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Chemistry: Post your doubts here!

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F(i) basically means a compound that has the structural formula as the original compound but the arrangement of species differ e.g optical isomers. Spital arrangement simply means arrangement of species in space
 
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F(i) basically means a compound that has the structural formula as the original compound but the arrangement of species differ e.g optical isomers. Spital arrangement simply means arrangement of species in space
Thanks but I would appreciate some detailed answer
Thanks though.
 
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At a pressure of 1.50 × 105Pa, 1.00mol of sulfur dioxide gas, SO2, was mixed with 1.00mol of oxygen gas, O2. The final equilibrium mixture formed was found to contain 0.505mol of O2. (2SO2(g) + O2(g) 2SO3(g)) (i) Calculate the amount, in mol, of SO2 and SO3 in the equilibrium mixture. SO2 = .............................. mol------- SO3 = .............................. mol

What is the way of solving this problem thanks
 
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At a pressure of 1.50 × 105Pa, 1.00mol of sulfur dioxide gas, SO2, was mixed with 1.00mol of oxygen gas, O2. The final equilibrium mixture formed was found to contain 0.505mol of O2. (2SO2(g) + O2(g) 2SO3(g)) (i) Calculate the amount, in mol, of SO2 and SO3 in the equilibrium mixture. SO2 = .............................. mol------- SO3 = .............................. mol

What is the way of solving this problem thanks
Okay so the equation is as follows
2SO2 + O2 ---> 2SO3
So we have 2 moles of SO2, 1 mole of O2 and 2 moles of SO2.
We are also given the final reading of O2.
1- Make a table as follows
upload_2018-4-22_7-12-26.png
2- Start with filling out the inital and final readings.
Now always remember that THE CHANGE IS ALWAYS WITH RESPECT TO THE NUMBER OF MOLES IN THE EQUATION.
So if you are denoting the change in O2 by 'x' it should be 2x for SO2 as it has 2 moles.
And don't forget to put the '-' sign as they decrease when more product is made.
3- Find the change in O2 by using final - inital = 1 - 0.505 = 0.495.
So if x = 0.495, then 2x = 2 x 0.495.
Now add the value of change in initial (REMEMBER FOR REACTANTS THE READING IS NEGATIVE)
So you will get
Final SO3 = 0 + (+2 x 0.495) = 0.99
Final SO2 = 1 + (-2 x 0.495) = 0.01
 

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lol this very concept was hard for me to grasp too
Well for part (ii), just focus on part (i)
And that's true it is the spatial arrangement of groups in a molecule.
Just consider the question above.
I'll start with part (ii) that way it'll be easier to understand.
Start with placing the three fingers (index, middle and ring) of one of your hands in a manner that they look like stands of a tripod.
The ring and index should be at the bottom at the same level with middle a little above.
Now as you can see in F, the H group is the nearest, OH the farthest and CH3 is in plane with the C-C bond.
We are only considering the right side of the chain.
On the tip of the nearest finger (index I suppose or depends on how u position ur hand), write H.
On the middle finger (the one that's not too near not too far) write CH3.
On the finger farthest from you, write OH.
Now just 'rotate' your hand so one of the other groups (let's say OH) is nearest to u.
Now see the arrangement. Does it look like the arrangement on isomer H or J?
You can do the same for the groups on the right and see for G I and J.
The one in which rotating your fingers position clockwise or anticlockwise, causes the same arrangement, THAT IS NOT AN OPTICAL ISOMER RATHER IT IS THE SAME COMPOUND.
So you see, we might think the bonds are arranged the same way but that is not the case.
The other two molecules have groups arranged differently and that's why on rotation we don't get the same arrangement as H and I.
Because they are differently arranged in space. So spatial arrangement is different.
 
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lol this very concept was hard for me to grasp too
Well for part (ii), just focus on part (i)
And that's true it is the spatial arrangement of groups in a molecule.
Just consider the question above.
I'll start with part (ii) that way it'll be easier to understand.
Start with placing the three fingers (index, middle and ring) of one of your hands in a manner that they look like stands of a tripod.
The ring and index should be at the bottom at the same level with middle a little above.
Now as you can see in F, the H group is the nearest, OH the farthest and CH3 is in plane with the C-C bond.
We are only considering the right side of the chain.
On the tip of the nearest finger (index I suppose or depends on how u position ur hand), write H.
On the middle finger (the one that's not too near not too far) write CH3.
On the finger farthest from you, write OH.
Now just 'rotate' your hand so one of the other groups (let's say OH) is nearest to u.
Now see the arrangement. Does it look like the arrangement on isomer H or J?
You can do the same for the groups on the right and see for G I and J.
The one in which rotating your fingers position clockwise or anticlockwise, causes the same arrangement, THAT IS NOT AN OPTICAL ISOMER RATHER IT IS THE SAME COMPOUND.
So you see, we might think the bonds are arranged the same way but that is not the case.
The other two molecules have groups arranged differently and that's why on rotation we don't get the same arrangement as H and I.
Because they are differently arranged in space. So spatial arrangement is different.
Thanks anastasia grey113.
It's hard to get though but still bundle of thanks!
 
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The enthalpy change of the neutralisation given below is –114 kJ mol–1.
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)
By using this information, what is the most likely value for the enthalpy change of the following
neutralisation?
Ba(OH)2(aq) + 2HCl(aq) → BaCl2(aq) + 2H2O(l)
A –57 kJ mol–1 B –76 kJ mol–1 C –114 kJ mol–1 D –228 kJ mol–1


HELP NEEDED
 
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Okay so the equation is as follows
2SO2 + O2 ---> 2SO3
So we have 2 moles of SO2, 1 mole of O2 and 2 moles of SO2.
We are also given the final reading of O2.
1- Make a table as follows
View attachment 63308
2- Start with filling out the inital and final readings.
Now always remember that THE CHANGE IS ALWAYS WITH RESPECT TO THE NUMBER OF MOLES IN THE EQUATION.
So if you are denoting the change in O2 by 'x' it should be 2x for SO2 as it has 2 moles.
And don't forget to put the '-' sign as they decrease when more product is made.
3- Find the change in O2 by using final - inital = 1 - 0.505 = 0.495.
So if x = 0.495, then 2x = 2 x 0.495.
Now add the value of change in initial (REMEMBER FOR REACTANTS THE READING IS NEGATIVE)
So you will get
Final SO3 = 0 + (+2 x 0.495) = 0.99
Final SO2 = 1 + (-2 x 0.495) = 0.01
Thanks for the continuous help
 
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Why do you multiply the conc. of HCOO- ions by 25? Anyone?
View attachment 63336

View attachment 63337
Because they are asking for concentration of SATURATED solution
And the concentration calculated in previous part is of dilute solution
and saturated solution is 50 times saturated so 50 times r the moles of Mg(HCOO-)2
but u see there r 2 moles of HCOO- in 1 mole of Mg(HCOO-)2
so just multiply it by 50/2 = 25
 
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Hi again in the data booklet states

MnO4- + e --> MnO4 2-

MnO4- + 4H +3e --> MnO2 +2H20

Mn04- +8H +5e -->mn 2+ +4h20

What are the difference between all 3 reactions
And when do we use it in equation?
 
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