VCE Chemistry Question Thread

[keltingmeith]

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

Well, we know it's in water because it's in a SOLUTION. We factored in water because the concentration was just so low.

If we had a higher concentration, we could factor in water (say, if [HCl]=10^(-2)), and still get the right answer. However, for the higher concentrations you can get away without considering water - in lower concentrations (like these) you HAVE to consider the water.

[Eiffel]

could someone please explain 4b in extended. http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2011chem1-w.pdf

i know the moles of the MgNH4PO4 (DENOTED BY x - m(6H20)) is same as MgNH4PO4.6H20 (x), however the moles of phosphorus will be fixed, as from part a. So now i find the % by mass of P2O5 in MgNH4PO4 which should be higher because i am dividing a x - m(6H20) number by a smaller number, hence higher percentage? Could someone give me there interpretation. The examiners report says.

also, why does hydrogen carbonate ion have the chemical formula . The H is and carbonate is hence H2CO ?

[Chang Feng]

Hello.
I'm currently struggling a bit with doing electrochemical questions= mainly the ones involving calorimetery. Specifically the formulas used in solving these problems= is their any intuition behind these formulas.
And also for faradays law. I'm struggling to grasp the concept of what is a charge, voltage, current and their relation in the formulas of e=vit and q=e/v. Thanks.

[cooldude123]

So now i find the % by mass of P2O5 in MgNH4PO4 which should be higher because i am dividing a x - m(6H20) number by a smaller number, hence higher percentage? Could someone give me there interpretation.

However, the molar mass is lower (due to the loss of the H2O) from the compound, so it balances out in the end- same percentage of P2O5

also, why does hydrogen carbonate ion have the chemical formula . The H is and carbonate is hence H2CO ?

The ion is the actually the intermediate form of the deprotonation of - which is formed by carbon dioxide dissolving in water +
( loses to become , which loses a proton to become

[cooldude123]

Hello.
I'm currently struggling a bit with doing electrochemical questions= mainly the ones involving calorimetery.

Are there any specific problems you had difficulty with?

I found that most of the equations used link together (it's possible to derive them yourself if you understand what each unit means), and it helps to know the definition of each unit - such as 1A = 1C/s, or 1V = 1J/C etc

[Eiffel]

if a sample of permethrin, C_21_H_20_CL_2_O_3, contains 0.480g of carbon then the mass of chlorine in the sample is...

i can do this question fine except one thing. When finding the mols of carbon, shouldnt it be 0.480/(21x12) , for the molar mass, rather than 0.480/12. The answer has the latter but why? theres 21 carbon atoms in in mol of permethrin.

[ras]

if a sample of permethrin, C_21_H_20_CL_2_O_3, contains 0.480g of carbon then the mass of chlorine in the sample is...

i can do this question fine except one thing. When finding the mols of carbon, shouldnt it be 0.480/(21x12) , for the molar mass, rather than 0.480/12. The answer has the latter but why? theres 21 carbon atoms in in mol of permethrin.

I think a small error you may be making is thinking in a mol of permethrin there are 21 carbon atoms. In a mol of permethrin, there are actually 21 moles of carbon atoms, and 2 moles of Cl atoms (which will be useful soon). In this question we don't know the moles of permethrin, but we are able to calculate the moles of C, the mole ratio of C to Cl, and thus the mass of Cl.

So, step 1 would be discovering how many moles there are in total of C, which is of course done through our favourite n=m/M, or n=0.480/12.0. So there are 0.0400 moles of C in the sample of permethrin.

Next, you would use the mole ratio of n(Cl)/n(C) in permethrin. In one mole of permethrin, there are 21 moles of C and 2 moles of Cl. Thus, the mole ratio of Cl to C is 2/21, or n(Cl)/n(C)=2/21

With some rearrangement, this equation becomes n(Cl)=n(C) x (2/21).  As we know n(C), we can then discover n(Cl)=0.0400 x (2/21) = 0.00381.

Then, of course, m=n(Cl) x M=0.135 g.

Hope this makes sense and helps (and is right... I'm not entirely certain that I haven't forgotten the whole chem course over 2 months)

[Eiffel]

However, the molar mass is lower (due to the loss of the H2O) from the compound, so it balances out in the end- same percentage of P2O5

The ion is the actually the intermediate form of the deprotonation of - which is formed by carbon dioxide dissolving in water +
( loses to become , which loses a proton to become

The mass of the phospurous is not affected by driving off the water. So therefore the ma's of the p2o5 is always fixed, but I am now dividing by an amount which has a smaller molar mass so hence smaller quantity of gram?
Therefore dividing the foxed number of p2o5 over this rather than the one which is hydrated it is higher??.

