Chemistry 3/4 2013 Thread

[Bad Student]

Hey everyone, I've just got a couple of questions about a prac we're doing to determine the content of ethanol and sulfur dioxide in wine.

Why may the SO2 value differ from what is stated on the packaging?

We had to add 2M H2SO4 to a solution that had had NaOH added to it, why was it necessary to add the H2SO4 to only one flask, titrate, and then move on to the next?

Thank you.

The SO2 value may differ from what is stated on the packaging because the value on the package is an average value of SO2 so that particular bottle of wine might be slightly different. Also, the winery might have crap quality control.

The H2SO4 is only added to one flask at a time because it probably absorbs water from the air or something so you have to keep the H2SO4 inside the container for as long as possible to avoid being diluted by water absorbed from the air.

[9_7]

The mercury level in oyster is determined by dissolving 2g sample of oyster in 50mL sulfuric acid. This solution is then diluted to 400mL and a 10mL sample of the diluted solution is detected by AAS. The absorbance gave a reading of 0.45. from the calibration curve, this reading corresponded to a concerntration of 0.05ppm. Find the amount of mercury(g) in the 2g oyster sample.

anyone?

[Reckoner]

From the information given we know that the concentration of mercury in the 10ml sample is 0.05ppm. We know that 0.05ppm is the same concentration as 0.05 mg/L; which therefore is the same as 0.05 X 10^-3 g/L.

So we know that the concentration of mercury in the 10ml sample is 0.00005 g/L. Can you figure out the rest from here? Further explanation is below.

Spoiler

The 10ml sample came from the 400ml solution, so therefore the concentration of the 2 solutions is the same.

So we can then calculate the mass of mercury in this 400ml solution

(remember, we need the 400ml to be in Litres)
of mercury in the 400ml solution

Now, all of this mercury came from the 2g oyster sample (which was then diluted to become 400ml

So we have our answer of 2 x 10^-5 grams of mercury

The reading of 0.45 was not necessary!

[9_7]

From the information given we know that the concentration of mercury in the 10ml sample is 0.05ppm. We know that 0.05ppm is the same concentration as 0.05 mg/L; which therefore is the same as 0.05 X 10^-3 g/L.

So we know that the concentration of mercury in the 10ml sample is 0.00005 g/L. Can you figure out the rest from here? Further explanation is below.

Spoiler

The 10ml sample came from the 400ml solution, so therefore the concentration of the 2 solutions is the same.

So we can then calculate the mass of mercury in this 400ml solution

(remember, we need the 400ml to be in Litres)
of mercury in the 400ml solution

Now, all of this mercury came from the 2g oyster sample (which was then diluted to become 400ml

So we have our answer of 2 x 10^-5 grams of mercury

The reading of 0.45 was not necessary!

yay thank you!!

[Will T]

One propene molecule will react with excess oxygen to produce three molecules of water and three molecules of carbon dioxide.
Balancing the reaction:
2C3H6 + 9O2 = 6CO2 + 6H2O.
I understand it says O2 is an excess reagent, but I still don't quite get how they can combine so that the reaction is C3H6 + nO2 = 3CO2 + 3H2O.

[Edward21]

My calculator seems to neglect what it already has worked out when I try and add more to it:

Question: Calculate the conc. of H+ ions in the final solution when 600mL of 0.10M HCl is mixed with 200mL of 0.10M NaOH

I understand that in the HCl there's 0.06mol of H+, and in the NaOH solutions there's 0.02mol of OH- ions, but when working back to find the H+ ions in the NaOH, I try and use the traditional [H+]=10^-14/[OH-] and it gives me an answer of 5X10^-13 mol of H+ in the NaOH solutions, which is fine. So logic would progress that I just add the mol of both H+ ion solutions, and divide it into the number of L of both solutions (0.8L), BUT when I try and add the 0.06mol I found in the HCl, the 5X10^-13 mol that I had just worked out disappears, eg. when I try to add the 0.06 figure, what comes out is just the 0.06mol!? where did the other figure go!? I know how to work this out, but my calculator isn't letting me!? Is there another way to work this out? HELP 

[hardworker]

Can some one please help me with balancing the electrons for the final part of this question

oxidation of SO3 2 to SO4 2-  I can work out redox half equations all the way to balancing it with H and H20 but I dont know what elements to calculate fir  balancing the electrons in the final part of the answer.

