# chem revision help

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exam paper link

http://www.ocr.org.uk/Images/144762-...d-elements.pdf

mark scheme link

http://www.ocr.org.uk/Images/142258-...ts-january.pdf

so the last part of the last question on this paper...

I understood by reading the mark scheme the v2+ :KMnO4 ratio is 5:3. However, I don't understand why the charge of the vanadium is 2+.. could someone explain this pls?

http://www.ocr.org.uk/Images/144762-...d-elements.pdf

mark scheme link

http://www.ocr.org.uk/Images/142258-...ts-january.pdf

so the last part of the last question on this paper...

I understood by reading the mark scheme the v2+ :KMnO4 ratio is 5:3. However, I don't understand why the charge of the vanadium is 2+.. could someone explain this pls?

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also need help with question 4dii of

http://www.ocr.org.uk/Images/131288-...d-analysis.pdf

mark scheme @

http://www.ocr.org.uk/Images/135165-...lysis-june.pdf

thanks in advance. i just swear there's nothing about metals with organic stuff in the textbook...

http://www.ocr.org.uk/Images/131288-...d-analysis.pdf

mark scheme @

http://www.ocr.org.uk/Images/135165-...lysis-june.pdf

thanks in advance. i just swear there's nothing about metals with organic stuff in the textbook...

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#3

(Original post by

exam paper link

http://www.ocr.org.uk/Images/144762-...d-elements.pdf

mark scheme link

http://www.ocr.org.uk/Images/142258-...ts-january.pdf

so the last part of the last question on this paper...

I understood by reading the mark scheme the v2+ :KMnO4 ratio is 5:3. However, I don't understand why the charge of the vanadium is 2+.. could someone explain this pls?

**Maths help**)exam paper link

http://www.ocr.org.uk/Images/144762-...d-elements.pdf

mark scheme link

http://www.ocr.org.uk/Images/142258-...ts-january.pdf

so the last part of the last question on this paper...

I understood by reading the mark scheme the v2+ :KMnO4 ratio is 5:3. However, I don't understand why the charge of the vanadium is 2+.. could someone explain this pls?

Mn reduced from +7 to +2 a change of 5e

V changes from +2 to +5 a change of 3e

For the numbers of e to balance the Mn reaction occurs three times for every five times the V reaction occurs. Total exchange of e is 15

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#4

(Original post by

also need help with question 4dii of

http://www.ocr.org.uk/Images/131288-...d-analysis.pdf

mark scheme @

http://www.ocr.org.uk/Images/135165-...lysis-june.pdf

thanks in advance. i just swear there's nothing about metals with organic stuff in the textbook...

**Maths help**)also need help with question 4dii of

http://www.ocr.org.uk/Images/131288-...d-analysis.pdf

mark scheme @

http://www.ocr.org.uk/Images/135165-...lysis-june.pdf

thanks in advance. i just swear there's nothing about metals with organic stuff in the textbook...

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(Original post by

True, alkyl metals are not mentioned in the textbook but in the question they tell you how they can be used so you are meant to read this and use the information to solve the question.

**TeachChemistry**)True, alkyl metals are not mentioned in the textbook but in the question they tell you how they can be used so you are meant to read this and use the information to solve the question.

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(Original post by

Oxiadtion state changes must be equal.

Mn reduced from +7 to +2 a change of 5e

V changes from +2 to +5 a change of 3e

For the numbers of e to balance the Mn reaction occurs three times for every five times the V reaction occurs. Total exchange of e is 15

**TeachChemistry**)Oxiadtion state changes must be equal.

Mn reduced from +7 to +2 a change of 5e

V changes from +2 to +5 a change of 3e

For the numbers of e to balance the Mn reaction occurs three times for every five times the V reaction occurs. Total exchange of e is 15

What I also dont understand is how we know that H2O is on the left side of the equation and H+ is on the right.. as in the equation given, it's the other way round.

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#7

(Original post by

would you mind explaining the specifics of how to answer the question and exactly what info would be used and guide me through? Thanks

**Maths help**)would you mind explaining the specifics of how to answer the question and exactly what info would be used and guide me through? Thanks

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#8

(Original post by

So are you saying the total of the oxidation numbers of either side of the equation must be equal?

What I also dont understand is how we know that H2O is on the left side of the equation and H+ is on the right.. as in the equation given, it's the other way round.

