Chemistry: Post your doubts here!

[littlecloud11]

salam
can anyone please explain q1(c) partts (i) and (ii) from May June 2012 paper 42

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_42.pdf

ci) The enthalpy change of solution represents the change of state of an ionic compound from solid to aqueous form. This change corresponds to the difference in energy values for ΔHf MgCl2(s) and ΔHf mgCl2 (aq) as it represents the exact same state change.
So, ΔHsol = -801 - (-641) = -160 KJ/mol

ii) Use the answer form (i) for the formula in (ii). The hydration enthalpy of Cl- should be multiplied by 2 as one mole MgCl2 gives 2 mols of Cl- ions.
ΔHsol (MgCl2) = {2*ΔHhyd(Cl-) + ΔHhyd(Mg2+)} - ΔHlatt(MgCl2)
-160 = 2* ΔHhyd(Cl-) + (-1890) - (-2526)
Therefore, ΔHhyd(Cl-) = (-160 + 1890 - 2526)/2 = -398 Kj/mol

 

[Joeylicious =)]

Does anyone know where I can find the whole chemistry syllabus??

 

[Ahmedraza73]

Does anyone know where I can find the whole chemistry syllabus??

Go To Xtreme papers and choose your Level (O OR A)
The syllabus will be Qouted like:9701_y13_sy.pdf
Sy=Syllabus

 

[!>@/^~^\@<!]

Re: All Chemistry help here!! Stuck somewhere? Ask here!

tonnes of CHEMISTRY WORKSHHETS
http://www.chemactive.com/a_level_chemi ... tions.html

Metal hydroxide reacts with water to give???

 

[hela]

what are the reactions in a catalytic converter
thanks

 

[hela]

can you plaese give me equations in a catalytic converter

 

[PhyZac]

Assalamu Alikum Wa Rahmatullah Wa Barakatoho..... Amy Bloom scouserlfc
Can someone explain this question.

A current of 0.500 A is passed through the electroplating cell. Calculate the time required to deposit a mas of 0.500 g of copper on to the ornament.

 

[littlecloud11]

Assalamu Alikum Wa Rahmatullah Wa Barakatoho..... Amy Bloom scouserlfc
Can someone explain this question.

A current of 0.500 A is passed through the electroplating cell. Calculate the time required to deposit a mas of 0.500 g of copper on to the ornament.

.5 g copper = .5/63.5 = 7.87 *10^-3 mol
Cu^2+ +2e- = Cu
To deposit one mole of copper you need 2 mols of electron, so to deposit 7.87*10^-3 mols of Cu you need = (7.87*10^-3)*2 = 0.0157 mols of electron.
The charge on one mole of electron is 1 Farad i.e. 96500 Coulomb. So the charge on .0157 mols of electron is = 0.0157* 96500 =1519.7 C
So, you need 1519.7 C to deposit .5 g of copper.
Now use the formula Q= It to find the time
1518.7 = .5 * t
t= 3039 sec = 50.7 mins.

 

[PhyZac]

.5 g copper = .5/63.5 = 7.87 *10^-3 mol
Cu^2+ +2e- = Cu
To deposit one mole of copper you need 2 mols of electron, so to deposit 7.87*10^-3 mols of Cu you need = (7.87*10^-3)*2 = 0.0157 mols of electron.
The charge on one mole of electron is 1 Farad i.e. 96500 Coulomb. So the charge on .0157 mols of electron is = 0.0157* 96500 =1519.7 C
So, you need 1519.7 C to deposit .5 g of copper.
Now use the formula Q= It to find the time
1518.7 = .5 * t
t= 3039 sec = 50.7 mins.

Ma Shaa Allah!! Jazaki Allah khairan...! Thank you so much! May Allah provide with the highest grades in this life and hereafter...!! Thank you for explaining each step!

 

[Gémeaux]

Could anyone please explain the trends in hydration energies of Group II sulfates, plus the changes in ΔH(sol)? Like what is more soluble, more exothermic or less exothermic?

 

[PhyZac]

Could anyone please explain the trends in hydration energies of Group II sulfates, plus the changes in ΔH(sol)? Like what is more soluble, more exothermic or less exothermic?

Asslamu Alikum Wa Rahmatullahi Wa Barakatoho..

I was about to ask the same question....please, Can anyone too explain the lattice energy and how it affect solubility !

 

[Gémeaux]

Asslamu Alikum Wa Rahmatullahi Wa Barakatoho..

I was about to ask the same question....please, Can anyone too explain the lattice energy and how it affect solubility !

w.a.salam

Down the group lattice energy decreases as the size of the cation increases, so the solubilty decreases.

 

[PhyZac]

w.a.salam

Down the group lattice energy decreases as the size of the cation increases, so the solubilty decreases.

Jazaki Allah khairan!! Thank you so much! it is a confusing chapter...! May Allah provide you with highest grades in this life and hereafter, Ameen!!

 

[Gémeaux]

Jazaki Allah khairan!! Thank you so much! it is a confusing chapter...! May Allah provide you with highest grades in this life and hereafter, Ameen!!

Wa eyyaki. Indeed it is, which is why I just don't get it :/
Aaameeenn and JazakAllah

 

[littlecloud11]

Could anyone please explain the trends in hydration energies of Group II sulfates, plus the changes in ΔH(sol)? Like what is more soluble, more exothermic or less exothermic?

