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omg how noww??It’s one
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omg how noww??It’s one
Its fine i got it....the ratio between expt 2 and 3omg how noww??
n(KOH) = 0.001 mol; n(CH3COOH)=0.01025 mol. They react in 1:1 ratio, so after neutralisation moles of ethanoic acid left = 0.01025 - 0.001=0.00925 mol.Re: All Chemistry help here!! Stuck somewhere? Ask here!
jaldi btao....i calculated bt not confident wid ma anser
If the molecule only had one primary and one tertiary H atoms, the ratio would be 21:1, but since there are 9 primary H atoms for every one tertiary, that increases its relative rate by 9 times, 21:9.Need help with this also the ans is 21:9View attachment 64677
Oh Thank you!!If the molecule only had one primary and one tertiary H atoms, the ratio would be 21:1, but since there are 9 primary H atoms for every one tertiary, that increases its relative rate by 9 times, 21:9.
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http://www.xtremepapers.com/CIE/Interna ... 4_qp_1.pdf
q. 20 and 26
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I am going to give it a try.
Moles HBr = 0.020 L x 0.200 M =0.0040
Moles NaOH = 0.015 L x 0.200 =0.0030
Moles H+ in excess = 0.0010
Total volume = 0.015 L
Concentration H+ = 0.0010 / 0.015 =0.0667
pH =-log ( 0.06667)
=0.176
9701/41/M/J/17 Q6(d)(i)
In the Proton NMR spectrum of 1-phenyl-ethan-1-ol, C6H5CH(OH)CH3 would the -CH- have a peak at 2.7 ppm or at 4.1 ppm?
In this compound -CH- is bonded to an aromatic ring at one end and to an electronegative O atom at the other end. So if -CH- is considered as an alkyl next to aromatic ring, it can have a peak at delta value of 2.7 ppm.
But if -CH- is considered as an alkyl next to electronegative atom it can have a peak at delta value of 4.0 ppm.
So what should be the correct option?
Please please help me!
What is the answer?help ASAP
I think S in an amide then it is reduced to an amine.help ASAP
For biii) why do we divide the moles we got in bii) with 3?
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