Chemistry 3/4 2013 Thread

[Mr Keshy]

Went really really really bad on my first Chem SAC. Did not expect it at all. Teacher still has high hopes though

[Jeggz]

Oh dear! What happened? 

[hungover]

on my first chem SAC i got 31/35. could have got that full 35 (only made silly mistakes) but we only had 45 minutes of time, many students didnt even finish. I am still happy as i am the RANK 1.

[Mr Keshy]

Just a huge mind blank, and misunderstandings really. My teacher didn't seem disappointed but just told me ways I can improve (ie do more questions etc..). Awesome teacher!

It does suck a bit since this was a rather large SAC too!

[hungover]

ours was quite long as well. most people got 18-21.

[Edward21]

Standard Question: basically AAS of Pb in mussels, max allowance is 2.5ppm, there's a 1.50g sample, made up to 100mL, and through the calibration curve a value of 2.5ug/L, what is the ppm?

If there's 2.5ug/L, then X10^-6 = 2.5X10^-6g/L, but there's only 100mL so multiply by 0.1L = 2.5X10^-7g of Pb in 1.50g

I know there's a ppm formula: g of solute/g total solution X 1,000,000 so 2.5X10^-7g/1.50g X 1,000,000 = 0.17 ppm??? is that right? 

[Stick]

My teacher gave us this worksheet for the holidays but didn't provide us with any solutions. I'd really appreciate it if someone could do these so I can check my work.

[Limista]

Standard Question: basically AAS of Pb in mussels, max allowance is 2.5ppm, there's a 1.50g sample, made up to 100mL, and through the calibration curve a value of 2.5ug/L, what is the ppm?

If there's 2.5ug/L, then X10^-6 = 2.5X10^-6g/L, but there's only 100mL so multiply by 0.1L = 2.5X10^-7g of Pb in 1.50g

I know there's a ppm formula: g of solute/g total solution X 1,000,000 so 2.5X10^-7g/1.50g X 1,000,000 = 0.17 ppm??? is that right? 

Yeah that's what I got too!

But I didn't know that a ppm formula existed, nor do I believe one needs to use it. So what I did was pretty much the same as you:

m(undiluted extract) = 2.5 x 0.1 = 0.25ug = 0.25 x 10^(-6) g
 1.5g has 0.25 x 10^(-6) g
 10^6g has (0.25 x 10^(-6) x 10^6)/1.5 = 0.17ppm (w/w)

[Cappuccinos]

(Image removed from quote.)

My teacher gave us this worksheet for the holidays but didn't provide us with any solutions. I'd really appreciate it if someone could do these so I can check my work.

disclaimer:Bit rusty, so there could be mistakes
Q1
a) 0.15 ppm (approximately, guessing the gradient on the line is 0.3)

b) Dilulation factor = 500/5 = 100
    Therefore [Na+] in body fluid = 0.15ppm x 100 = 15 ppm

c) 15 ppm = 15 g Na+ in 1,000,000 ml of body fluid
     Hence in 1000mL (1L) = 1000/1,000,000  x 15g
     = 1.5 x 10^-2 g/L
     = 15mg/L
     No, contains 15 mg/L which is less than 319.7mg/L

Q2
a) 0.66 ppm (from graph again)

b) Dilulation factor = 500/5 = 100
    Therefore [Na+] in body fluid = 0.66ppm x 100 = 66 ppm

c) 66 ppm = 66g Na+ in 1,000,000 ml of body fluid
     Hence in 1000mL (1L) = 1000/1,000,000  x 66g
     = 6.6 x 10^-2 g/L
     = 66mg/L
     No, contains 66 mg/L which is  less than 253 mg/L

d)  In 0.4mL of body fluids = 0.4/1,000,000 x 66ppm
     = 2.6 x 10^-5 g in 0.4mL of  body fluid
     = 26 μg in 0.4mL of body fluid

Q3) 
a) Maximum mecury allowed in fish = 500 ppm (w/w)  i.e 500 g in 1,000,000 g of fish
Hence in 200 g = 200/1,000,000 x 500g
           = 0.1g

b) 0.4ppm (off the graph)

c) Dilution factor = 4
    Therefore [Hg] in fish = 0.4ppm x 4
     =2 ppm

d) 2 ppm = 2g in 1,000,000, g
    Hence in whole fish (200g)  = 200/1,000,000  x 2
    = 0.04 g
     Which is less than 0.1g, so yes it is safe to eat.

[Stick]

I got a different answer for the very last question: 0.0004g. I think it's just a typo on your part since your arithmetic is correct. Thanks so much for taking the time to do that for me.

[tae]

Can someone please help?
Need help with c. ii.

[Limista]

c. i) 400mg/L

c. ii) There have been a total of 2 dilutions. When extract dissolved in nitric acid, this does not count as a dilution.

Concentration of sample after 1st dilution = 400 x (100/25) = 1600mg/L

Mass of undiluted extract = 1600 x 0.5 = 800mg = 0.8g

Percentage purity of copper in anode = 0.8/0.855 x 100% = 93.6% (w/w)

[Homer]

how would i name and draw (CH3)2CHCH2CH3

[jgoudie]

CH₃
       \
        CH-CH₂-CH₃
       /
CH₃

Naming:
Longest chain is a 4 chain hydrocarbon  (butane)
there is a is a methyl group (CH₃) coming of the second carbon in it however this is the only possible carbon that can come off therefor the number is not needed.
-- Methyl Butane.


It would be better to draw bonds to each of the hydrogen also.

how would i name and draw (CH3)2CHCH2CH3

[Homer]

perfect thanks