Maths 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?
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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...
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username986184
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(Original post by 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?
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
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(Original post by 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...
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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Maths help
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(Original post by 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.
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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(Original post by 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
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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(Original post by 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
Did you do (d) (i)?
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(Original post by 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.
Did you construct a half equation for the V oxidation step?
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(Original post by 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.
See attached. I did this for one of my students last year.
Attached files
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(Original post by TeachChemistry)
See attached. I did this for one of my students last year.
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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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
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richpanda
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Why are you saying this is organic? do you mix organic and redox in ocr or something?
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(Original post by 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
V goes from 2+ to 5+
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(Original post by 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..
That's not complex.

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

https://en.wikipedia.org/wiki/Paclitaxel
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(Original post by richpanda)
Why are you saying this is organic? do you mix organic and redox in ocr or something?
If you read the whole thread you will see we are talking about two completely separate questions.
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(Original post by TeachChemistry)
V goes from 2+ to 5+
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?
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(Original post by 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?
Acid + Base --> Salt + Water

This is not an equilibrium
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(Original post by TeachChemistry)
Acid + Base --> Salt + Water

This is not an equilibrium
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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(Original post by 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-?
There's an excess of the acid present.
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