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Biology; Chemistry; Physics: Post your doubts here!

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About the last line, the statement is incorrect. We have added the KMnO4 in excess (KMnO4 is purple in colour), and as a result, the solution has turned purple
it is added till the "end point",which i think makes KMnO4 not in excess......
but do tell me how it is in excess if i am incorrect.......:)
 
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End point is the exact point when the indicator changes color,so in this case at the end point,it would be in excess and show its color
I don't really get this thing. End point is the exact point when the acid/ alkali has been neutarlised but when we write the colour we consider it in excess. :/
 
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green/yellow/colourless- to- purple
as the solution(P) in flask is very dilute it will show a very light colour (a mixture of the fe 2+ and fe 3+ ions) its not red-brown or dirty green as solution DILUTE.
when potassium manganate(VII) is added, the solution changes colour as potassium manganate in excess.
to yellow kaisay hua?
 
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I don't really get this thing. End point is the exact point when the acid/ alkali has been neutarlised but when we write the colour we consider it in excess. :/
Yes,it changes color as acid/alkali goes into excess.We cant stop at the exact moment when its neutralized,there is still a small amount that is present and even that small amount would be in excess.
 
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C ---H
12 + 2 =14
always divide by 14
Theres ALWAYS a CH2 group difference so CxHy will ALWAYS be 14 hence H will be twice of C.. the two types are :
e.g 1) CxHy = 128
x will be 8 H will be 16
e.g 2) H2NCxHyCOOH = 145
CxHy= 145 - (mass of H2NCOOH)
x=6 H=12
ALWAYS divide by 14 when it comes to CxHy
we even use this technique to find number of carbon atoms

*CH2 sorry my bad .. It's CH2
FUNKY BRAT! See we have to take CH2 . But I dont get why! :'(
 
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