Chemistry: Post your doubts here!

[ashcull14]

HELP PLEASE EXPLAIN !!!!

 

[qwertypoiu]

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HELP PLEASE EXPLAIN !!!!

I'm confused about the first one, which year / paper is it from?
As for the second question, is it D?

 

[ashcull14]

 

[ashcull14]

I'm confused about the first one, which year / paper is it from?
As for the second question, is it D?

its http://freeexampapers.com/A-Level/Chemistry/CIE/2006-Nov/9701_w06_qp_1.pdf
yes hows it D?
cn u xplain the one above aswell?

 

[ashcull14]

 

[qwertypoiu]

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HELP PLEASE EXPLAIN !!!!

H2SO4 is an acid, and it is considered to be more acidic than HCl because it can dissociate and form 2H+ ions.
When pure H2SO4 liquid is put in water, the first H+ released is like this:
H2SO4(l) + aq = H+(aq) + HSO4-(aq)
When they say that H2SO4 is a strong acid, it means this equilibrium is shifted towards the RIGHT side, meaning a lot of H+ ions are released, only few H2SO4 remain as molecules. (equilibrium shifted to right side.)
The dissociation can continue further, to remove another H+ ion:
HSO4-(aq) = H+(aq) + SO4 2-(aq)
However, this only happens to a slight extent, most of the HSO4- ions remain as they are and only a few dissociate to form H+ ions, this is what they mean when they say HSO4- is a WEAK acid. (equilibrium shifted to left)
So for 1 mol/dm^3 of H2SO4:
1. True, concentration of H+ ions is high, since, as we saw in the first equilibrium above, the H2SO4 dissociates to form lots of H+ ions
2. False, concentration of SO4 2- ions must be low, since the second equilibrium above is shifted to left side.
(Over here you can already conclude D as answer)
3. False, how can the concentration of HSO4- ions be equal to that of SO4 2- ions, seeing as the second equilibrium is shifted to left side?
So your answer is D.

 

[qwertypoiu]

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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."

 

[qwertypoiu]

View attachment 52194

Write out the reactions of the series:
CH4 + 2O2 --> CO2 + 2H2O
C2H6 + 3.5O2 --> 2CO2 + 3H2O
C3H8 + 5O2 --> 3CO2 + 4H2O
C4H10 + 7.5O2 --> 4CO2 + 5H2O
In the first case:
CH4 + 2O2 --> CO2 + 2H2O
10 20 10 20
10 of methane used up, 20 of oxygen used up, 10 of CO2 produced, (water is not a gas)
since 70 of oxygen was there at first, 50 remain, plus 10 CO2 to make 60cm^3 in total of gases remaining in the mixture, This already means answer is D. I tried with other equations and they all fit the graph, so answer is D, you may try for yourself and see it's true.

 

[princess Anu]

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."

But why would the catalyst not be effective :/
I have not been taught about any conditions required for functioning of catalyst :s and if that's the case then in haber's process too at low temperatures the catalyst would not be effective?

 

[princess Anu]

h bond

but how? CO2 oxygen is not bonded with an H :/

 

[qwertypoiu]

But why would the catalyst not be effective :/
I have not been taught about any conditions required for functioning of catalyst :s and if that's the case then in haber's process too at low temperatures the catalyst would not be effective?

You see the way chemical catalysts work is that they react with one of the reactants of the original reaction, and make a product that reacts with the second reactant from the original reaction, in a way that the overall reaction still occurs, and the catalyst is regenerated . In other words, it provides an alternative reaction route. Something like this:
X + C → XC (1)
Y + XC → XYC (2)
XYC → CZ (3)
CZ → C + Z (4)
Although the catalyst is consumed by reaction 1, it is subsequently produced by reaction 4, so for the overall reaction:

X + Y → Z

"Generic potential energy diagram showing the effect of a catalyst in a hypothetical exothermic chemical reaction X + Y to give Z. The presence of the catalyst opens a different reaction pathway (shown in red) with a lower activation energy. The final result and the overall thermodynamics are the same."

Basically a catalyst lowers the activation energy, but it does not make it zero. Therefore, you still need some energy for the reaction to start. So every catalyst can become ineffective at a certain temperature, a temperature so low that the activation energy of the alternative chemical reaction route is not exceeded.
Technically at Absolute Zero (0 kelvins), nothing can react, so use of catalysts are futile.

 

[Xylferion]

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

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

 

[qwertypoiu]

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

Hey can you help with this?

 

[qwertypoiu]

The part on the cis-trans is incorrect. We are talking of cycloalkenes, for ring structures less than 8 carbons, we will not expect to see the trans isomer due to the strain on the bonds.

bro can u help with this

 

[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]

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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?