# Edexcel Chemistry A2 Unit 5 ~ Wednesday 19th June 2013 (Now Closed) Watch

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#1561

(Original post by

if we were asked to show how zinc hydroxide is amphoteric by equations are these correct:

(Zn(OH)2(H2O)4] + H+ --> [Zn(OH)(H2O)5]+1

[Zn(OH)2(H2O)4] + OH- --> [Zn(OH)3(H2O)3]-1 +H2O

**LeaX**)if we were asked to show how zinc hydroxide is amphoteric by equations are these correct:

(Zn(OH)2(H2O)4] + H+ --> [Zn(OH)(H2O)5]+1

[Zn(OH)2(H2O)4] + OH- --> [Zn(OH)3(H2O)3]-1 +H2O

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#1563

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#1564

(Original post by

Why is the answer B? Isnt 1moldm-3 the requirement for a standard reaction?

Posted from TSR Mobile

**freakynerdlol**)Why is the answer B? Isnt 1moldm-3 the requirement for a standard reaction?

Posted from TSR Mobile

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#1566

can someone please explain combustion analysis calculations? i've always struggled with knowing with diatomic molecules and their Mr value.

i know for working out masses of example C you do (12/44) X mass of CO2. but if we are given a mass of oxygen do we divide it by 16 or 32? also if we're given moles do we times by 16 or 32 to get mass?

i know for working out masses of example C you do (12/44) X mass of CO2. but if we are given a mass of oxygen do we divide it by 16 or 32? also if we're given moles do we times by 16 or 32 to get mass?

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#1567

(Original post by

can someone please explain combustion analysis calculations? i've always struggled with knowing with diatomic molecules and their Mr value.

i know for working out masses of example C you do (12/44) X mass of CO2. but if we are given a mass of oxygen do we divide it by 16 or 32? also if we're given moles do we times by 16 or 32 to get mass?

**LeaX**)can someone please explain combustion analysis calculations? i've always struggled with knowing with diatomic molecules and their Mr value.

i know for working out masses of example C you do (12/44) X mass of CO2. but if we are given a mass of oxygen do we divide it by 16 or 32? also if we're given moles do we times by 16 or 32 to get mass?

_{2}produced and mass of H

_{2}O produced:

1 – Calculate the mass of C and H produced.

C is 12/44 x mass of CO2

H is 2/18 x mass of H20

2 – Subtract the mass of C & H from initial mass to find the mass of Oxygen produced.

3 – Calculate moles of C, H & O produced.

Mass of C / 12

Mass of H / 1

Mass of O / 16

4 – Find the empirical formula by putting moles in a ratio and dividing through by the smallest value.

There was also 1 question where the compound had N in it too, you had to work this out in a different way and then take it away from the initial mass along with C & H to find the mass of O.

If they gave you the mass of Oxygen you'd just divide by 16 although this is unlikely, it's more likely if they gave you anything that it'd be volume and you could use moles = volume / 24

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#1568

Hi All, I'm struggling a little bit on a 'Part C' question.

I've worked it through, and have found the moles of thiosulfate, and then the moles of iodine - but I'm unsure as to why the markscheme then suggests dividing the moles of iodine by three to get the moles of dichromate unreacted?

EDIT: This is June 2011

I've worked it through, and have found the moles of thiosulfate, and then the moles of iodine - but I'm unsure as to why the markscheme then suggests dividing the moles of iodine by three to get the moles of dichromate unreacted?

EDIT: This is June 2011

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#1570

(Original post by

You will be given the mass of the sample initially, mass of CO

1 – Calculate the mass of C and H produced.

C is 12/44 x mass of CO2

H is 2/18 x mass of H20

2 – Subtract the mass of C & H from initial mass to find the mass of Oxygen produced.

3 – Calculate moles of C, H & O produced.

Mass of C / 12

Mass of H / 1

Mass of O / 16

4 – Find the empirical formula by putting moles in a ratio and dividing through by the smallest value.

There was also 1 question where the compound had N in it too, you had to work this out in a different way and then take it away from the initial mass along with C & H to find the mass of O.

If they gave you the mass of Oxygen you'd just divide by 16 although this is unlikely, it's more likely if they gave you anything that it'd be volume and you could use moles = volume / 24

**GeorgeL3**)You will be given the mass of the sample initially, mass of CO

_{2}produced and mass of H_{2}O produced:1 – Calculate the mass of C and H produced.

C is 12/44 x mass of CO2

H is 2/18 x mass of H20

2 – Subtract the mass of C & H from initial mass to find the mass of Oxygen produced.

