Chemistry: Post your doubts here!

Metanoia

Metanoia

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qwertypoiu said:
I think it should be hexane.
1-chloro-4-ethyl-5-methylhexane.
However, counting carbon the other way it could also be:
6-chloro-3-ethyl-2-methylhexane.
I think the first one is more correct but not sure why.
Click to expand...

asadalam said:
Hexyl?Isnt it a chloroalkane?

1-chloro-4-ethyl-5-methylhexane i think?Cuz we call them 1-bromopentane etc
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1-chloro-4-ethyl-5-methylhexane is correct. The smallest number assigned to the halogen group.

Last time I mentioned that "halogen is lower priority than alkane" , I realized that can be misinterpreted.

Between halogen and alkyl group, we assign a lower number to the halogen group, but we still end the molecule with -ane.
 
princess Anu

princess Anu

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Metanoia said:
1-chloro-4-ethyl-5-methylhexane is correct. The smallest number assigned to the halogen group.

Last time I mentioned that "halogen is lower priority than alkane" , I realized that can be misinterpreted.

Between halogen and alkyl group, we assign a lower number to the halogen group, but we still end the molecule with -ane.
Click to expand...
what does that mean? when assaigning numbers do we consider the priorities or not?
 
nehaoscar

nehaoscar

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princess Anu said:
what does that mean? when assaigning numbers do we consider the priorities or not?
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Don't you usually name it alphabetically, and the priority is usually to the functional group right... Like the functional group should get the least number...
unless they tell you otherwise...
 
mehria

mehria

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The Sarcastic Retard there u go... :)
these notes explains y conc n pressure dnt have any effect while temp does effect kp n kc...
n just ignre the designs n shayari on these pages.. :p
hope it helps... (y)
 

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Nowrin Yasmin

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mehria

mehria

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DeViL gURl B) said:
http://www.franklychemistry.co.uk/GCE_Papers/CIE/2010/November/9701_w10_qp_43.pdf

CAN SOMEONE PLEASE EXPLAIN ME THW WHOLE OF QUESTION 5. ESP. PART a and b.
PLEASEEEEE ASAP!!!
Thank you. :)
Click to expand...

5
(a) In neutral NaOH(aq) the Na + and Cl- dissociates without causing any chnge in the pH and there will be no reaction between NaCl and H2O so there is H2O present in the solution instead of H+ and OH - so here H2O will be oxidised at anode instead of OH-

(i) 2H2O - 4e- -----------> 4 H+ + O2
(ii) 2Cl- - 2e- ----------> Cl2

(b) the equation for Ecell = Eleft - Eright

Eright is the voltage in -ve electrode i.e cathode where reduction takes place
Eleft is the voltage in +ve electrode i.e anode where oxidation takes place

from the equations in (a) and the data given to us in the question we cn now easily find the Ecell using the data booklet

Screenshot (6736).png
here we wont use the value of the third equation as we didnt use this equation for the reaction at anode

(i) Ecell= 1.23 - (- 0.83)
= + 2.06
(ii) Ecell = 1.36 - ( - 0.83)
= + 2.19

(c)
(i) for Eanode when H2O is oxidised then there won't be any chnge by increasing the conc of NaCl cuz only conc of NaCl is chnged not the conc of water
for Eanode when Cl- is oxideised then by increasing conc of NaCl the Eanode will be smaller as more Cl2 will be oxidised. so it will be less positive.

(ii) as mentioned above more n more Cl- will oxidise to more Cl2 will be formed than O2 that's y the ratio increases.
so to explain ths in terms of Ecell we have to write that as the Eanode for Cl2 decreases and Eanode for O2 remains the same.

(d)
(i) for ths part the first thing we have to do is to balance the e- in both cathode n anode reactions.
Cl- + 6OH- – 6e- -----------> ClO3- + 3H2O(l)
2H2O + 2e- -----------> H2 + 2OH-
as there r 6e- in the anode equation we will multiply the equation of cathode with 3 to balance the e-
3 x (2H2O + 2e- -----------> H2 + 2OH-)
6H2O + 6e- --------------> 3H2 + 6OH-

we will write the combined equation
Cl- + 6OH- – 6e- + 6H2O + 6e- -----------> ClO3- + 3H2O(l) + 3H2 + 6OH-
so the overall equation will be
Cl- + 3H2O -----------> ClO3- + 3H2


(ii)
first we will find the charge
Q = It
= (250) ( 3600)
= 900,000 C

no. of moles e- => (900000) / (96500) = 9.33
no. of moles NaClO3 => 9.33/6 = 1.55
Mr of NaClO3 = 106.5
mass of NaClO3 = (106.5) x (1.55) = 165.5 g
 
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