Chemistry - Got Some Questions (1 Viewer)

[Life'sHard]

1)

2)


3)

 

[jazz519]

1) It's a Le Chatelier's Principle question as there is like an equilibrium. More solvent in the X part means concentration of I2(x) decreases as in c=n/v the volume has gone up. This means the equil. is disturbed and shifts right to minimise change. Leading to increase in I2(x) and decrease I2(Y). So B. C and D, dont make sense for this as temperature changing is what changes the K value

 

[Life'sHard]

1) It's a Le Chatelier's Principle question as there is like an equilibrium. More solvent in the X part means concentration of I2(x) decreases as in c=n/v the volume has gone up. This means the equil. is disturbed and shifts right to minimise change. Leading to increase in I2(x) and decrease I2(Y). So B. C and D, dont make sense for this as temperature changing is what changes the K value

Ohhh I see I didn't even relate the volume increase with c=n/v. Tyvm.

 

[jazz519]

2) Whenever questions have things about concentrations or values in general, it is a good idea to write some formulas or equations. Here there is dissolving equations:
Ba(OH)2(s) --> Ba2+(aq) + 2OH-(aq)
NaOH(s) --> Na+(aq) + OH-(aq)

So in total we have 1Ba2+, 1Na+ and 3OH-

[Ba2+] = [Na+]
3[Ba2+] = [OH-] (as there's 3 times as much of OH-)

If [Na+] = [Ba2+], then arrange that above eqn a little to:
[Ba2+] + 2[Ba2+] = [OH-]
and sub the [Na+] = [Ba2+]
[Na+] + 2[Ba2+] = [OH-]

 

[jazz519]

3) Under same temp and conditions, equal volumes of gases have the same moles. So when you write the equations for combustion the CO2 and H2O should have same molar ratio. Go through and do fuel + O2 --> CO2 + H2O and balance all of them. The one that has same balancing for the CO2 and H2O is correct

 

[Life'sHard]

2) Whenever questions have things about concentrations or values in general, it is a good idea to write some formulas or equations. Here there is dissolving equations:
Ba(OH)2(s) --> Ba2+(aq) + 2OH-(aq)
NaOH(s) --> Na+(aq) + OH-(aq)

So in total we have 1Ba2+, 1Na+ and 3OH-

[Ba2+] = [Na+]
3[Ba2+] = [OH-] (as there's 3 times as much of OH-)

If [Na+] = [Ba2+], then arrange that above eqn a little to:
[Ba2+] + 2[Ba2+] = [OH-]
and sub the [Na+] = [Ba2+]
[Na+] + 2[Ba2+] = [OH-]

3[Ba2+] = [OH-] (as there's 3 times as much of OH-)

How did you get to this line? Just a little confused about this part.

Is it more so the ratio how [Ba2+] : [OH-] is 3:1?
Also, if it wasn't a multiple choice would the reverse also be true whereby 2[Na+] + [Ba2+] = [OH-]

 

[CM_Tutor]

View attachment 31186

Another perspective:

Let 1 L of the solution contain mol of sodium hydroxide and mol of barium hydroxide, so that its composition is:







And, since the volume is 1 L, we know that







Using these values, it is clear that:

Option (A) requires which can only be true if .

Option (B) requires which can only be true if .

Option (C) requires which can only be true if and thus if .

Option (D) requires and thus is true for any values of and .

 

[Life'sHard]

Just part b) is what i need help for
Why is it that when finding the mass of tannic acid, you divide the mols of gallic acid/5

I get you do m=n*MM
but why is n gallic acid/5?

@CM_Tutor or @jazz519 I trust your proffesional opinion hahaha.

 

[jazz519]

It's like doing a molar ratio like you would with a normal equation.
The equation for polymer formation is:
5 galic acid ---> tannic acid + 4H2O

so from part (a) if it is known: n(gallic acid) = 0.0001092 mol

n(tannic acid) = 1/5 n(gallic acid)

using a molar ratio for that equation

 

[Life'sHard]

It's like doing a molar ratio like you would with a normal equation.
The equation for polymer formation is:
5 galic acid ---> tannic acid + 4H2O

so from part (a) if it is known: n(gallic acid) = 0.0001092 mol

n(tannic acid) = 1/5 n(gallic acid)

using a molar ratio for that equation

Thank you so much that made a lot more sense

 

[Life'sHard]

Can someone reason out the answer. Thanks

 

[Life'sHard]

A little confused as to why there's a 4 infront of x^3 for the first line of working.

I get ksp = [Ag+]^2 * [CO3^2-] . But where does the 4 come from?

 

[jazz519]

Use RICE table

Ag2CO3(s) < -- > 2Ag+(aq) + CO3 2-(aq)

Ksp = [Ag+]^2 [CO3 2-]

	Ag2CO3	Ag+	CO32-
R	1	2	1
I	/	0	0
C	/	+2x	+x
E	/	2x	x

Sub the 2x and x

gives the Ksp = (2x)^2 (x) = 4x^3

 

[Life'sHard]

Thank you.

 

[Life'sHard]

Am I blind I can't see anything wrong with any of the other choices.

 

[Life'sHard]

damn really am blind hahaha ty

 

[Life'sHard]

1)

Probably haven't got up to this part of the module at school yet. But need help with this.

2)


Also I need clarification. I had thought that addition reaction were slow reactions and required a catalyst? Someone clarify this for me please.

 

[CM_Tutor]

Esterification is an equilibrium phenomenon, and so does not go to completion.

Do a mole calculation on the reaction

methanol + ethanoic acid ---> methyl ethanoate + water

where methyl ethanoate is CH₃COOCH₃. Work out the theoretical yield of the ester based on the whichever of the two reagants is limiting, then find 24% of that.

As for additions, hydrogenation definitely needs a catalyst. Hydration and hydrohalogenation are acid catalysed, though in the case of hydrohalogenation the catalyst comes from the HX itself and the reaction will certainly occur in gas phases. Halogenation of an alkene does not require a catalyst though substitution side reactions are possible if there is UV light around. I would call it a fairly rapid addition and the only product will be 1,2-dichloroethane.

 

[Life'sHard]

Esterification is an equilibrium phenomenon, and so does not go to completion.

Do a mole calculation on the reaction

methanol + ethanoic acid ---> methyl ethanoate + water

where methyl ethanoate is CH₃COOCH₃. Work out the theoretical yield of the ester based on the whichever of the two reagants is limiting, then find 24% of that.

As for additions, hydrogenation definitely needs a catalyst. Hydration and hydrohalogenation are acid catalysed, though in the case of hydrohalogenation the catalyst comes from the HX itself and the reaction will certainly occur in gas phases. Halogenation of an alkene does not require a catalyst though substitution side reactions are possible if there is UV light around. I would call it a fairly rapid addition and the only product will be 1,2-dichloroethane.

Thanks man. Really appreciate your activity in these forums. You've really helped!

 

[Life'sHard]

Need some help with Q12.