VCE Chemistry Question Thread

[lzxnl]

When it says 'nucleotides', it means base+deoxyribose sugar. So, if you want to separate a polynucleotide into its constituent nucleotide monomers, you are essentially breaking the existing bond between the phosphate group and the deoxyribose sugar molecule. If they question asked for how many waters to completely dissociate the molecule (i.e. including nucleotide hydrolysis), you would then to the 16 water molecules, 2 per nucleotide (needed to break the 2 covalent bonds between the sugar and the base, and the sugar and the phosphate group).

A nucleotide consists of a base + (deoxy)ribose + a phosphate.

Need I say any more?

[kawfee]

Before I explain it, is the answer C?

It's D

[kawfee]

- Also, do we need to know how to name and draw cycloalkanes?

- And for catalysts,

It increases the no./proportion of successful collisions BUT does NOT change/increase the TOTAL no. of collisions ?
And does this also apply for increase in temperature ?

- When stating the Ea/ Activation energy , do we need to include a positive sign e.g +350 kJ/mol OR just simply leave it as  350 kJ/mol

- And regardless of being the activ. energy of forward or reverse reaction, the activ. energy will always be positive?

Thank you.

[Robert123]

- Also, do we need to know how to name and draw cycloalkanes?

- And for catalysts,

It increases the no./proportion of successful collisions BUT does NOT change/increase the TOTAL no. of collisions ?
And does this also apply for increase in temperature ?

- When stating the Ea/ Activation energy , do we need to include a positive sign e.g +350 kJ/mol OR just simply leave it as  350 kJ/mol

- And regardless of being the activ. energy of forward or reverse reaction, the activ. energy will always be positive?

Thank you.

For a reaction to be successful, there are 3 criteria. Firstly, they have to collide, secondly they had to collide with sufficient amount of energy to break the bonds and thirdly they have to collide at the right angle. Now to increase the rate of reaction, we can either increase the amount of collision per unit time or increase the amount of energy of the particle so that a greater proportion of particles would have  amount of energy to react. What a catalyst does it lower the activation energy and thus a greater proportion of particles would   sufficent amount of energy to react.
An analogy of how a catalyst work can be thought of as climbing over a wall. If you have a high wall, not much people would be able to climb over it however, if you use a 'catalyst' to chop the wall in half, a greater proportion of people would be able to get over it compare to it initial state.

For activation energy, you don't need to include a sign however, the units are just kJ not Kj/mol.
Activation energy is always postive since you can't give a system negative energy to react
edit: actually the units are kJ/mol not just kJ, my bad

[lzxnl]

- Also, do we need to know how to name and draw cycloalkanes?

- And for catalysts,

It increases the no./proportion of successful collisions BUT does NOT change/increase the TOTAL no. of collisions ?
And does this also apply for increase in temperature ? nope, temeprature increase increases the number of collisions too as the particles move faster. Catalysts only decrease activation energy

- When stating the Ea/ Activation energy , do we need to include a positive sign e.g +350 kJ/mol OR just simply leave it as  350 kJ/mol should be fine

- And regardless of being the activ. energy of forward or reverse reaction, the activ. energy will always be positive? if you had a negative activation energy, it would imply that you need to lose energy to cause a reaction

Thank you.

[lzxnl]

Hello,

For question 15, I'm not too sure why the answer is (A). As the cell is 'generating electricity' isn't chemical energy -> electrical energy (discharging process). Thus, as the metal alloy acts as the negative electrode, the 2nd equation they've given us should represent the oxidation reaction where Ni(OH)2(s) + OH-(aq)->NiO(OH)(s) + H2O(l) + e-?

Could someone please point out any faults inherent within my explanation.

Thanks.

Generating electricity => discharging => more favourable reaction occurs
Hence, negative electrode = anode
Although I don't know what question you're referring to :\

[Bestie]

Hello
for the attached question below:
why can't acid base titration be used to determine the amount of aspirin produced?

the ans says: Acid–base titration could not be used as there will be unused salicylic acid in the reaction vessel, resulting in
an incorrect equivalence point.

Why would there be excess?

can someone please explain???

[bts]

how does a secondary alkanol oxidise? I know it doesn't produce a carboxylic acid?