[Daenerys Targaryen]

One you get to the electrons you just add elections denoted by . You should only add them to one side of the equation, it is generally the side that has a higher charge and you want to bring it down to make it even to the opposite side.

[Reckoner]

My calculator seems to neglect what it already has worked out when I try and add more to it:

Question: Calculate the conc. of H+ ions in the final solution when 600mL of 0.10M HCl is mixed with 200mL of 0.10M NaOH

I understand that in the HCl there's 0.06mol of H+, and in the NaOH solutions there's 0.02mol of OH- ions, but when working back to find the H+ ions in the NaOH, I try and use the traditional [H+]=10^-14/[OH-] and it gives me an answer of 5X10^-13 mol of H+ in the NaOH solutions, which is fine. So logic would progress that I just add the mol of both H+ ion solutions, and divide it into the number of L of both solutions (0.8L), BUT when I try and add the 0.06mol I found in the HCl, the 5X10^-13 mol that I had just worked out disappears, eg. when I try to add the 0.06 figure, what comes out is just the 0.06mol!? where did the other figure go!? I know how to work this out, but my calculator isn't letting me!? Is there another way to work this out? HELP 

That's because 5x10^-13 is 0.0000000000005. So adding this onto 0.06 moles would would make next to no difference.

I think you're over complicating the question. Think of what might happen when you mix these two solutions togethor.
The equation H⁺ + OH^- ---> H₂O may help too

Further explanation below

Spoiler

we have 0.060 mol of H+ in the 600ml solution, and 0.020 mol of OH- in the other. So they can then react to form H₂O,with an excess of 0.040 moles of H+.

Therefore...
[H₃O⁺] = n/v
     = 0.040/0.800     (the volume is now 800ml)
     =0.50 M

 

[Edward21]

That's because 5x10^-13 is 0.0000000000005. So adding this onto 0.06 moles would would make next to no difference.

I think you're over complicating the question. Think of what might happen when you mix these two solutions togethor.
The equation H⁺ + OH^- ---> H₂O may help too

Further explanation below

Spoiler

we have 0.060 mol of H+ in the 600ml solution, and 0.020 mol of OH- in the other. So they can then react to form H₂O,with an excess of 0.040 moles of H+.

Therefore...
[H₃O⁺] = n/v
     = 0.040/0.800     (the volume is now 800ml)
     =0.50 M

Omg that makes sense, argh why didn't I think of the simplest equation in the book!?  thanks!

[Stick]

I don't know if this is just me, but I don't think I'm enjoying chromatography very much. I really liked the straightforward nature of the other analytical techniques discussed up until this point, but it seems to me that chromatography is very much so a theory-based topic with quite challenging problem-solving questions. Perhaps it's because I'm not learning it well, or because it's a new concept. Does anyone understand me?

[Daenerys Targaryen]

Chromotography sucks flopppppppp
I'm finding it really hard!!

[Jeggz]

Totally, totally understand you! I was enjoying the calculations and all, and now we've even finished the whole topic of Chromatography and I'm still quite clueless 

[Bad Student]

I hate the questions where there are the different dilution factors and different units of concentration like ppm and M. I wanna just plug numbers into formulas not think.

[Daenerys Targaryen]

Finding iron sample in 10.0g of ceral. It burnt to ash, then the ash was dissolved in nitric acid and diluted to 100mL with water.

The results were:
Sample
Cereal Solution Conc ?? absorbance: 0.410
1. conc 0 absorbance: 0.030
2. conc 2.5 absorbance: 0.158
3. conc 5.0 absorbance: 0.342
4. conc 7.5 absorbance: 0.451
5. conc 10.0 absorbance: 0.595

Concentration was measured in where

What mass of iron was present in the ashes of the 10.0g of ceral?
What is te concentration of iron, in mg per 100g, in the ceral?