Chemistry: Post your doubts here!

[qwertypoiu]

lol i dont even know how to tag ppl
I have to reply to their messages instead.

 

[My Name]

lol i dont even know how to tag ppl
I have to reply to their messages instead.

just put a @ before the members name if you want to tag

 

[qwertypoiu]

just put a @ before the members name if you want to tag

Thanks

 

[My Name]

Thanks

Welcome

 

[Xylferion]

View attachment 52189 View attachment 52190
HELP PLEASE EXPLAIN !!!!

The answer to question 37 is D and here's why.

Each of the 3 hydrocarbons contain double bonds, the addition of hot conc. kmno4 reiterates the fact that these are alkenes.

Hot conc. kmno4 has alkenes undergo strong oxidization, that is after the diols place themselves across the double bonds, they can be oxidized to COOH, CHO, CO or CO2.

For these specific 3 hydrocarbons you want to be able to figure out what products will form after bond cleavage at the double bonds.

Here's how you're supposed to approach it:

Hydrocarbon 1 -->

Hydrocarbon 2 -->

Hydrocarbon 3 -->

If the carbon partaking in the double bond is bonded to one hydrogen, it will oxidize completely to an aldehyde. This question specifically states that we want ketones in " both " which means there has to be two products with a ketone group. Hydrocarbon 1 is the only one that supports what the question is asking, so your answer has to be D.

Hope that helped.

 

[princess Anu]

Hydrogen bonding only takes place with Fluorine, Oxygen and Nitrogen.

Right, but then what could it be?

 

[Xylferion]

Right, but then what could it be?

CO2 + H2O = H2CO3

That is H+ added to [CO3]^2-

These undergo ionic bonding to form H2CO3 since 2 H+ ions are needed to balance the -2 charge on CO3.

 

[qwertypoiu]

The answer to question 37 is D and here's why.

Each of the 3 hydrocarbons contain double bonds, the addition of hot conc. kmno4 reiterates the fact that these are alkenes.

Hot conc. kmno4 has alkenes undergo strong oxidization, that is after the diols place themselves across the double bonds, they can be oxidized to COOH, CHO, CO or CO2.

For these specific 3 hydrocarbons you want to be able to figure out what products will form after bond cleavage at the double bonds.

Here's how you're supposed to approach it:

Hydrocarbon 1 --> View attachment 52200

Hydrocarbon 2 --> View attachment 52201

Hydrocarbon 3 --> View attachment 52203

If the carbon partaking in the double bond is bonded to one hydrogen, it will oxidize completely to an aldehyde. This question specifically states that we want ketones in " both " which means there has to be two products with a ketone group. Hydrocarbon 1 is the only one that supports what the question is asking, so your answer has to be D.

Hope that helped.

Wow thanks a lot for this, it really helped me.
I have a question though. Wouldn't the aldehydes formed me further oxidised to carboxylic acids?
Also you said, "they can be oxidized to COOH, CHO, CO or CO2." Under what conditions is CO formed? Thanks.

 

[Xylferion]

Wow thanks a lot for this, it really helped me.
I have a question though. Wouldn't the aldehydes formed me further oxidised to carboxylic acids?
Also you said, "they can be oxidized to COOH, CHO, CO or CO2." Under what conditions is CO formed? Thanks.

Oh sorry CO was meant to indicate a ketone formed, and yes the aldehydes can be oxidized further to COOH.

 

[qwertypoiu]

Oh sorry CO was meant to indicate a ketone formed, and yes the aldehydes can be oxidized further to COOH.

Alright thank you

 

[ashcull14]

TBH I wasn't sure about this but this is how I thought through this:
A is definitely wrong due to Le Chatelier's principle.
C seems to be wrong, we never studied Contact Process involving such catalysts.
D seemed wrong too, we've been taught the process is carried out at 450 degrees.
So I would select B.
Examiner Report on this Question:
"A question on the reaction conditions of the Contact process has not previously been asked, and Question 19 was set to test this. The reason why a temperature as high as 450 o C is used was only properly understood by 39% of candidates (the key B): at lower temperatures the V2O5 catalyst is ineffective. Curiously, given that the reaction is an exothermic one such that lower temperatures could be expected to move the equilibrium in the direction of the SO3 product, 32% chose distractor A believing that this would not be the case."

very well explained thnk u so mch

 

[ashcull14]

The answer to question 37 is D and here's why.

Each of the 3 hydrocarbons contain double bonds, the addition of hot conc. kmno4 reiterates the fact that these are alkenes.

