Chemistry: Post your doubts here!

[Ericyong]

where is this ques? can you tell me?

http://www.scribd.com/mobile/doc/73549537

Page two first question

 

[Thought blocker]

http://www.scribd.com/mobile/doc/73549537

Page two first question

Vishal

 

[♣♠ Magnanimous ♣♠]

http://www.scribd.com/mobile/doc/73549537

Page two first question

Umm, idk
But we just have to do it by calculations as we calculate relative atomic mass. It's just we have to put x instead of percentage abundance and calculate it. It's all algebra

 

[♣♠ Magnanimous ♣♠]

Vishal

yar tuje pata hai me maths me bohot kaccha hu and isme calculations he aur wo bhi algebra to probs he

 

[Thought blocker]

yar tuje pata hai me maths me bohot kaccha hu and isme calculations he aur wo bhi algebra to probs he

RIP.

 

[♣♠ Magnanimous ♣♠]

http://www.scribd.com/mobile/doc/73549537

Page two first question

what's the answer?
I think it might be C or D coz the Si28 should be the highest in percentage abundance(relative abundance)

 

[♣♠ Magnanimous ♣♠]

kar raha hu ruk muje answer to pata chale kyuki i am a bit unsure in it

RIP.

 

[Ericyong]

what's the answer?
I think it might be C or D coz the Si28 should be the highest in percentage abundance(relative abundance)

The answer is B. I've done some calculations and still couldn't find it.

 

[♣♠ Magnanimous ♣♠]

The answer is B. I've done some calculations and still couldn't find it.

First i thought of B but i didn't have any reason
may be some other person help you as idk this

 

[Thought blocker]

Lemme see.

 

[♣♠ Magnanimous ♣♠]

Lemme see.

yeah see it as this is a type of algebra ques

 

[♣♠ Magnanimous ♣♠]

Naturally occuring silicon is a mixture of three isotopes, 28Si, 29Si, and 30Si. The relative atomic mass of silicon is 28.109. What could be the relative abundance of each of the three isotopes?

Help pls I can't figure it out

http://www.scribd.com/mobile/doc/73549537

Page two first question

ZaqZainab do solve this

 

[AbbbbY]

ZaqZainab do solve this

A: (91.1*28 + 7.9*29+30)/100 = 28.009
B: (92.2*28 + 4.7*29 + 3.1*30)/100 = 28.109

So, B.

The idea is to multiply the relative abundance with the isotopic mass and take the average of the abundances.

Do note, though, that sometimes the three or four or however many abundance percentages are given DO NOT add up to 100%. You've to be careful what you divide with!

 

[AbbbbY]

Hi can anyone please help me with the following questions-
View attachment 46260

The answer is B, if possible can someone please highlight the chiral centres cause i thought it should be C

View attachment 46261
The answer is D, i don't get this one at all.

View attachment 46262
The answer is B though I thought it can equally be A or B.

View attachment 46264
The answer is C

View attachment 46265
The answer is B

View attachment 46267
The answer is C, that is option 2 and 3 is correct.

View attachment 46268
The answer is D

Thank you so much for your effort and time!

I'm a bit late to the party, but I hope this helps
I've only answered the questions that the other guys didn't. Do let me know if I missed anything.

21:

A chiral center is one which has FOUR DIFFERENT groups attached to the central Carbon.

Citric acid has no carbon with 4 entirely different groups.

Isocitric acid, though, has 2 carbons with 4 entirely different groups.

As shown, lets call the top most 1 and the bottom most 3.

C1 has 2 Hydrogens so that's out.
C2 has H, COOH, CH(OH)COOH and CH2COOH. That's 4 different groups. Chiral.
C3 has H, OH, COOH and R (i.e the C(CH2COOH)HCOOH branch). Again, 4 different groups so chiral.

Let me know if this doesn't make sense and I'll draw out the entire acid in a fully displayed form that makes more sense.

20:
7. Simple really. Draw them out. Barely takes a minute if you've practiced chem enough.

In questions like this, your thought process should be something like this:
1- C3X6. 1 double bond.
2- 3 Carbon chain, so double bond position can only be one.
3- No branching possible
4- Make all possible structural isomers first by changing the position of Cl
5- Once you've drawn all structural isomers, identify the isomers which can have cis-trans.

You just have to add the number of isomers that can show cis-trans to the basic structural isomers. In our case, 2 show cis-trans and there are 5 basic, so 5+2 = 7 isomers.

I've also drawn them out for you incase you can't figure them out.

8:

Merely a game of numbers. Focus on the question, don't lose track and it's a freebie, really.

XCl2. Of this, X is 2.920
So, Cl2 = 5.287-2.920 = 2.367g

The idea is to get the mass of X which reacts with 1 mol of Cl2. That Mr = the Ar of X.
2.367 : 2.920
71 : y
y = 87.58 so, Sr.

 

[Ericyong]

A: (91.1*28 + 7.9*29+30)/100 = 28.009
B: (92.2*28 + 4.7*29 + 3.1*30)/100 = 28.109

So, B.

The idea is to multiply the relative abundance with the isotopic mass and take the average of the abundances.

Do note, though, that sometimes the three or four or however many abundance percentages are given DO NOT add up to 100%. You've to be careful what you divide with!

I know how to do it if i were to calculate the answers one by one, but i was actually looking for an alternative way to find it without doing that.

But i guess it's impossible since there's 3 unknown and 1 equation.

Thanks btw!

 

[AbbbbY]

I know how to do it if i were to calculate the answers one by one, but i was actually looking for an alternative way to find it without doing that.

But i guess it's impossible since there's 3 unknown and 1 equation.

Thanks btw!

What you're asking for is only possible when there are 2 isotopes. This question was merely testing your concept of finding the average atomic mass via the percentage abundance approach.

 

[UVYEAH]

I need help with the following question:
9701_w12_qp_23.pdf
Q5

The molecular formula C4H8O can represent a number of compounds which have different
functional groups and which show different types of isomerism.
Compounds H, J and K each have the molecular formula C4H8O.
In each of the molecules of H, J and K,
● the carbon chain is unbranched and the molecule is not cyclic,
● no oxygen atom is attached to any carbon atom which is involved in π bonding.
When compound H is reacted with sodium metal, a colourless fl ammable gas is produced.
Both J and K give an orange-red precipitate when reacted with 2,4-dinitrophenylhydrazine
reagent but only K reacts with Fehling’s solution.

Clearly, J and K are carbonyl groups. But that would indicate a C=O double bond exists, and that violates the condition they give us: "no oxygen atom is attached to any carbon atom which is involved in π bonding." because the C=O bond involves one π bond.
What am I doing/thinking wrong here?

 

[Nigarm]

Please help me! I have a problem solving these 2 questions :
1. Calculate the mass of water produced in the reaction between 0.1 L of hydrogen gas and
20 g of copper oxide (II).
2. What volume (in normal conditions) is occupied by a gas produced in the reaction
between 30 g of ethylene and the excess of hydrogen gas?

 

[Zepudee]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_42.pdf

Q3 (C) can someone explain?

 

[Zepudee]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
Q3(e) (iv) why is [CH3CO2Na] = 0.05 and [CH3CO2H]= 0.075 ?