Chemistry: Post your doubts here!

[drowning-in-studies]

The only difficult thing here is to form the equations, which u have to do by balancing the oxidation numbers.
Once the equation is there, find the moles of Sn2+, and then do the ratio thing.

Thank you so much!

 

[Saad the Paki]

Thank you That was a nice explanation! Do you know if this is out of the syllabus?

Your welcome, and no we still have this in the AS syllabus
Actually this was in the igcse syllabus so that's probably why it isn't mentioned in the AS syllabus

 

[shehroze iqbal]

In an experiment various masses of the sodium salt of the acid, NaX, are added to separate portions of 100 cm3 of HX with stirring. After each addition the pH of the solution obtained is measured.

The graph shows that a pH of 3.86 is obtained when 1.12 g of NaX is added to 100 cm3 of HX. Remember that pKa of HX is also 3.86. Use this information to calculate the relative molecular mass, Mr , of HX. Show your working. [Ar : H, 1.0; C, 12.0; O, 16.0; Na, 23.0]

 

[qwertypoiu]

How do you do Q 5 and 6?
View attachment 59613

Q6. In any hydrolysis reaction, the catalyst that can be used are of two types: acid and base.

Answer should therefore be C, to increase reaction rate by catalysis.

 

[holoholo]

When NH3(aq) is added to a green solution containing Ni2+(aq) ions, a grey-green precipitate is formed. This precipitate dissolves in an excess of NH3(aq) to give a blueviolet solution. Suggest an explanation for these observations, showing your reasoning and including equations for the reactions you describe.

The green ppt. is Ni(OH)2. How are we supposed to get that ?

 

[qwertypoiu]

In an experiment various masses of the sodium salt of the acid, NaX, are added to separate portions of 100 cm3 of HX with stirring. After each addition the pH of the solution obtained is measured.

The graph shows that a pH of 3.86 is obtained when 1.12 g of NaX is added to 100 cm3 of HX. Remember that pKa of HX is also 3.86. Use this information to calculate the relative molecular mass, Mr , of HX. Show your working. [Ar : H, 1.0; C, 12.0; O, 16.0; Na, 23.0]

You should have mentioned that the concentration of HX is 0.1 mol/dm^3.

Remember the formula for finding pH in a buffer solution?

pH = pKa + log([salt]/[acid])

If the pH is equal to pKa, like in this case, then:

[salt] = [acid]
[NaX] = [HX]

The question paper mentioned [HX] = 0.1mol/dm^3, so:
[NaX] = 0.1mol/dm^3
moles of NaX in 100cm^3 = 0.1 * 100/1000 = 0.01mol
Mr of NaX = 1.12g / 0.01mol = 112g/mol.
So Mr of HX = 112 - 22 = 90g/mol

 

[qwertypoiu]

When NH3(aq) is added to a green solution containing Ni2+(aq) ions, a grey-green precipitate is formed. This precipitate dissolves in an excess of NH3(aq) to give a blueviolet solution. Suggest an explanation for these observations, showing your reasoning and including equations for the reactions you describe.

The green ppt. is Ni(OH)2. How are we supposed to get that ?

This is similar to how Cu reacts with NH3.

On adding drops of NH3, it ionizes:

NH3 + H2O <---> NH4+ + OH-

The Ni2+ (aq) is in fact [Ni(H2O)6]2+ (aq). A ligand exchange reaction occurs upon addition of OH-:

[Ni(H2O)6]2+ (aq) + 2OH-(aq) <----> Ni(H2O)4(OH)2 (s) + 2H2O

The solid produced is the ppt. Sometimes we ignore the water ligands and simply write Ni(OH)2. Same thing applies for copper. Only the color differs.