Hot conc. kmno4 has alkenes undergo strong oxidization, that is after the diols place themselves across the double bonds, they can be oxidized to COOH, CHO, CO or CO2.

For these specific 3 hydrocarbons you want to be able to figure out what products will form after bond cleavage at the double bonds.

Here's how you're supposed to approach it:

Hydrocarbon 1 --> View attachment 52200

Hydrocarbon 2 --> View attachment 52201

Hydrocarbon 3 --> View attachment 52203

If the carbon partaking in the double bond is bonded to one hydrogen, it will oxidize completely to an aldehyde. This question specifically states that we want ketones in " both " which means there has to be two products with a ketone group. Hydrocarbon 1 is the only one that supports what the question is asking, so your answer has to be D.

Hope that helped.

nice diagrams THNK U VERRY MUCH

 

[DeadlYxDemon]

May 2011, Q1) part c)

 

[abdul moeed rana]

Need A2 INORGANIC chemistry notes.ASAP!!!

 

[Metanoia]

The answer to question 37 is D and here's why.

Hydrocarbon 3 --> View attachment 52203

If the carbon partaking in the double bond is bonded to one hydrogen, it will oxidize completely to an aldehyde. This question specifically states that we want ketones in " both " which means there has to be two products with a ketone group. Hydrocarbon 1 is the only one that supports what the question is asking, so your answer has to be D.

Hope that helped.

Nicely done diagrams and explanation.

I'll just like to add on that the 2nd product of the 3 compound would eventually be oxidized to HOOCCOOH and then to CO2.

 

[Metanoia]

Also,
What could be the bonding between C02 and H20 in fizzy drinks? is it Hydrogen bonding?

This is an instance where there could be confusion between the limited "school definition" and the wider scientific definition.

H-bond donor : H attached to N, O, F
H-bond acceptor: highly electronegative atom with partial negative charge N, O, F

In this case, we do have H-bonding.
H-bond donor is water, with H attached to O
H-bond acceptor is CO2, with the highly electronegative atom O.

Of course, by default of their electron clouds, there is also VDW forces of attraction among CO2 and H2O.

 

[Metanoia]

Equilibrium means that the rate of forward reaction is equal to the rate of backward reaction.
So they increased the temperature and still want it to be in Equilibrium?
When the temperature is increased the temperature more of products will be formed since forward reaction is exothermic....but since they want equilibrium both sides increase...
Did I get it right ?

Why are BOTH increasing?

Thank you for your help!

A <--> B

One way to think of it is that at higher temperatures, A and B both experience an increase in energies.
So a larger portion of A is able to overcome the activation energy of the forward reaction to form B.
At the same time, a larger portion of B is able to overcome the activation energy of the backward reaction to form A.

Therefore, both forward and backward reaction rate increases when temp increases. But they increase to different extents (depending on which direction is exo or end0) , leading to equilibrium shifts.

 

[Metanoia]

May 2011, Q1) part c)
View attachment 52224

i) mass of kerosene used = 8195 x 10.8 = 88506 kg = 88.506 tonnes

ii) Form the equation for complete combustion of kerosene
C14H30 + 21.5O2 --> 14CO2 + 15H2O

Hint: to save time, a balanced equation is not required, as it can be observed that 1 mole of C14H30 produces 14 moles of CO2

Moles of C14H30 burnt = 88 506 000/198 = 447 000

Moles of CO2 produced = (88 506 000/198) x 14 = 6 258 000

Mass of CO2 produced = 6 258 000 x 44 g = 275 tonnes

 

[My Name]

A <--> B

One way to think of it is that at higher temperatures, A and B both experience an increase in energies.
So a larger portion of A is able to overcome the activation energy of the forward reaction to form B.
At the same time, a larger portion of B is able to overcome the activation energy of the backward reaction to form A.

Therefore, both forward and backward reaction rate increases when temp increases. But they increase to different extents (depending on which direction is exo or end0) , leading to equilibrium shifts.

I'm getting a bit confused I think.
This is what I understood so far....

When the temperature increases both( A and B side) have an increase in energies.
So both the forward and backward reaction rate increases. But they increase to different extents (depending on which direction is exo or end0).

OR

When the temperature is increased and the forward reaction is exothermic the backward reaction increases but since the system needs to be in Equilibrium the forward reaction also increases(to retain the equilibrium).

What's right and what's not?
Please and Thank you!

 

[My Name]

Need A2 INORGANIC chemistry notes.ASAP!!!

Check here:
https://www.xtremepapers.com/community/threads/some-different-notes-website-are-available.10423/
might help?