Visual please?

thank you

[Yacoubb]

- Also, do we need to know how to name and draw cycloalkanes?

- And for catalysts,

It increases the no./proportion of successful collisions BUT does NOT change/increase the TOTAL no. of collisions ?
And does this also apply for increase in temperature ?

- When stating the Ea/ Activation energy , do we need to include a positive sign e.g +350 kJ/mol OR just simply leave it as  350 kJ/mol

- And regardless of being the activ. energy of forward or reverse reaction, the activ. energy will always be positive?

Thank you.

Yeah well a cycloalkane has the general formula CnH2n. If it's a cyclic molecule, count the number of carbon atoms, and if there are only single C-C bonds and only carbon and hydrogen consistutes the molecule, it's a cycloalkane.

What a catalyst does is it essentially lowers the activation energy of the reaction. Establishing what activation energy is, is the first key. Now, the activation energy is the MINIMUM amount of energy needed to break existing bonds within the reactant molecules. When the catalyst binds to the reactant molecule substrate, the bonds within the reactants are weakened. So, less energy is needed to disrupt and break these bonds, in order to reach the activated complex before new bonds are formed to give you products.

So, adding a catalyst decreases the activation energy, and a greater proportion of molecules now have enough energy to be able to react (since the energy of the system has been reduced). Thus, given that the reactant molecules collide with enough force in the correct alignment, you'll get a fruitful (or successful) collision occuring. In this way, the reaction rate increases.

You don't need to provide a sign for the activation energy. It'll always be positive anyway. The units are still kJ/mol though.

[Aurelian]

how does a secondary alkanol oxidise? I know it doesn't produce a carboxylic acid?

Visual please?

thank you

You will get a ketone. See here for a visual example.

[Yacoubb]

You will get a ketone. See here for a visual example.

We aren't required to know this though, right?

We're only required to know oxidation of primary alcohols to form carboxylic acids.

[bts]

thank you!

I'm not sure it was on the NEAP  trial?

just another question: with the one attached, if i washed the conical flask with water, shouldn't it give me a higher conc of acid becasue i diluted the c204 concerntration with the addition of water, so less mno4- required and i would think that its cause of a higher conc of mno4?

[allstar]

Explain how it can be demonstrated that chemical reactions are still occurring during a state of
equilibrium by introducing into this system a trace amount of the radioactive isotope of
hydrogen known as deuterium.

help please !

[Robert123]

Explain how it can be demonstrated that chemical reactions are still occurring during a state of
equilibrium by introducing into this system a trace amount of the radioactive isotope of
hydrogen known as deuterium.

help please !

Ok, so let look at both scenarios. If A system at equilibrium wasn't reacting, then all the deuterium would stay in the same form as the product (I need the equation to see where the hydrogen would be replaced).
Now if the equilibrium is still reacting (which it is is), then after sometime, there would be deuterium in both the reactant and the product (well, whether there is hydrogen) and the concentration wouldn't of change. You could then isolate the product and then get a fancy machine to confirm for the presence of deuterium.
Does that clarify it?

thank you!

I'm not sure it was on the NEAP  trial?

just another question: with the one attached, if i washed the conical flask with water, shouldn't it give me a higher conc of acid becasue i diluted the c204 concerntration with the addition of water, so less mno4- required and i would think that its cause of a higher conc of mno4?

It doesn't matter what the concentration is in the flask since the amount of moles reacting would be the same since all of it would be reacting.
Hypothetically, the conical flask could be half filled with distilled water before adding the c2O4 in, so long as it's not too dilute that you can't see the end point

We aren't required to know this though, right?

We're only required to know oxidation of primary alcohols to form carboxylic acids.

You don't need to understand anything involving the function or naming of ketones. Only thing they could give is draw the structure of a molecule given all this spectroscopoic data which they had done on a previous year (2009 I think it was)

[allstar]

thank you robert123

Has anyone done the VCAA sample?
for ii) the ans says
NaOH(aq) + HA(aq) ⇌ NaA(aq) + H2O(l)
n( HA) = n ( NaOH) = 14.4/1000 x 0.100 = 1.44 x 10‐3mol.
Conc. (HA) = 1.44 x 10‐3 x 1000/20.0 = 0.0720M

where they get the 20L?
I got 0.144M?