of carbon : 4 moles of hydrogen

But, if we were to use weigh (say, grams), we'd get:

12 grams of carbon : 4 grams of hydrogen

And all of a sudden, our ratio is off. So, we use the mole to find our molecular formula.

[cosine]

Think of it like this:

1 carbon atom : 4 hydrogen atoms

But, an atom and a molecule are almost interchangable, so:

1 carbon molecule : 4 hydrogen molecules

And we just said that the mole and molecule are basically the same thing, so:

1 mole of carbon : 4 moles of hydrogen

But, if we were to use weigh (say, grams), we'd get:

12 grams of carbon : 4 grams of hydrogen

And all of a sudden, our ratio is off. So, we use the mole to find our molecular formula.

Thanks Euler, I sort of understand it now. Still suck at everything else though :/ Do you have any tips to really improve, I mean, I am willing to work for it but i have absolutely no idea where to start. Thanks man

[keltingmeith]

Thanks Euler, I sort of understand it now. Still suck at everything else though :/ Do you have any tips to really improve, I mean, I am willing to work for it but i have absolutely no idea where to start. Thanks man

Conceptual understanding. All. The. Way.

To be perfectly honest, I hated and sucked at chemistry. I did great in year 11, but then came year 12, and bad things happened. Then came uni, and I decided to take chemistry because "it was a good idea". The first thing we covered was the structure of an atom, and what an atom looked like. Once I got that down, all of a sudden, things became so easy.

Remember, chemistry is essentially the study of the atom, and so everything you do, you must be thinking of that atom. If you can picture that atom and what it's doing in everything you do in chemistry, things will become a lot easier.

[cosine]

Conceptual understanding. All. The. Way.

To be perfectly honest, I hated and sucked at chemistry. I did great in year 11, but then came year 12, and bad things happened. Then came uni, and I decided to take chemistry because "it was a good idea". The first thing we covered was the structure of an atom, and what an atom looked like. Once I got that down, all of a sudden, things became so easy.

Remember, chemistry is essentially the study of the atom, and so everything you do, you must be thinking of that atom. If you can picture that atom and what it's doing in everything you do in chemistry, things will become a lot easier.

Got it. Concepts, atoms, basics, will do!

Thank you, reassuring that you also endured difficulty in Chem in year 12.

I might be pushing your patience now, sorry. What would you recommend me to do in order to get down the basic things, atoms and stuff? Watch YouTube videos? Thanks!

[keltingmeith]

Got it. Concepts, atoms, basics, will do!

Thank you, reassuring that you also endured difficulty in Chem in year 12.

I might be pushing your patience now, sorry. What would you recommend me to do in order to get down the basic things, atoms and stuff? Watch YouTube videos? Thanks!

Whatever works for you, really. Watch YouTube videos, play with molecular modelling kits, or even go on to learn about the atom beyond VCE level (in particular, looking at electron orbitals and learning MO theory) if that works for you.

[MNM101]

Need help on question 10 on the attached photo

[Champ101]

(Image removed from quote.)

Need help on question 10 on the attached photo

n(CaCO3) in dolomite = n(CaO) produced
 = 1.24 g / 56.1 g mol-1
 = 0.0221 mol
m(CaCO3) in dolomite = 0.0221 mol x 100.1 g mol-1
 = 2.21 g
% CaCO3 in dolomite = [m(CaCO3) / m(dolomite)] x 100
 = (2.21 / 3.72) x 100
 = 59.5 %
Therefore the correct answer is D

[MNM101]

n(CaCO3) in dolomite = n(CaO) produced
 = 1.24 g / 56.1 g mol-1
 = 0.0221 mol
m(CaCO3) in dolomite = 0.0221 mol x 100.1 g mol-1
 = 2.21 g
% CaCO3 in dolomite = [m(CaCO3) / m(dolomite)] x 100
 = (2.21 / 3.72) x 100
 = 59.5 %
Therefore the correct answer id D

Thanks Champ

[Chang Feng]

Heyy I was just wondering how do you know the states of an element when your writing chemical equation. I know for obvious one of if the compound dissolves in water it is aqueous, but what about compound such as some hydrocarbon's= is there a way of knowing the states??
Thanks

[--whiteskies]

Heyy I was just wondering how do you know the states of an element when your writing chemical equation. I know for obvious one of if the compound dissolves in water it is aqueous, but what about compound such as some hydrocarbon's= is there a way of knowing the states??
Thanks

Someone here can probably give a better response but I'm pretty sure for hydrocarbons: methane/ ethane (usually gaseous), propane/ butane (is gaseous but can be liquid) and everything after that is pretty sure liquid? The longer the hydrocarbon chain, the more intermolecular forces there are and thus exhibits higher melting/ boiling temperatures (in this case, most likely a liquid at room temperature).

If a substance is insoluble, it would be a solid (indicating that it's a precipitate). Or if you're actually working with it in powder/ granular-form, obviously it's solid.

From experience, details such as states are really tricky and can be confusing. It just "depends" on the situation. I just remember the teachers saying something about VCAA would only expect you to know the easy ones and any weird/ difficult ones they'd usually give the state in the question.