When excess ammonia is added, more ligands are replaced, this time by molecular NH3:

Ni(H2O)4(OH)2 (s) + 4NH3(aq) ------> [Ni(NH3)4(H2O)2]2+ (aq) + 2H2O + 2OH- (aq)

 

[The Sarcastic Retard]

When NH3(aq) is added to a green solution containing Ni2+(aq) ions, a grey-green precipitate is formed. This precipitate dissolves in an excess of NH3(aq) to give a blueviolet solution. Suggest an explanation for these observations, showing your reasoning and including equations for the reactions you describe.

The green ppt. is Ni(OH)2. How are we supposed to get that ?

Colors of complexes.
Ni2+(aq) + NH3(aq) ------> Ni(OH)2 (precipitated form is x(OH)2 If u are asked to write about formula of precipiate it will have (OH)2 as ligand with some metal X) When u add excess of NH3 you will get Ni(H2O)2(NH3)4 which will give blue violet solution.

 

[The Sarcastic Retard]

Colors of complexes.
Ni2+(aq) + NH3(aq) ------> Ni(OH)2 (precipitated form is x(OH)2 If u are asked to write about formula of precipiate it will have (OH)2 as ligand with some metal X) When u add excess of NH3 you will get Ni(H2O)2(NH3)4 which will give blue violet solution.

qwertypoiu mine is correct?

 

[Rizwan Javed]

Can anyone tell me what will be the structural formula of the product, if CH3CH=CHCHO is reacted with H2 under nickel catalyst?

 

[qwertypoiu]

qwertypoiu mine is correct?

Yes sort of
It's based on what has been assumed. For example, saying:

Ni2+(aq) + NH3(aq) ------> Ni(OH)2

is a bit strange given there is NH3 on left side and suddenly OH- on right side. But of course, if the reader knows NH3 ionises to form OH- ions, then it may be fine. Also you said X(OH)2 for some metal X, again this is true if we assume X to be X2+ ion and not some other valency.

 

[The Sarcastic Retard]

Yes sort of
It's based on what has been assumed. For example, saying:

is a bit strange given there is NH3 on left side and suddenly OH- on right side. But of course, if the reader knows NH3 ionises to form OH- ions, then it may be fine. Also you said X(OH)2 for some metal X, again this is true if we assume X to be X2+ ion and not some other valency.

wuhhuuuuuuuuuuuu!! Bingo

 

[The Sarcastic Retard]

Can anyone tell me what will be the structural formula of the product, if CH3CH=CHCHO is reacted with H2 under nickel catalyst?

According to me its REduction and Hydrogenation takes place. CH3CH2CH2CH2OH

 

[qwertypoiu]

Can anyone tell me what will be the structural formula of the product, if CH3CH=CHCHO is reacted with H2 under nickel catalyst?

According to me its REduction and Hydrogenation takes place. CH3CH2CH2CH2OH

I think aldehyde group will also be reduced.
So
CH3CH2CH2CH2OH

Edit: Sorry The Sarcastic Retard I think you edited your answer

 

[The Sarcastic Retard]

I think aldehyde group will also be reduced.
So
CH3CH2CH2CH2OH

I just edited my post

 

[Rizwan Javed]

I think aldehyde group will also be reduced.
So
CH3CH2CH2CH2OH

Thanks. I didn't notice that the aldehyde group will also reduced.

 

[Rizwan Javed]

I just edited my post

Thanks

 

[The Sarcastic Retard]

Thanks. I didn't notice that the aldehyde group will also reduced.

It reduces due to addition of H2 (Addition of H2 causes reduction) *Phew*

 

[holoholo]

Yes sort of
It's based on what has been assumed. For example, saying:

is a bit strange given there is NH3 on left side and suddenly OH- on right side. But of course, if the reader knows NH3 ionises to form OH- ions, then it may be fine. Also you said X(OH)2 for some metal X, again this is true if we assume X to be X2+ ion and not some other valency.

So it is not (OH)2(H2O)4 ?

 

[The Sarcastic Retard]

Can anyone draw structure of Fe(EDTA)
qwertypoiu