ΔH(sol) = ΔH(hyd) - ΔH(latt)

Sulphate ion is a relatively large anion compared to the sizes of the group 2 cations, so sulphate ion effects the lattice enthalpy more than the cation. As the size of the cation increases down the group the lattice enthalpy decreases down the group (becomes less exothermic), but sulphate is large enough to even be slightly polarised by barium and elements below it, so the decrease in lattice enthalpy is relatively small down the group.

For the hydration of the ion, the enthalpy of hydration remains constant for the sulphate ion but as the cation size increases the hydration enthalpy decreases (becomes less exothermic), this decrease is relatively large as the cation size increases quite drastically, and is independent of the anion size.

So, For example if the initial ΔH(hyd) and ΔH(latt) were -20 and -15 respectively ΔH(sol) = -20 - (-15) = -5
there is a greater decrease for ΔH(hyd) than for ΔH(latt), so the new values for ΔH(hyd) and ΔH(latt) would be -10 and -12 respectively. so ΔH(sol) = -10 -(-12) = +2
These are just random values to show the trend, you can see that ΔH(sol) gradually becomes more endothermic, so the salts becomes less soluble.

 

[Gémeaux]

ΔH(sol) = ΔH(latt) - ΔH(hyd)

Sulphate ion is a relatively large anion compared to the sizes of the group 2 cations, so sulphate ion effects the lattice enthalpy more than the cation. As the size of the cation increases down the group the lattice enthalpy decreases down the group (becomes less exothermic), but sulphate is large enough to even be slightly polarised by barium and elements below it, so the decrease in lattice enthalpy is relatively small down the group.

For the hydration of the ion, the enthalpy of hydration remains constant for the sulphate ion but as the cation size increases the hydration enthalpy decreases (becomes less exothermic), this decrease is relatively large as the cation size increases quite drastically, and is independent of the anion size.

So, For example if the initial ΔH(latt) and ΔH(hyd) were -20 and -15 respectively ΔH(sol) = -20 - (-15) = -5
there is a greater decrease for ΔH(hyd) than for ΔH(latt), so the new values for ΔH(latt) and ΔH(hyd) would be -25 and -30 respectively. so ΔH(sol) = -25 -(-30) = +10
These are just random values to show the trend, you can see that ΔH(sol) gradually becomes more endothermic, so the salts becomes less soluble.

Thankyou so much!!
One more thing, ΔH(sol) increases down the group?

 

[littlecloud11]

Thankyou so much!!
One more thing, ΔH(sol) increases down the group?

For group 2 sulphates, yes.

 

[PhyZac]

ΔH(sol) = ΔH(latt) - ΔH(hyd)

Sulphate ion is a relatively large anion compared to the sizes of the group 2 cations, so sulphate ion effects the lattice enthalpy more than the cation. As the size of the cation increases down the group the lattice enthalpy decreases down the group (becomes less exothermic), but sulphate is large enough to even be slightly polarised by barium and elements below it, so the decrease in lattice enthalpy is relatively small down the group.

For the hydration of the ion, the enthalpy of hydration remains constant for the sulphate ion but as the cation size increases the hydration enthalpy decreases (becomes less exothermic), this decrease is relatively large as the cation size increases quite drastically, and is independent of the anion size.

So, For example if the initial ΔH(latt) and ΔH(hyd) were -20 and -15 respectively ΔH(sol) = -20 - (-15) = -5
there is a greater decrease for ΔH(hyd) than for ΔH(latt), so the new values for ΔH(latt) and ΔH(hyd) would be -25 and -30 respectively. so ΔH(sol) = -25 -(-30) = +10
These are just random values to show the trend, you can see that ΔH(sol) gradually becomes more endothermic, so the salts becomes less soluble.

Jazak Allah khiaran....thank you soo much !!!!!!!!!!! Really wonderful explanation ! May Allah reward you and your family with Jannah tul firdous in hereafter. Ameen

 

[hela]

ΔH(sol) = ΔH(latt) - ΔH(hyd)

Sulphate ion is a relatively large anion compared to the sizes of the group 2 cations, so sulphate ion effects the lattice enthalpy more than the cation. As the size of the cation increases down the group the lattice enthalpy decreases down the group (becomes less exothermic), but sulphate is large enough to even be slightly polarised by barium and elements below it, so the decrease in lattice enthalpy is relatively small down the group.

For the hydration of the ion, the enthalpy of hydration remains constant for the sulphate ion but as the cation size increases the hydration enthalpy decreases (becomes less exothermic), this decrease is relatively large as the cation size increases quite drastically, and is independent of the anion size.

So, For example if the initial ΔH(latt) and ΔH(hyd) were -20 and -15 respectively ΔH(sol) = -20 - (-15) = -5
there is a greater decrease for ΔH(hyd) than for ΔH(latt), so the new values for ΔH(latt) and ΔH(hyd) would be -25 and -30 respectively. so ΔH(sol) = -25 -(-30) = +10
These are just random values to show the trend, you can see that ΔH(sol) gradually becomes more endothermic, so the salts becomes less soluble.

i think that ΔH(sol) =ΔH(hyd) _ ΔH(latt) and not the opposit

how to explain the solubility of the group 2 hydroxide
t

 

[Alice123]

what are the reactions in a catalytic converter
thanks

the reactions that take place in the catalylic converter:
*oxidation of CO to form CO2
*the reduction of nitrogen oxides to form nitrogen
*the oxidation of unburnt hydrocarbon to produce CO2 n H2O
HOPE THIS HELPS