Answer:
The concentration c is equal to Ka
Explanation:
The acid will ionize as observed in the following reaction:
HA = H+ + A-
H+ is the proton of the acid and A- is the conjugate base
. The equation to calculate the Ka is as follows:
Ka = ([H+]*[A -])/[HA]
Initially we have to:
[H+] = 0
[A-] = 0
[HA] = c
During the change we have:
[H+] = +x
[A-] = +x
[HA] = -x
During balance we have:
[H+] = 0 + x
[A-] = 0 + x
[HA] = c - x
Substituting the Ka equation we have:
Ka = ([H+]*[A-])/[HA]
Ka = (x * x)/(c-x)
x^2 + Kax - (c * Ka) = 0
We must find c, having as [H+] = 1/2c. Replacing we have:
(1/2c)^2 + (Ka * 1/2 * c) - (c * Ka) = 0
(c^2)/2 + Ka(c / 2 - c) = 0
(c^2)/2 + (-Ka * c/2) = 0
c^2 -(c*Ka) = 0
c-Ka = 0
Ka = c
Answer:
A) The reaction is exothermic reaction
B) 46 gm CH3CH2OH = 1236KJ
15.3 gm ch3ch2OH = 1236/46 x 15.3
= 411.10 KJ..........released
therefore 1 KJ = 0.239 K cal
so, 411.10 x 0.239 kcal
= 98.2529 .................. released
c) 54 gm of H2O produced = 1236KJ
so, 42.7 gm H2O produced = 1236/54 x 42.7KJ
= 977.35 KJ released.
Explanation:
Answer:
5)HOCH2CH2OH
Explanation:
This is also known as ethylene glycol. An increase in hydrogen bonds of a compound means an increase in the viscosity. Hydrogen bonds occur as a result of bonding with electronegative elements such as Oxygen, Nitrogen etc.
The compounds with the highest amount of Hydrogen bond represents the one with the highest viscosity which is B) HOCH2CH2OH
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