Last minute preparation on MCQS 9701 and 9702. Discussion here.

[awesomaholic101]

Q7 n 34 i dnt get it
ty

Q7: calculating the pressure that each gas would contribute (mole fraction * its pressure) and adding them gives the total pressure ... I don't know if this is a concept we should know or sth, I just did it by logic.

Q34: The exothermic reaction between H2SO4(strong acid) and Mg will be faster than the same with ethanoic acid. So,
H2SO4 will reach a higher temperature and also a higher yield of H2 gas at 2 mins.
After 20 mins, H2SO4 would have completely reacted to give maximum H2 while ethanoic acid would have only reached equilibrium. So, H2 produced by H2SO4 is greater.

 

[The Sarcastic Retard]

ty

Q7: calculating the pressure that each gas would contribute (mole fraction * its pressure) and adding them gives the total pressure ... I don't know if this is a concept we should know or sth, I just did it by logic.

Q34: The exothermic reaction between H2SO4(strong acid) and Mg will be faster than the same with ethanoic acid. So,
H2SO4 will reach a higher temperature and also a higher yield of H2 gas at 2 mins.
After 20 mins, H2SO4 would have completely reacted to give maximum H2 while ethanoic acid would have only reached equilibrium. So, H2 produced by H2SO4 is greater.

 

[saturn21]

Ans is C..how? Shouldn't it be 6?

 

[The Sarcastic Retard]

View attachment 57298

Ans is C..how? Shouldn't it be 6?

There is a formula, 2^n. My teacher never taught me this, but once I had a same doubt. So 2^n is the formula used for number of isomers. Where n is the number of double bonds within the molecule. Hence 2^3 = 8

 

[awesomaholic101]

There is a formula, 2^n. My teacher never taught me this, but once I had a same doubt. So 2^n is the formula used for number of isomers. Where n is the number of double bonds within the molecule. Hence 2^3 = 8

but can that be generalised for every molecule coz what if both Carbons in the double bond have the same groups attached to them ... then cis-trans isomers won't esixt ... ?

 

[The Sarcastic Retard]

but can that be generalised for every molecule coz what if both Carbons in the double bond have the same groups attached to them ... then cis-trans isomers won't esixt ... ?

idk

 

[awesomaholic101]

idk

thers no general formula actually ... atleast not yet, so we need to suck it up and deal with it

 

[The Sarcastic Retard]

thers no general formula actually ... atleast not yet, so we need to suck it up and deal with it

 

[DarkEclipse]

but can that be generalised for every molecule coz what if both Carbons in the double bond have the same groups attached to them ... then cis-trans isomers won't esixt ... ?

2^n is for the number of isomers. They can be cis-trans, chain, position, blah blah blah. I think this formula is dependant upon the nature of the compound, and I have tried it with various compounds, and it seems to work so far.

Let me elaborate. In this case, since the compound is symmetrical and contains double bonds, 'n' becomes the number of double bonds. In a symmetrical compound with double bonds, we know it must have cis-trans isomerism. Therefore, raising 2 to the power of the number of double bonds; i.e. 2^3, we get the number of cis-trans isomers.

Let me use another example. In a molecule, there may be 'n' number of chiral carbons. Therefore, the number of enantiomers in the molecule is 2^n, where 'n' is the number of chiral carbons.

As far as I've seen, 2^n is a formula restricted to only stereoisomers. What I've written is based solely on my experience, so use with caution

 

[awesomaholic101]

2^n is for the number of isomers. They can be cis-trans, chain, position, blah blah blah. I think this formula is dependant upon the nature of the compound, and I have tried it with various compounds, and it seems to work so far.

Let me elaborate. In this case, since the compound is symmetrical and contains double bonds, 'n' becomes the number of double bonds. In a symmetrical compound with double bonds, we know it must have cis-trans isomerism. Therefore, raising 2 to the power of the number of double bonds; i.e. 2^3, we get the number of cis-trans isomers.

Let me use another example. In a molecule, there may be 'n' number of chiral carbons. Therefore, the number of enantiomers in the molecule is 2^n, where 'n' is the number of chiral carbons.

As far as I've seen, 2^n is a formula restricted to only stereoisomers. What I've written is based solely on my experience, so use with caution

nice ... thanks for explaining, should save some time for me.

 

[DarkEclipse]

nice ... thanks for explaining, should save some time for me.

No problem

 

[Shujaat Khan]

There is a formula, 2^n. My teacher never taught me this, but once I had a same doubt. So 2^n is the formula used for number of isomers. Where n is the number of double bonds within the molecule. Hence 2^3 = 8

I also have another one for you incase you need:

Question is from p4 application part, Asks for how many different group of tri-aminoacids can be formed with val, ser and another(asp?)
answer is n! where n is number of names given, its simple maths.
Let's say now it asks how many di-amino(peptide) you can form from this.
Answer is 3P2 where you choose 2 from 3 and arrange in 2! ways

You should always find the logical way to do things in either chemistry or physics!

 

[The Sarcastic Retard]

Thanks all!
9701_w14_qp_12
Q4

 

[The Sarcastic Retard]

DarkEclipse
awesomaholic101
FranticAmaze

 

[DarkEclipse]

DarkEclipse
awesomaholic101
FranticAmaze

Since this is an EM wave, its speed is 3x10^8 ms^-1. Frequency is 1GHz= 1 000 000 000 Hz.
Use v = fλ.
Therefore, λ = 0.3m.
Distance = 45 cm = 0.45 m.
Periods present b/w the transmitter and the plate = 0.45/0.3 = 1.5.

In a stationary wave, 1 period contains 2 antinodes. Therefore, half a period contains 1 antinode.
Therefore, 2 + 1 = 3, option C.

 

[The Sarcastic Retard]

Since this is an EM wave, its speed is 3x10^8 ms^-1. Frequency is 1GHz= 1 000 000 000 Hz.
Use v = fλ.
Therefore, λ = 0.3m.
Distance = 45 cm = 0.45 m.
Periods present b/w the transmitter and the plate = 0.45/0.3 = 1.5.

In a stationary wave, 1 period contains 2 antinodes. Therefore, half a period contains 1 antinode.
Therefore, 2 + 1 = 3, option C.

0.45/0.3 y?

 

[DarkEclipse]

0.45/0.3 y?

To find the number of periods between two points, you have to divide the total distance by the wavelength. In this case, the wavelength is 0.3 m, and the distance is 0.45 m.

 

[The Sarcastic Retard]

To find the number of periods between two points, you have to divide the total distance by the wavelength. In this case, the wavelength is 0.3 m, and the distance is 0.45 m.

Is there a formula like this??

 

[DarkEclipse]

Is there a formula like this??

No, it's just logic.

 

[The Sarcastic Retard]

No, it's just logic.

I dont understand this logic?
How did this clicked to ur mind?