3 – Calculate moles of C, H & O produced.

Mass of C / 12

Mass of H / 1

Mass of O / 16

4 – Find the empirical formula by putting moles in a ratio and dividing through by the smallest value.

There was also 1 question where the compound had N in it too, you had to work this out in a different way and then take it away from the initial mass along with C & H to find the mass of O.

If they gave you the mass of Oxygen you'd just divide by 16 although this is unlikely, it's more likely if they gave you anything that it'd be volume and you could use moles = volume / 24

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#1571

Hey guys!

I really need some help with the January 2012 paper question 21(b)(ii)...how is it that you select a suitable metal for the reaction by considering the standard electrode potential? Can someone please explain? Thanks in advance

I really need some help with the January 2012 paper question 21(b)(ii)...how is it that you select a suitable metal for the reaction by considering the standard electrode potential? Can someone please explain? Thanks in advance

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#1572

(Original post by

Hi All, I'm struggling a little bit on a 'Part C' question.

I've worked it through, and have found the moles of thiosulfate, and then the moles of iodine - but I'm unsure as to why the markscheme then suggests dividing the moles of iodine by three to get the moles of dichromate unreacted?

EDIT: This is June 2011

**Hunarench95**)Hi All, I'm struggling a little bit on a 'Part C' question.

I've worked it through, and have found the moles of thiosulfate, and then the moles of iodine - but I'm unsure as to why the markscheme then suggests dividing the moles of iodine by three to get the moles of dichromate unreacted?

EDIT: This is June 2011

Hope it helps

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#1573

(Original post by

Hey guys!

I really need some help with the January 2012 paper question 21(b)(ii)...how is it that you select a suitable metal for the reaction by considering the standard electrode potential? Can someone please explain? Thanks in advance

**imasha.sj**)Hey guys!

I really need some help with the January 2012 paper question 21(b)(ii)...how is it that you select a suitable metal for the reaction by considering the standard electrode potential? Can someone please explain? Thanks in advance

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#1574

(Original post by

CR3+ turns into cr2+, the E cell = -0.41, and its on right hand side since it gets reduced, so the left hand cell must have a E value more negative than -0.41 so the overall E is positive.

**clydeshen411**)CR3+ turns into cr2+, the E cell = -0.41, and its on right hand side since it gets reduced, so the left hand cell must have a E value more negative than -0.41 so the overall E is positive.

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#1575

(Original post by

Thank you but why can't we use any metal other than the ones mentioned in the markscheme?

**imasha.sj**)Thank you but why can't we use any metal other than the ones mentioned in the markscheme?

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#1576

**GeorgeL3**)

You will be given the mass of the sample initially, mass of CO

_{2}produced and mass of H

_{2}O produced:

1 – Calculate the mass of C and H produced.

C is 12/44 x mass of CO2

H is 2/18 x mass of H20

2 – Subtract the mass of C & H from initial mass to find the mass of Oxygen produced.

3 – Calculate moles of C, H & O produced.

Mass of C / 12

Mass of H / 1

Mass of O / 16

4 – Find the empirical formula by putting moles in a ratio and dividing through by the smallest value.

There was also 1 question where the compound had N in it too, you had to work this out in a different way and then take it away from the initial mass along with C & H to find the mass of O.

If they gave you the mass of Oxygen you'd just divide by 16 although this is unlikely, it's more likely if they gave you anything that it'd be volume and you could use moles = volume / 24

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#1577

(Original post by

umm, i guess it's because metals in items 1-5 are too reactive? So it's difficult to get a solid electrode of pure Na for example.

**clydeshen411**)umm, i guess it's because metals in items 1-5 are too reactive? So it's difficult to get a solid electrode of pure Na for example.

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#1578

(Original post by

Hi George, the notes u shared are amazing! but i do not understand how reducing a nitrile then add a weak acid can produce amines? i thought reduce nitrile will give aldehyde.

**clydeshen411**)Hi George, the notes u shared are amazing! but i do not understand how reducing a nitrile then add a weak acid can produce amines? i thought reduce nitrile will give aldehyde.

When you add LiAlH

_{4 }and a dilute acid, it reduces the nitrile group down to a primary amine.

As always,

**chemguide**can probably explain it better than I can.

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#1579

(Original post by

Remember a nitrile group is -C≡N

When you add LiAlH

As always,

**GeorgeL3**)Remember a nitrile group is -C≡N

When you add LiAlH

_{4 }and a dilute acid, it reduces the nitrile group down to a primary amine.As always,

**chemguide**can probably explain it better than I can.
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