Chemistry: Post your doubts here!

[asma tareen]

I could too say that.

But we can right ??

 

[Thought blocker]

But we can right ??

 

[sitooon]

By adding water ._.

I want the procedure

 

[asma tareen]

I want the procedure

Sorry m In AS._.

 

[AbbbbY]

0.2 mol of aluminium is burned in oxygen and the product reacted with 2.00 mol/dm-3 HCL. What is the minimum volume of acid required for a complete reaction?

A. 15 cm3
B. 20 cm3
C. 30 cm3
D. 60 cm3

Can someone please explain this to me? :/

Are you sure it's not 0.02 mol of Al? Doesn't make sense otherwise...

___

2Al + 3/2 O2 -> Al2O3

Al2O3 + 6HCl -> 2AlCl3 + 3H2O


0.2 mol of Al will give 0.1 mol of Al2O3.

1 mol of Al2O3 reacts completely with 6 mol of HCl
0.1 mol will react with 0.6mol of HCl.


conc = mol/vol
2.00=0.6/vol
vol = 0.3dm3 = 300cm3

Could you please recheck to confirm if the question is correct?

 

[AbbbbY]

paper5 :
how you would measure the independent variable ( concentration of HCl )
markscheme says : "Dilutes a range of volumes of HCl" how ?

Since we know that conc = mol/vol, take a fixed concentration of acid, lets say 0.2mol HCl and then dilute it like this.

1st Titre -> 50 cm3 acid 0 cm3 water
2nd Titre -> 40 cm3 acid 10 cm3 water
3rd Titre -> 30 cm3 acid 20 cm3 water
4th Titre -> 20 cm3 acid 30 cm3 water
5th Titre -> 10 cm3 acid 40 cm3 water

Conc of acid in exp 2:
moles of acid : conc = mol/vol => 0.2 = mol/(40/1000)= 0.008. Solution conc = 0.008/(50/1000) = 0.16 mol/dm3

Conc of acid in exp 3:
moles of acid: conc = mol / vol => 0.2 = mol/(30/1000) = 0.006. Solution concentration = mol / (50/1000) = 0.12 mol/dm3

And so on. You don't need to explain the working. Just boldly how you'd find the concentration of the acid in each titre
Get it?

 

[AbbbbY]

Can I please get the summary of organic chemistry..for AS in an easy form? Thank you

http://i.imgur.com/G2akAHB.jpg


Try zooming in to see the reactants/conditions. If they're not visible quote me or send me a VM and let me know I'll scan it and reupload. Idk if it's clear enough took a pic from my phone.

Also note of the of arrows is the other way around.. Primary Alcohol -> Aldehyde ->Carboxylic
The Aldyhyde -> Carboxylic arrows are the other way around.

Aldehyde -> Carboxylic reagents are K2Cr2O7 + Acid under reflux
and
Carboxylic -> Aldehyde reagents are LiAlH4 or NaBH4

 

[robinhoodmustafa]

Kindly help me in this.

 

[AbbbbY]

View attachment 38209

Kindly help me in this.

D

CxHy + O2 -> CO2 + H2O
0.2 mol -> 0.8 mol + 0.8 mol

Ratio wise, it's

1 : 4 : 4

so we need a hydrocarbon whose 1 mole will give you 4 moles of CO2 and H2O each.

Since we need 4 moles of CO2, options A and B are eliminated. Since we need 4 moles of H2O and not 2, option C is eliminated. If you're not comfortable with eliminating it like this, just combust equation C and D and see.

C4H4 + 5O2 -> 4CO2 + 2H2O
C4H8 + 6O2 -> 4CO2 + 4H2O

 

[AbbbbY]

Can someone please help me with bond angles? AB2/AB3 systems etc I've got absolutely no clue what it is. asma tareen

http://www.scribd.com/doc/140353885/A-Level-Chemistry-Factsheets

Page 16 of the document. Helped me a lot.

 

[robinhoodmustafa]

D

CxHy + O2 -> CO2 + H2O
0.2 mol -> 0.8 mol + 0.8 mol

Ratio wise, it's

1 : 4 : 4

so we need a hydrocarbon whose 1 mole will give you 4 moles of CO2 and H2O each.

Since we need 4 moles of CO2, options A and B are eliminated. Since we need 4 moles of H2O and not 2, option C is eliminated. If you're not comfortable with eliminating it like this, just combust equation C and D and see.

C4H4 + 5O2 -> 4CO2 + 2H2O
C4H8 + 6O2 -> 4CO2 + 4H2O

Thankyou.


How to take out moles from volume?
Check this q.

 

[AbbbbY]

Thankyou.


How to take out moles from volume?
Check this q.
View attachment 38211

C

Mass of 1cm3 of H2O(g) = 1g
Mol of H2O = 1/18 = 0.0556
1 mol occupies 24dm3
0.0556 will occupy 1.33 dm3

Volume of steam at 298K = 1.33 dm3
Volume of steam at 598K = 2.67 dm3

 

[halloffame]

Assalam Alaikum,

The A2 level exams are on the way ... plz do someone upload the A2 chemistry APPLICATIONS section notes.



I'll glad to appreciate you and THANKS !!!! in advanced

 

[halloffame]

Assalam Alaikum,

The A2 level exams are on the way ... plz do someone upload the A2 chemistry APPLICATIONS section notes.



I'll glad to appreciate you and THANKS !!!! in advanced

 

[sitooon]

Assalam Alaikum,

The A2 level exams are on the way ... plz do someone upload the A2 chemistry APPLICATIONS section notes.



I'll glad to appreciate you and THANKS !!!! in advanced

 

[AbbbbY]

Don't bother with this ancient stuff. Read the stuff at chemguide.co.uk.
The cambridge booklet is so horribly mundane and full of crap.

 

[sitooon]

Can someone tell me why
1- why they used 7.20 as pka in q 2c(iii) > http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_4.pdf

2- Find %by mass of CU in question 2(c) http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_43.pdf

 

[AbbbbY]

Can someone tell me why
1- why they used 7.20 as pka in q 2c(iii) > http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_4.pdf

2- Find %by mass of CU in question 2(c) http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_43.pdf

2-
Mols of S2O3 = Mols of Cu.
Mols of S2O3= 0.02 = mol/(19.5/1000)
= 0.00039 mol

0.00039 mol in 50cm3
so [Cu2+] = 0.00039/(50/1000) = 0.0078 mol/dm3

mols of copper in 50cm3 = 0.00039
mols of copper in 100cm3 = 0.00078
mass of copper in 100cm3 = 0.00078*63.5 = 0.0495 (this value is mass / 100cm3 thus percentage)

So 0.0495% copper. If you don't get this method, let me know I'll tell you the alternative method.

 

[periyasamy]

Hello guys ,really need help here.Can someone explain both the parts for me.Thanks a lot.......

 

[Thought blocker]

Full paper solved ---> 9701_wO7_qp_22
POST #30