VCE Chemistry Question Thread

[J_Rho]

A simple definition is that polar molecules have charges that's not evenly distributed while nonpolar molecules are evenly distributed.

(Image removed from quote.)

Thanks! That image really helped! Polar molecules are like uneven/lopsided whereas nonpolar are even/same on each side?

[colline]

Thanks! That image really helped! Polar molecules are like uneven/lopsided whereas nonpolar are even/same on each side?

I can't remember where I got this from, but the rule of thumb is "symmetry prevents polarity" -- so if it's symmetrical, then the molecule is nonpolar. I think there are also instances where the molecule is asymmetrical but the charges are so weak it can be considered nonpolar, but those are exceptions.

[Sine]

Thanks! That image really helped! Polar molecules are like uneven/lopsided whereas nonpolar are even/same on each side?

The symmetry argument would mainly be for molecules with polar bonds in order to make sure the electron attractions cancel each other out.

Molecules can be asymmetrical and still non-polar e.g. any hydrocarbon chain that has branches

Back to the roots,
0 00000000000000000000

[student1081]

If n=cv is the formula to find the molarity of a solution, why can it be used in equilibrium equations to the find the concentration of gases? What's the difference between a solution and a gas? Is a gas a solution? Any help would be greatly appreciated!

[Chocolatemilkshake]

Another question...

For 3/4 Chem, do we have to know the difference between cells under alkaline conditions versus acidic conditions?

EDIT; also, is oxygen always used as the oxidant in fuel cells?

  Thanks!

[angrybiscuit]

Another question...

For 3/4 Chem, do we have to know the difference between cells under alkaline conditions versus acidic conditions?

EDIT; also, is oxygen always used as the oxidant in fuel cells?

  Thanks!

1) Yes! The pH of the electrolyte is very very important as it will change both half equations completely. If it's an acidic electrolyte H⁺ will be exchanged and if it's alkaline it's going to be OH^-

2) Also yes

[Chocolatemilkshake]

1) Yes! The pH of the electrolyte is very very important as it will change both half equations completely. If it's an acidic electrolyte H⁺ will be exchanged and if it's alkaline it's going to be OH^-

2) Also yes

Thanks angrybiscuit! Any chance you could direct me to a resource that explains this well as I'm not sure I fully understand everything I need to know about the differences between the two? (or could someone explain it?) Thank you!

[thatdumbstudent]

When calculating the voltage of a galvanic cell, why don't you double the E oxidant/E reductant value if you doubled the half equation to make a full equation?

this was a question someone asked in the atarnotes lecture today and it made me curious too 

[angrybiscuit]

Thanks angrybiscuit! Any chance you could direct me to a resource that explains this well as I'm not sure I fully understand everything I need to know about the differences between the two? (or could someone explain it?) Thank you!

Unfortunately, the way I learnt this was through VCAA questions during my time of doing practice exams. But let me try to explain it. _{(Sorry if there's any mistakes, I forgot most of it and stole this from Google whoops)}

Let's take an alkaline battery with zinc and manganese. You have a zinc (Zn) anode and a manganese dioxide (MnO₂)  cathode. The electrolyte is potassium hydroxide (KOH).

The half equations would be:   
Zn_(s) + 2OH^-_(aq) --> ZnO_(s) + H₂O_(l) + 2e^-
2MnO_2(s) + H₂O_(l) + 2e^- --> Mn₂O_3(s) + 2OH^-_(aq)

The overall equation is:
Zn_(s) + 2MnO_2(s) --> ZnO_(s) + Mn₂O_3(s)

Let's take an acidic battery with zinc and manganese. You have a zinc (Zn) anode and a manganese dioxide (MnO₂)  cathode. The electrolyte is acidic (at this moment I can't think of a specific electrolyte)

The half equations would be:   
Zn_(s) + H₂O_(l) --> ZnO_(s) + 2H⁺_(aq) + 2e^-
2MnO_2(s) + 2H⁺_(aq) + 2e^- --> Mn₂O_3(s) + H₂O_(l)

The overall equation is:
Zn_(s) + 2MnO_2(s) --> ZnO_(s) + Mn₂O_3(s)

So in both instances, the overall equation is the same. But the difference lies in the reduction and oxidation half equations. In the alkaline half cells, there's OH^- ions and in the acidic cells, there's H⁺

When dealing with alkaline cells, instead of H⁺ ions you write OH^- ions and then do what you usually do when balancing out the equations. Encountering these different type of cells will hopefully help you out better

[Chocolatemilkshake]

Unfortunately, the way I learnt this was through VCAA questions during my time of doing practice exams. But let me try to explain it. _{(Sorry if there's any mistakes, I forgot most of it and stole this from Google whoops)}

Let's take an alkaline battery with zinc and manganese. You have a zinc (Zn) anode and a manganese dioxide (MnO₂)  cathode. The electrolyte is potassium hydroxide (KOH).