I think a small error you may be making is thinking in a mol of permethrin there are 21 carbon atoms. In a mol of permethrin, there are actually 21 moles of carbon atoms, and 2 moles of Cl atoms (which will be useful soon). In this question we don't know the moles of permethrin, but we are able to calculate the moles of C, the mole ratio of C to Cl, and thus the mass of Cl.

So, step 1 would be discovering how many moles there are in total of C, which is of course done through our favourite n=m/M, or n=0.480/12.0. So there are 0.0400 moles of C in the sample of permethrin.

Next, you would use the mole ratio of n(Cl)/n(C) in permethrin. In one mole of permethrin, there are 21 moles of C and 2 moles of Cl. Thus, the mole ratio of Cl to C is 2/21, or n(Cl)/n(C)=2/21

With some rearrangement, this equation becomes n(Cl)=n(C) x (2/21).  As we know n(C), we can then discover n(Cl)=0.0400 x (2/21) = 0.00381.

Then, of course, m=n(Cl) x M=0.135 g.

Hope this makes sense and helps (and is right... I'm not entirely certain that I haven't forgotten the whole chem course over 2 months)

I see there is 21 mol of carbon therefore wouldn't the molar mass be 12x21?? If it was C2 we would've done 12x2? But its 21 hence 12x21

[ras]

I see there is 21 mol of carbon therefore wouldn't the molar mass be 12x21?? If it was C2 we would've done 12x2? But its 21 hence 12x21

The thing is that if it was C2, and you used 12x2, you would be finding the moles of C2, not the moles of C. However, we want to find the amount of total C atoms in this question so that we can later use the mole ratio (which is when the 21 becomes a factor), thus we use the molar mass of C and not C21.

This is not making much sense, is it... Anyone else have a coherent and simple way of putting it, other than my rambling response?

[Eiffel]

That makes sense now. I thought they could only ask it one way (e.g. mols of C rather then differentiating with C21). So when they ask mols of C, we only use C and C21, we use C21, just to make sure?

Thanks

[ras]

That makes sense now. I thought they could only ask it one way (e.g. mols of C rather then differentiating with C21). So when they ask mols of C, we only use C and C21, we use C21, just to make sure?

Thanks

Exactly! Hardest part for me with these types of questions was determining which substance we needed to find the moles of, either C or C21, and then making sure that for C we use the molar mass of C and for C21 we use the molar mass of C21, as you've highlighted

[IndefatigableLover]

Scenario:
You're heating three aspirin tablets in 50.00mL of NaOH at 0.5190M
After cooling, the solution is then transferred to a 100.00 mL volumetric flask and the volume is made up to exactly the 100.00 mL mark. Aliquots of 20.00 mL of this solution were then titrated against HCl of concentration at 0.1232M.

I'm all good with the calculations part but then I don't know what to say for this:
Identify two safety precautions that would have been needed for this procedure and state your reasons.

[pi]

Just logically, you're using heat and some acid, you'd want to protect yourself against them, no?

[keltingmeith]

Scenario:
You're heating three aspirin tablets in 50.00mL of NaOH at 0.5190M
After cooling, the solution is then transferred to a 100.00 mL volumetric flask and the volume is made up to exactly the 100.00 mL mark. Aliquots of 20.00 mL of this solution were then titrated against HCl of concentration at 0.1232M.

I'm all good with the calculations part but then I don't know what to say for this:
Identify two safety precautions that would have been needed for this procedure and state your reasons.

Should I be wearing safety goggles? (*hint hint*: you never say no to this, if Mary-Rose hasn't already taught you this. )
Secondly, how are you heating the tablets? Is there a possibility of burns?
What if I have a headache? Can I take the aspirin?
And that's pretty low concentrated HCl! I won't need gloves for that, will I?

No matter how obvious it might seem, it's appropriate to put there. In one of my lab reports, the demonstrator took off marks for everyone who did not mention lab coat/safety goggles. (everyone in the lab lost one mark LOL)

[IndefatigableLover]

Just logically, you're using heat and some acid, you'd want to protect yourself against them, no?

Should I be wearing safety goggles? (*hint hint*: you never say no to this, if Mary-Rose hasn't already taught you this. )
Secondly, how are you heating the tablets? Is there a possibility of burns?
What if I have a headache? Can I take the aspirin?
And that's pretty low concentrated HCl! I won't need gloves for that, will I?

No matter how obvious it might seem, it's appropriate to put there. In one of my lab reports, the demonstrator took off marks for everyone who did not mention lab coat/safety goggles. (everyone in the lab lost one mark LOL)

LOLOL oh wow and it was right in front of me... over-complicating the question too much >.<
Thanks pi and EulerFan101!