Answer:
Methanol is more acidic than the alkyne and will be deprotonated instead.
Explanation:
of methanol is around 15 and of terminal alkyne is around 26.
, where is acid dissociation constant
So, higher the acidity of an acid, higher will its value and thereby lower will be its value.
So, methanol is certainly stronger acid than terminal alkyne.
Hence sodium amide preferably deprotonates methanol instead of terminal alkyne.
Hence, option (A) is correct.
<span>The pointer will be above the zero mark</span>
The molar mass of the unknown compound is calculated as follows
let the unknown gas be represented by letter Y
Rate of C2F4/ rate of Y = sqrt of molar mass of gas Y/ molar mass of C2F4
= (4.6 x10^-6/ 5.8 x10^-6) = sqrt of Y/ 100
remove the square root sign by squaring in both side
(4.6 x 10^-6 / 5.8 x10^-6)^2 = Y/100
= 0.629 =Y/100
multiply both side by 100
Y= 62.9 is the molar mass of unknown gas
Answer:
We can distinguish between ethanol and ethanoic acid by reacting them with a metallic carbonate like Na2CO3 or a bicarbonate like NaHCO3 ....... ethanoic acid reacts to give to give CO2 and no reaction takes place in case of ethanol.