VCE Chemistry Question Thread

[IndefatigableLover]

How would you do this question?

If atoms of calcium have a mass of 40.1 g,what is the mass of one calcium atom?

Take the mass of calcium and divide it by the number of atoms and you'll get the mass of one calcium atom

Essentially your final answer should be 6.68x10^-23 g

EDIT: Nice pick up for my silly calculation mistake xD

[knightrider]

Take the mass of calcium and divide it by the number of atoms and you'll get the mass of one calcium atom

Essentially your final answer should be 6.67x10^-23 g

Thanks IndefatigableLover   

i am getting 6.68x10^-23g on my calculator? not  6.67x10^-23 g

[knightrider]

How would you do this question?

What mass of iron (Fe) would contain as many iron atoms as there are molecules in 20.0 g water (?

[IndefatigableLover]

Thanks IndefatigableLover   

i am getting 6.68x10^-23g on my calculator? not  6.67x10^-23 g

Sorry my bad read it incorrectly (nice pick up)!

How would you do this question?

What mass of iron (Fe) would contain as many iron atoms as there are molecules in 20.0 g water (?

So here we'll dealing with the amount of atoms and molecules being the same and we know that we measure this in moles (that is they want to make it so that the moles for both are the same).

You're given the information to work out the holes for Water and after you work out the moles for that, you know that the amount for both Fe and Water are the same so the moles of Water is the same for Fe (and then you just plug it back into the formula n=m/M to work out the mass)!

[knightrider]

Sorry my bad read it incorrectly (nice pick up)!
So here we'll dealing with the amount of atoms and molecules being the same and we know that we measure this in moles (that is they want to make it so that the moles for both are the same).

You're given the information to work out the holes for Water and after you work out the moles for that, you know that the amount for both Fe and Water are the same so the moles of Water is the same for Fe (and then you just plug it back into the formula n=m/M to work out the mass)!

Thanks so much IndefatigableLover
How did you know that the amount of moles for both fe and water are the same?

[IndefatigableLover]

Thanks so much IndefatigableLover
How did you know that the amount of moles for both fe and water are the same?

Well if we go back to your question:
What mass of iron (Fe) would contain as many iron atoms as there are molecules in 20.0 g water (?
So essentially thorugh "as many", I interpreted it as the same amount of atoms and molecules for both Fe and Water. In order for us to attain the same amount of both, we use moles since they're measured on the same scale and can be compared to as a whole

[warya]

Why does the stationary phase in GC have to have a certain boiling point?

[IndefatigableLover]

Why does the stationary phase in GC have to have a certain boiling point?

I wouldn't say a "certain" boiling point but rather just a high boiling point (so that it won't vaporise) aka a non-volatile liquid. If the stationary phase were to have a low boiling point then when heated up to high temperatures, then the stationary phase would turn into a gas and may decompose (which wouldn't be good because then you'd have no stationary phase to adsorb/desorb onto)!

[Redoxify]

I wouldn't say a "certain" boiling point but rather just a high boiling point (so that it won't vaporise) aka a non-volatile liquid. If the stationary phase were to have a low boiling point then when heated up to high temperatures, then the stationary phase would turn into a gas and may decompose (which wouldn't be good because then you'd have no stationary phase to adsorb/desorb onto)!

Isn't the stationary in GC the silica gel or something similar located throughout column in the column oven, which allows for the gases that are vaporized to adsorb and desorb

[IndefatigableLover]

Isn't the stationary in GC the silica gel or something similar located throughout column in the column oven, which allows for the gases that are vaporized to adsorb and desorb

Well yes that's what you want your stationary phase to do when you inject your sample into the column coated with a liquid with a high boiling point (keeping in mind this is GLC we're referring to and not GSC).

[cosine]

Well yes that's what you want your stationary phase to do when you inject your sample into the column coated with a liquid with a high boiling point (keeping in mind this is GLC we're referring to and not GSC).

Can someone explain back titration to me, Im having a hard time understanding it, thanks.

[IndefatigableLover]

Can someone explain back titration to me, Im having a hard time understanding it, thanks.

What aspect of Back Titration do you not understand? Do the diagrams in the Heinemann textbook and Chemguide not help?

Basically in short, if you have a substance when reacted won't give you a sharp end point (due to a weak acid/base interaction) or is insoluble or volatile, then you can react the substance which behaves like a weak acid/base with a known concentration of a strong acid/base in excess and then with the unreacted amount of the strong acid/base left, you neutralise it with a strong acid/base in a normal titration to which you can then go back and work the amount reacted overall in the first titration.

[Redoxify]

How do we find the amount in mol of sodium hydroxide that does not react with the fertiliser solution, and why do we use this answer to work out the amount reacted, how does it work?

[paper-back]

With the molecule:
CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3
Do the middle: "-CH2-CH2-CH2-CH2-CH2-" have the same proton environment?

[IndefatigableLover]

How do we find the amount in mol of sodium hydroxide that does not react with the fertiliser solution, and why do we use this answer to work out the amount reacted, how does it work?

The amount reacted is simply the amount in excess (that is that does not react with the fertiliser solution) minus the initial amount. You can work out the amount that does not react with the fertiliser solution by reacting it with a strong acid/base with known concentration until they neutralise (which from there you can work out )

With the molecule:
CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3
Do the middle: "-CH2-CH2-CH2-CH2-CH2-" have the same proton environment?

Yeah they do! So in all you should have three different proton environments
Worse case scenario is to draw it out and just circle environments that are similar/dissimilar