Hopefully that helped!

[Kel9901]

Heyy I was just wondering how do you know the states of an element when your writing chemical equation. I know for obvious one of if the compound dissolves in water it is aqueous, but what about compound such as some hydrocarbon's= is there a way of knowing the states??
Thanks

General breakdown (generally works for VCE)
Metals
@ room temperature- all solid except mercury (liquid)
@ high temperatures for electrolysis- either molten or solid (generally Q will tell you, expect group 1/2/3 metals to be liquid, and remember the electrode is always solid)

Ionic salts (including common bases)- at room temperature, solid unless water is present, in which case look at/remember your solubility rules (most important things to remember are that nitrates, ammonium and group 1 salts are always soluble, that sulfates and halogen salts are usually soluble, and some exceptions like silver chloride and barium sulfate)
at high temperatures for electrolysis- liquid if it is the electrolyte

Carbon-containing non-metal molecules- this depends on the polarity+size of the molecule. Eg methane is gas at room temp, octane is liquid and 'really big hydrocarbon' is solid. In the presence of water, hydrocarbons that have undergone addition/substitution reactions and now have a -Cl, -Br, -OH, -NH2, etc functional group are often aqueous, but hydrocarbons that haven't are insoluble. 'Small' carboxylic acids are soluble in water, but 'large' carboxylic acids (eg fatty acids) are insoluble, and can be liquid or solid. Fats are generally solids, monosaccharides/disaccharides are soluble in water but otherwise solid, starch/cellulose is insoluble and a solid, amino acids are all soluble in water, but some more than others depending on their Z-group, proteins are insoluble (I think) and a solid. Esters are considered insoluble (generally liquid), but the alcohol and carboxylic acid reacting to form it are in the liquid state as very little water should be present (you'll learn about this in unit 4)

Generally high temperatures won't be used in these reactions, however a common exception is when ethene reacts with steam to produce ethanol. In this case everything is in the gaseous state. The other exception which you may be expected to know is fractional distillation, but they won't ask you to write states for that as they change (from liquid to gas).

Other stuff:
H2O depends on the temperature (you're expected to know boiling/melting points!)
Common acids (HCl HNO3 etc) are generally in the aqueous state
NH3 is soluble in water and is a gas at room temperature but at about -30-ish degrees it becomes liquid
Let me know if I missed anything

[Eiffel]

the pH of a 10^-9 M solution of HCL is ?

since c(h+) = c(hcl) = 10^-9M

-log(10^-9) = 9, but since its an acid has to be under 7, the answer is 7 (because the original question ask "is closest to") how so?

[keltingmeith]

the pH of a 10^-9 M solution of HCL is ?

since c(h+) = c(hcl) = 10^-9M

-log(10^-9) = 9, but since its an acid has to be under 7, the answer is 7 (because the original question ask "is closest to") how so?

Note that 10^-9 is an incredibly small number. In decimal form, it is 0.0000000001 (might have one too many zeroes). Since we have such a small amount of HCl in there, we must also consider the hydrogen concentration of the other chemicals in there - essentially, water. Remember the dissociation of water:



So, [H⁺]=[OH^-]. Since so much of what we have is just water, we can also use the identity:



Now, remember that we have two things in here - HCl and H2O, both of these giving us protons, so we get:



Subbing this into the above, we get:



Where x is the concentration of protons donated by water. Solving this for x, we get x=9.95*10^(-8), and now we can use pH=-log(H+)=-log(9.95*10^(-8))=7.002, which we round down to 7. Note that this is a slightly dodgy question, since we've had to assume that the solution is at 25 degrees.

[Kel9901]

Note that 10^-9 is an incredibly small number. In decimal form, it is 0.0000000001 (might have one too many zeroes). Since we have such a small amount of HCl in there, we must also consider the hydrogen concentration of the other chemicals in there - essentially, water. Remember the dissociation of water:



So, [H⁺]=[OH^-]. Since so much of what we have is just water, we can also use the identity:



Now, remember that we have two things in here - HCl and H2O, both of these giving us protons, so we get:



Subbing this into the above, we get:



Where x is the concentration of protons donated by water. Solving this for x, we get x=9.95*10^(-8), and now we can use pH=-log(H+)=-log(9.95*10^(-8))=7.002, which we round down to 7. Note that this is a slightly dodgy question, since we've had to assume that the solution is at 25 degrees.

just to emphasise the point, for fairly concentrated acids (c>10^-5 ish), we ignore the H+/OH- ions that is 'in'/due to the water as it is virtually negligible. However, where c<10^-5 ish, the naturally present H+/OH- ions become a significant factor and so must be taken into account. On the other hand, when c<10^-9 ish, the acid is what is negligible, so only the water needs to be taken into account and so the pH is about 7

[Eiffel]

Hey, was wondering if you could clarify what you said again. I've never come across needing to factor in water when doing these pH's, usually i get given the concentration , or work it out, then use mole ratios for the number of OH- or H+ or use the formula. Where (or why) did you get/need water from?

Thanks