Chemistry: Post your doubts here!

[anastasia grey113]

Hi. I just want to ask.

In the data booklet there's so many standard electrode potential equation specifically for Mn04-. So which one should I use?

When the question is related to reduction of acidified KMnO4, use
MnO4 – + 8H+ + 5e ---> Mn2+ + 4H2O

 

[Thought blocker]

Hi. I just want to ask.

In the data booklet there's so many standard electrode potential equation specifically for Mn04-. So which one should I use?

These are the three electrode potentials given to us in data booklet for MnO4-.
You can see on RHS of each equation, there are different species mentioned, namely MnO4(-2), MnO2 and Mn(+2), so now depending on to the data given to you in the question, you use respective electrode potentials.. I hope that made some sense.

 

[mehran46]

Well first write down the volumes of each gas.
1- The initial mixture with H2O + CO2 + O2 has a total volume of 250cm3 but removing H2O gave a volume of 170cm3.
So VOLUME OF H2O = 250cm3 - 170cm3 = 80cm3.

2- The remaining mixture of CO2 + O2 = 170cm3. Removing CO2 with alkali gives 110cm3.
So VOLUME OF CO2 = 170cm3 - 110cm3 = 60cm3.

3- The final mixture only contains O2 which is 110cm3. So out of 2o0cm3, 110cm3 did not react.
so VOLUME of reacted O2 = 200cm3 - 110cm3 = 90cm3.

4- Volume of alcohol = 20cm3.

Now since all are gases (remember water is actually water vapour so its a gas too) , we can take their ratios to find no. of moles.
so
CxHyO + zO2 -> xCO2 + y/2H2O
20 : 90 : 60 : 80
Now in order to find the value of x, we are supposed to take only 1 mole of the alcohol. So we will reduce the ratios down in such a way that the no. of moles of alcohol are 1.
So it'll be like
CxHyO : O2 : CO2 : H2O
1 : 4.5 : 3 : 4
The no. of moles of CO2 r 3 so x = 3.

thanks a looot

 

[Raheel616]

Which process could be used to calculate the bond energy for the covalent bond X-Y by dividing its ΔH by n?
A. XYn (g) ---> X (g) +nY (g)
B. 2XYn (g) ---> 2XYn-1 (g) +Y2 (g)
C. Y (g) + XYn-1 (g) --->XYn (g)
D. nXY (g) ---> nX (g) + n/2Y2 (g)

 

[Hamnah Zahoor]

Which process could be used to calculate the bond energy for the covalent bond X-Y by dividing its ΔH by n?
A. XYn (g) ---> X (g) +nY (g)
B. 2XYn (g) ---> 2XYn-1 (g) +Y2 (g)
C. Y (g) + XYn-1 (g) --->XYn (g)
D. nXY (g) ---> nX (g) + n/2Y2 (g)

Consider a similar reaction e.g CH4------> C + 4H
For case of option A when we have to find the average bond energy of C-H we will divide ΔH by 4 which in this case we will divide it by n. (thus option A is correct)
In case of second and third option X-Y have not been completely broken down thus we cannot divide it's ΔH directly by n.
For the Last option there is a formation of Y2 molecules which means bond formation occurred.
In case of ΔH of bond energy only bonds are broken (bond breaking) occurs and NO bonds are formed. In bond breaking atoms are formed not molecules.

 

[Holmes]

Help me in f (ii) & (iii)

 

[Mstudent]

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

 

[Holmes]

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.

 

[Unknown202]

Can some one solve tht question plz?

 

[Mstudent]

What's the result?

View attachment 63300
View attachment 63301
Can some one solve tht question plz?

 

[Unknown202]

 

[Mstudent]

Sorry for the terrible handwriting!

 

[mehran46]

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

 

[anastasia grey113]

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

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

 

[anastasia grey113]

View attachment 63298

View attachment 63299
Help me in f (i) & (ii)

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.

 

[Holmes]

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!

 

[anastasia grey113]

Thanks anastasia grey113.
It's hard to get though but still bundle of thanks!

even my friend had trouble helping me understand that
she had to see me in person in order to explain it :/
in short just imagine a 3D diagram n imagine rotating the groups around clockwise or anticlockwise

 

[ANAS_15_2002]

chem is so hard whyyy

 

[Mstudent]

Thanks anastasia grey113.
It's hard to get though but still bundle of thanks!

helps a bit better if you hold three pens in your hand

 

[MShaheerUddin]

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


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