Chemistry: Post your doubts here!

[PixelsLevls]

But i heard it should +/- 0.15 cm3 as we take readings three times, initial, final and titre is that correct?

The Uncertainty should be half of the smallest reading on your burette. So if the smallest reading is 0.1cm3, you uncertainty will +/- 0.05cm^3
So if you have a uncertainty of 0.15cm^3 then your smallest reading must've been 0.30cm^3! which is very unlikely.
Hope this helped

 

[Kanekii]

The Uncertainty should be half of the smallest reading on your burette. So if the smallest reading is 0.1cm3, you uncertainty will +/- 0.05cm^3
So if you have a uncertainty of 0.15cm^3 then your smallest reading must've been 0.30cm^3! which is very unlikely.
Hope this helped

Yes and thanks

 

[Hamnah Zahoor]

But i heard it should +/- 0.15 cm3 as we take readings three times, initial, final and titre is that correct?

Also you don't take readings three times it's only twice initial and final. Titre is the value you obtain via calculations.

 

[BlahBlahBlah_]

ANY IDEA REGARDING THE PAPER TOMORROW? PLEASE PLEASE TELL.

 

[Psyvlone007]

Any Guess paper for chem p33 tomorrow? or any individual guesses would be appreciated.

 

[Psyvlone007]

how was p33?

 

[Psyvlone007]

can it be that the value of Rough titration to be 0.1 cm lower than the original titration . Annd will we lose mark of Not showing working while taking the mean of best titration. Anyone?

 

[BlahBlahBlah_]

can it be that the value of Rough titration to be 0.1 cm lower than the original titration . Annd will we lose mark of Not showing working while taking the mean of best titration. Anyone?

Not sure about the mean thing, but the rough titration is supposed to be above the titre values. So you might lose a mark there.

 

[dr tahir]

any guess for p34

 

[BlahBlahBlah_]

any guess for p34

I heard organic is included, and the paper will be difficult than 33. Sorry.

 

[ShahdNaasan]

Can someone help with this question ?

 

[Hamnah Zahoor]

Can someone help with this question ?
View attachment 63679

Easier way to do these kind of questions is to make a fully displayed formula of the structure which will help you to recognise the answer more easily

Thus, CH2=CHCl is the answer. option B

 

[ShahdNaasan]

Easier way to do these kind of questions is to make a fully displayed formula of the structure which will help you to recognise the answer more easily
View attachment 63680
Thus, CH2=CHCl is the answer. option B

Thank you so much

 

[Hamnah Zahoor]

Thank you so much

You're welcome

 

[ShahdNaasan]

Can anyway explain this question or walk me through the answers ?

 

[aloo Kha]

any info on chem 34? is organic really included for qualitative analysis?

 

[Hamnah Zahoor]

Can anyway explain this question or walk me through the answers ?
View attachment 63684

I had the same problem in this question when I solved this paper but now after repeated tries I was able to solve it correctly Alhamdulillah.
First find the value of the constant K as the angle of deflection is directly proportional to the charge mass ratio
PROTON:
Angle=+15 Charge= + mass= 1
15= k(+/1) thus K=+15

Option 1 :
Charge= +1-2= -1
mass= (atomic mass) 1+2=3
k=+15
Angle= (+15)*(-1/3)= -5..............option 1 correct

Option 2 :
Charge = +3-5 = -2
mass=3+3 =6
K=+15
Angle = (+15)*(-2/6) = -5..................option 2 is also correct

Option 3 :
Charge :4-1 = +3
mass=4+5 =9
K=+15
Angle = (+15)*(+3/9) = +5.......................Option 3 is incorrect.

Ans: B

 

[Judy Jiang]

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A 0.005 mol sample of anhydrous calcium carbonate was completely thermally decomposed to give 100cm3 of gas measured at a certain temperature and pressure. In a separate experiment carried out at the same temperature and pressure, a 0.005mol sample of anhydrous calcium nitrate was completely thermally decomposed. The volume of gaseous products was measured. What total volume of gaseous products was produced from the calcium nitrate?


A 50cm3 B 100cm3 C 200cm3 D 250cm3

 

[FairyTail]

Guys does anyone have info on chem paper 34
If u do pls post

 

[Hamnah Zahoor]

p.p1 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica; -webkit-text-stroke: #000000} p.p2 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica; -webkit-text-stroke: #000000; min-height: 13.0px} span.s1 {font-kerning: none}

A 0.005 mol sample of anhydrous calcium carbonate was completely thermally decomposed to give 100cm3 of gas measured at a certain temperature and pressure. In a separate experiment carried out at the same temperature and pressure, a 0.005mol sample of anhydrous calcium nitrate was completely thermally decomposed. The volume of gaseous products was measured. What total volume of gaseous products was produced from the calcium nitrate?


A 50cm3 B 100cm3 C 200cm3 D 250cm3

First reaction :
CaCO3 ----------> CaO + CO2
0.05mol---------> 100cm3
1 mol of CaCO3 : 1 mol of gas
decomposes to give 100 cm3 of gas

Second reaction:
Ca(NO3)2 -----------> CaO + 2NO2 + 1/2O2
0.05 mol
1 mol of Ca(NO3)2 : 2.5 moles of gas ( NO2 & O2)
As the number of moles are the same thus,
1 mol --------100 cm3
2.5 mol ---------2.5*100 = 250 cm3