**Maths help**)So are you saying the total of the oxidation numbers of either side of the equation must be equal?

What I also dont understand is how we know that H2O is on the left side of the equation and H+ is on the right.. as in the equation given, it's the other way round.

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#9

(Original post by

So are you saying the total of the oxidation numbers of either side of the equation must be equal?

What I also dont usnderstand is how we know that H2O is on the left side of the equation and H+ is on the right.. as in the equation given, it's the other way round.

**Maths help**)So are you saying the total of the oxidation numbers of either side of the equation must be equal?

What I also dont usnderstand is how we know that H2O is on the left side of the equation and H+ is on the right.. as in the equation given, it's the other way round.

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(Original post by

See attached. I did this for one of my students last year.

**TeachChemistry**)See attached. I did this for one of my students last year.

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i hate organic chemistry!!! everything here seems like guesswork... some of the equations you have to figure out are confusing coz the structures are too complex..

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"I did this for one of my students last year"

Oh so you're a chemistry teacher? so stupid of me not to figure that from your username, haha

Oh so you're a chemistry teacher? so stupid of me not to figure that from your username, haha

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#13

Why are you saying this is organic? do you mix organic and redox in ocr or something?

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#14

(Original post by

the charge is 2+ because the electron change is 3? pls explain the logic..doesn't make sense to me. everything else was fine, mass etc. so thanks

**Maths help**)the charge is 2+ because the electron change is 3? pls explain the logic..doesn't make sense to me. everything else was fine, mass etc. so thanks

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#15

(Original post by

i hate organic chemistry!!! everything here seems like guesswork... some of the equations you have to figure out are confusing coz the structures are too complex..

**Maths help**)i hate organic chemistry!!! everything here seems like guesswork... some of the equations you have to figure out are confusing coz the structures are too complex..

Here's one I did some research on in USA.

https://en.wikipedia.org/wiki/Paclitaxel

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#16

(Original post by

Why are you saying this is organic? do you mix organic and redox in ocr or something?

**richpanda**)Why are you saying this is organic? do you mix organic and redox in ocr or something?

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(Original post by

V goes from 2+ to 5+

**TeachChemistry**)V goes from 2+ to 5+

http://www.ocr.org.uk/Images/131289-...d-elements.pdf

another question - 50cm^3 of 0.25moldm^-3 butanoic acid is added to 50cm^3 of 0.05moldm^-3 sodium hydroxide.

Ka of butanoic acid = 1.51 * 10^-5 so calculate the PH of the buffer solution

number of moles of CH3(CH2)2COO- is 0.0025, but why would the number of moles of this be equal to NaOH? How would you know?

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#18

(Original post by

kinda makes sense.. another way of looking at it could it be that if electrons are gained, they must be lost on the other side of the equation and vice-versa so the oxidation numbers on both sides must be equal - is this correct?

http://www.ocr.org.uk/Images/131289-...d-elements.pdf

another question - 50cm^3 of 0.25moldm^-3 butanoic acid is added to 50cm^3 of 0.05moldm^-3 sodium hydroxide.

Ka of butanoic acid = 1.51 * 10^-5 so calculate the PH of the buffer solution

number of moles of CH3(CH2)2COO- is 0.0025, but why would the number of moles of this be equal to NaOH? How would you know?

**Maths help**)kinda makes sense.. another way of looking at it could it be that if electrons are gained, they must be lost on the other side of the equation and vice-versa so the oxidation numbers on both sides must be equal - is this correct?

http://www.ocr.org.uk/Images/131289-...d-elements.pdf

another question - 50cm^3 of 0.25moldm^-3 butanoic acid is added to 50cm^3 of 0.05moldm^-3 sodium hydroxide.

Ka of butanoic acid = 1.51 * 10^-5 so calculate the PH of the buffer solution

number of moles of CH3(CH2)2COO- is 0.0025, but why would the number of moles of this be equal to NaOH? How would you know?

This is not an equilibrium

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#20

(Original post by

why isn't the number of moles of naoh equal to ch3(ch2)2cooh ? why is it just equal to ch3(ch2)2coo-?

**Maths help**)why isn't the number of moles of naoh equal to ch3(ch2)2cooh ? why is it just equal to ch3(ch2)2coo-?

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