The half equations would be:   
Zn_(s) + 2OH^-_(aq) --> ZnO_(s) + H₂O_(l) + 2e^-
2MnO_2(s) + H₂O_(l) + 2e^- --> Mn₂O_3(s) + 2OH^-_(aq)

The overall equation is:
Zn_(s) + 2MnO_2(s) --> ZnO_(s) + Mn₂O_3(s)

Let's take an acidic battery with zinc and manganese. You have a zinc (Zn) anode and a manganese dioxide (MnO₂)  cathode. The electrolyte is acidic (at this moment I can't think of a specific electrolyte)

The half equations would be:   
Zn_(s) + H₂O_(l) --> ZnO_(s) + 2H⁺_(aq) + 2e^-
2MnO_2(s) + 2H⁺_(aq) + 2e^- --> Mn₂O_3(s) + H₂O_(l)

The overall equation is:
Zn_(s) + 2MnO_2(s) --> ZnO_(s) + Mn₂O_3(s)

So in both instances, the overall equation is the same. But the difference lies in the reduction and oxidation half equations. In the alkaline half cells, there's OH^- ions and in the acidic cells, there's H⁺

When dealing with alkaline cells, instead of H⁺ ions you write OH^- ions and then do what you usually do when balancing out the equations. Encountering these different type of cells will hopefully help you out better

Thank you so much!

[Geoo]

These significant figures are giving me a headache.

So, here is the sum, 899 x 0.25. Now, from my knowledge 899 has 3 sig figures, and 0.25 has 2.
So my answer of 222.25, becomes 2.2 x 10^2. However this is what the answer is:



So who is right?

[sweetcheeks]

These significant figures are giving me a headache.

So, here is the sum, 899 x 0.25. Now, from my knowledge 899 has 3 sig figures, and 0.25 has 2.
So my answer of 222.25, becomes 2.2 x 10^2. However this is what the answer is:

(Image removed from quote.)

So who is right?

Where do the 899 and 0.25 come from? The reason I'm asking is that if the 0.25 is a definitive value (e.g. if you are asked to find 1/4 of a value), then you don't consider it when doing significant figures because that value is infinitely precise. If the 0.25 is the result of some sort of experimental or rounded value than you would consider it for significant figures.

[Geoo]

Where do the 899 and 0.25 come from? The reason I'm asking is that if the 0.25 is a definitive value (e.g. if you are asked to find 1/4 of a value), then you don't consider it when doing significant figures because that value is infinitely precise. If the 0.25 is the result of some sort of experimental or rounded value than you would consider it for significant figures.

Okay i'll just write out the question with my working outs.

So, calculate the energy that will be released from the combustion of 4.00g of methane. (the molar heat of combustion is given at 889 kj mol-1).
n(CH4) = 4.00/16.00 = 0.25 mol,   so 0.25/1.0=x/889,  this becomes 889 x 0.25 = x. x=unrounded 222.25, so to significant figures is 2.2 x 10^2. as there is only 2 significant figures in 0.25.

[whys]

Okay i'll just write out the question with my working outs.

So, calculate the energy that will be released from the combustion of 4.00g of methane. (the molar heat of combustion is given at 889 kj mol-1).
n(CH4) = 4.00/16.00 = 0.25 mol,   so 0.25/1.0=x/889,  this becomes 889 x 0.25 = x. x=unrounded 222.25, so to significant figures is 2.2 x 10^2. as there is only 2 significant figures in 0.25.

Correct me if I'm wrong. The number of significant figures in your answer should be the least amount of significant figures used in the question. In the question, both numerical values have 3 significant figures, so your answer should also be to 3 significant figures. The 0.25 was derived from your calculations, so is not in consideration when deciding how many sig figs to use (this is derived from the question itself).

[Geoo]

Correct me if I'm wrong. The number of significant figures in your answer should be the least amount of significant figures used in the question. In the question, both numerical values have 3 significant figures, so your answer should also be to 3 significant figures. The 0.25 was derived from your calculations, so is not in consideration when deciding how many sig figs to use (this is derived from the question itself).

I love how my chem teachers never tell me stuff, so your saying that since 0.25 is derived, that I should discount is and use the 889 which is three sig figs so it would be 222 instead of 2.2 x 10 ^2?
This is so confusing...