VCE Chemistry Question Thread

[knightrider]

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

Thanks IndefatigableLover   

[Zailiner]

I'm asked to write a balanced ionic equation descrbing the reaction between MnO₄^- and Fe²⁺.

Here's what I've got after combining the two half equations I had previously written up:
MnO₄^-+8H⁺+5Fe²⁺ -> Mn²⁺+5Fe³⁺+4H2O

However, the answer at the back of the book does not include the resulting manganese ion. Why is this, and is it regular practice?

[IndefatigableLover]

I'm asked to write a balanced ionic equation descrbing the reaction between MnO₄^- and Fe²⁺.

Here's what I've got after combining the two half equations I had previously written up:
MnO₄^-+8H⁺+5Fe²⁺ -> Mn²⁺+5Fe³⁺+4H2O

However, the answer at the back of the book does not include the resulting manganese ion. Why is this, and is it regular practice?

I'd say your answer is correct and it's just a typo error at the back of the book


Also my post got lost a couple of pages back:
So I completed a LisaChem Exam for Unit 3 (no answers unfortunately) but I just had a few queries:

1. For UV-Vis, length of light path can be essentially be interpreted as length of cell from the Beer-Lambert Law correct?

2. For pH curves, I was meant to work out the volume at the equivalence point (to work out the concentration for an acid) so I estimated it but my final answer was not one of the answers on MCQ but was between two values which were close enough.. would VCAA give something like this?

3. When writing the formula of the parent ion for mass spectrometry, do I need to bracket the ion and add the '+' outside or can I write it without the bracket?

4. Is there a method in working out m/e values for substances or do we just have to "mix and match" and get the right numbers? Essentially my m/e ratio for my parent ion was 88 but the highest relative abundance occurred at 43 so I played around and got an answer (which mathematically works) but it was more trial and error than anything.. any particular method I could use?

5. When drawing peaks for relative abundance, I was told to draw the parent ion peaks for the compound BrCH2CH2Br and I labelled my m/e axis however when I was confused whether there was a specific height I had to draw to for the relative abundance? I drew them in ratios (due to the percentage abundance given in the question) but not sure about whether the actual height matters when drawing...

6. The last question was pretty easy but it kind of threw me off LOL: "What causes the absorbance of light by a species in the UV-vis region of the spectrum" but just checking.. it's to do with the electron configuration and how the electrons are promoted absorbing the energy given off by the light (which is similar to AAS but got confused that's all)..

[knightrider]

Using as caffeine

How many molecules of caffeine are in 1.00 g caffeine?
For this part i got the answer of 3.10x10^21 molecules


How many atoms altogether are in 1.00 g caffeine?
How would you do this part?

[Splash-Tackle-Flail]

Using as caffeine

How many molecules of caffeine are in 1.00 g caffeine?
For this part i got the answer of 3.10x10^21 molecules


How many atoms altogether are in 1.00 g caffeine?
How would you do this part?

I'm assuming the first bit is correct. In one molecule of caffeine there are (4+5+2+1) 12 atoms, so the number of atoms altogether would be 12x3.10x10^21 (which I don't have a calculator on me for)

[knightrider]

I'm assuming the first bit is correct. In one molecule of caffeine there are (4+5+2+1) 12 atoms, so the number of atoms altogether would be 12x3.10x10^21 (which I don't have a calculator on me for)

Thanks Splash-Tackle-Flail   

I did the exact same thing as you  but the book has the answer 7.44 × 10^25 atoms altogether?

[cosine]

Using as caffeine

How many molecules of caffeine are in 1.00 g caffeine?
For this part i got the answer of 3.10x10^21 molecules


How many atoms altogether are in 1.00 g caffeine?
How would you do this part?

For the second part, we know that the mol of caffeine is 0.0103mol, so we have 0.0103*6.02*10^23 molecules of CAFFEINE. But in each molecule of caffeine, we have an additional 12 atoms. So simply 12*0.0103*6.02*10^23

[Splash-Tackle-Flail]

For the second part, we know that the mol of caffeine is 0.0103mol, so we have 0.0103*6.02*10^23 molecules of CAFFEINE. But in each molecule of caffeine, we have an additional 12 atoms. So simply 12*0.0103*6.02*10^23

Wait, so is there 0.0103*6.02*10^23 molecules of CAFFEINE in a gram, or  3.10x10^21 molecules? I'm confused haha

[keltingmeith]

Wait, so is there 0.0103*6.02*10^23 molecules of CAFFEINE in a gram, or  3.10x10^21 molecules? I'm confused haha

Check the scientific notation, they're the exact same number...

[knightrider]

For the second part, we know that the mol of caffeine is 0.0103mol, so we have 0.0103*6.02*10^23 molecules of CAFFEINE. But in each molecule of caffeine, we have an additional 12 atoms. So simply 12*0.0103*6.02*10^23

Thanks cosine Where did you get the 0.0103mol from?How did you work it out?

Wait, so is there 0.0103*6.02*10^23 molecules of CAFFEINE in a gram, or  3.10x10^21 molecules? I'm confused haha

Same here 

[cosine]

Thanks cosine Where did you get the 0.0103mol from?How did you work it out?

Same here 

n(caffeine) = m(caffeine)/M(caffeine)
We have 1.00grams of caffeine, and we know that 1 mole of it weighs (48+5+28+16 = 97grams) 1/97= 0.0103mol (sorry I confused you splash-tackle for not using scientific notation, I get confused with it )

I hope it makes sense knightrider, it's simply just working out mol as any other compound/element!

[Splash-Tackle-Flail]

Check the scientific notation, they're the exact same number...

n(caffeine) = m(caffeine)/M(caffeine)
sorry I confused you splash-tackle for not using scientific notation, I get confused with it )

Oh, my bad then Think I probably need caffeine rn to think aha.. while we're on the topic..

[knightrider]

n(caffeine) = m(caffeine)/M(caffeine)
We have 1.00grams of caffeine, and we know that 1 mole of it weighs (48+5+28+16 = 97grams) 1/97= 0.0103mol (sorry I confused you splash-tackle for not using scientific notation, I get confused with it )

I hope it makes sense knightrider, it's simply just working out mol as any other compound/element!

Thanks cosine 

But if the question tells you that there are 3.10x10^21 molecules in 1.00 g caffeine.

Wouldn't you just multiply this by 12 to get the atoms altogether that are in 1.00 g caffeine?
Feeling confused 

[cosine]

Thanks cosine 

But if the question tells you that there are 3.10x10^21 molecules in 1.00 g caffeine.

Wouldn't you just multiply this by 12 to get the atoms altogether that are in 1.00 g caffeine?
Feeling confused 

Yeah that's correct!

[lzxnl]

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 )
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

Wait wait wait what? The middle CH2 hydrogens all have different proton environments...for starters the CH3-CH2 protons (the CH2 ones) will be split into 12 peaks whereas the other CH2 ones will be split into 9.