@Cobalt I love how your practically our online teacher :'D
Don't forget to give credit to others though! I do make mistakes and there are other very helpful people here who can correct me and have a deeper knowledge because I don't have any further knowledge than AS materials I just happen to be online most of the time lol
If anyone knows why option A in the question is the answer, pls let me know. and please note, the question asks to put a cross in the INCORRECT option.
Also, someone pls tell me what the fingerprint region is O.o
If anyone knows why option A in the question is the answer, pls let me know. and please note, the question asks to put a cross in the INCORRECT option.
Also, someone pls tell me what the fingerprint region is O.o
Fingerprint region is the region on the right hand side of the spectrum with so many complicated peaks and they are completely different for every compound so can be used to distinguish two compounds like butan-1-ol and butan-2-ol. They will have similar pattern on the left but completely different on the right.
So the two compounds in this question will have different fingerprint region. The answer is A because the C=O peak is different in aldehyde and ketone I guess?
WAIT don't they have same molecular ion peak in mass spectra? Are you sure it's not D Posted from TSR Mobile
Also, someone pls tell me what the fingerprint region is O.o
The messy bit on the right hand side of the spectra, basically It's usually from about 1500-500cm-1, and it contains a complicated series of absorptions, due to the bending vibrations within the molecule (think about how difficult it would be to pick out individual bonds from this part of the spectra, compared to the LHS). It's called the fingerprint region because every compound produces a unique pattern of troughs.
Fingerprint region is the region on the right hand side of the spectrum with so many complicated peaks and they are completely different for every compound so can be used to distinguish two compounds like butan-1-ol and butan-2-ol. They will have similar pattern on the left but completely different on the right.
So the two compounds in this question will have different fingerprint region. The answer is A because the C=O peak is different in aldehyde and ketone I guess?
I'm pretty sure the answer is d as well. Especially since in a data booklet ketone stretch is at 1715 cm/s compared to 1720-1740cm/s for aldehyde. I'd say they're pretty similar
I'm pretty sure the answer is d as well. Especially since in a data booklet ketone stretch is at 1715 cm/s compared to 1720-1740cm/s for aldehyde. I'd say they're pretty similar
Need help with question 5(g) - AQA Unit 1 June 2013.
Which one of the first, second or third ionisations of thallium produces an ion with the election configuration [Xe] 5d10 6s1?
The answer is second - i understand why it is second but how can you to do without writing the whole configuration of thallium out as the question is only worth one mark.
Hi, Tl, thallium is in group 3 therefore it has 3 electrons in its outer S shell. The config you got given said S1 so you know its second as it lost two
I have it somewhere... I'll try and find it for you later
The messy bit on the right hand side of the spectra, basically It's usually from about 1500-500cm-1, and it contains a complicated series of absorptions, due to the bending vibrations within the molecule (think about how difficult it would be to pick out individual bonds from this part of the spectra, compared to the LHS). It's called the fingerprint region because every compound produces a unique pattern of troughs.
Hope this helps
Would you have the unit 2 paper june 2014? Because I really really need it.
Do you follow this? X is the percentage abundance of 10 and there is only 11 so 11 must have abundance of 100-x % You plug those into formula to find percentage abundance