andrew's really crappy chem thread

[nhmn0301]

hey thanks! how would i deduce the fact that K is a spectator ion in this case?

Sorry I made a huge mistake in the answer above by not balancing the reaction with MnO4(1-) right. (too sleepy at that time I guess). I've already edited it. Now, if I write down the balanced equation again but include the spectator ion, I should have:
KMnO4^1 +8H^1+ + 5e^1- => Mn^2+ + 4H2O & K+

In the reaction, it is the Mn atom that reacts. The K^+ ion remains in the same state. If you used, say, NaMnO4, you would get the same result, because it is the MnO4^- entirely that reacts, so the K^+ ion is not necessary. Btw, in this reaction, SO4 (2-) is a spectator ion as well, you shall use the same method and prove.
Hope this helps.

[soNasty]

Thank you mate !

[soNasty]

Galena is the naturally occurring ore containing lead(II) sulfide, PbS. It may also contain silver sulfide, Ag2S. A sample of galena of mass 50.00 g contains only lead and silver sulfides. Lead makes up 60.00% of the total mass. What mass of silver is in the sample?

help.. this looks so simple but i cant get the answer for some reason..
it's 13.36g

[lzxnl]

Galena is the naturally occurring ore containing lead(II) sulfide, PbS. It may also contain silver sulfide, Ag2S. A sample of galena of mass 50.00 g contains only lead and silver sulfides. Lead makes up 60.00% of the total mass. What mass of silver is in the sample?

help.. this looks so simple but i cant get the answer for some reason..
it's 13.36g

So...if lead is 60% of the total mass and it is 207.21/(207.21+32.1) of the mass of lead sulfide, the lead sulfide mass must be 60% divided by the previous fraction to be 69.3 of the total mass.

30.7% is Ag2S. The proportion that is silver is 2*107.9/(2*107.9+32.1), so multiply this proportion by 30.7% of 50 g to get 13.36g

[soNasty]

OK so I have a SAC coming up and it involves a redox/back titration to determine the ethanol content in wine.
We were told to record the brand of the wine and the manufacturer's specified alcohol content before we began the SAC.
In this case, it was 11.5% Alc/Vol.

What exactly does this 11.5% mean? Is it 11.5% (v/v) where there is 11.5ml of ethanol (alcohol) per 1000ml of wine? Is that how it works? The casket of wine is 4L. We are taking 10 ml of the wine to use, then dilute to 250ml, then take 20ml aliquots of.
If so, and using that info, am I (using the reaction formulas and current information I have about aliquot and pipetting measurements) able to start my calculations using that? Even if I haven't started titrating let alone received an average titre yet?

[jgoudie]

Yes using the amounts you have been given you can calculate an estimated titre.  The only thing you need to be careful of is %alc/vol = ml/100ml not ml/1000ml as you said.   You will also need to factor in the density of ethanol to convert ml/100ml into g/100g.  I won't go through the calculations but this should give you somewhere to start.

OK so I have a SAC coming up and it involves a redox/back titration to determine the ethanol content in wine.
We were told to record the brand of the wine and the manufacturer's specified alcohol content before we began the SAC.
In this case, it was 11.5% Alc/Vol.

What exactly does this 11.5% mean? Is it 11.5% (v/v) where there is 11.5ml of ethanol (alcohol) per 1000ml of wine? Is that how it works? The casket of wine is 4L. We are taking 10 ml of the wine to use, then dilute to 250ml, then take 20ml aliquots of.
If so, and using that info, am I (using the reaction formulas and current information I have about aliquot and pipetting measurements) able to start my calculations using that? Even if I haven't started titrating let alone received an average titre yet?

[soNasty]

How would I even calculate an estimated titre?
so 11.5% would mean 11.5ml of ethanol per 100ml of wine
So in 10ml of wine I'd have 1.15ml of ethanol
But working out an average titre before even titrating? That's confusing.

[soNasty]

bump lol

[jgoudie]

So far so good.  That 1.15ml of ethanol is in the 250ml.  You are taking 20ml aliquots out of this.  Thus you should know the amount of ethanol in each aliquot.  (convert to grams using the density of ethanol).  If you know the concentrations of each of the other reactants you should be able to work through the steps of the titration to calculate the theoretical average titre.

If you need help with back titrations here is a video to help you through the basic theory:

How would I even calculate an estimated titre?
so 11.5% would mean 11.5ml of ethanol per 100ml of wine
So in 10ml of wine I'd have 1.15ml of ethanol
But working out an average titre before even titrating? That's confusing.

[soNasty]

to convert it to grams would i go: m(ethanol)=0.00115 x 0.789
which would then give me a mass of 0.00090275g of ethanol in 250ml of water?

[jgoudie]

Yep.

to convert it to grams would i go: m(ethanol)=0.00115 x 0.789
which would then give me a mass of 0.00090275g of ethanol in 250ml of water?

[soNasty]

and then i'd be able to calculate mole using n=m/M
and that gives me 1.9625 x10^-5 mol
then would i calculate the concentration using V= 1.9625x10^-5/0.01 (10ml original sample)
I still dont get how this will help me in estimating a titre lol

[jgoudie]

You need to know the concentration of the other reactants you are using and think about the reactions that happen during the method.  Use stoic to calculate the titre.

and then i'd be able to calculate mole using n=m/M
and that gives me 1.9625 x10^-5 mol
then would i calculate the concentration using V= 1.9625x10^-5/0.01 (10ml original sample)
I still dont get how this will help me in estimating a titre lol