How much energy is required to vaporize 14.27 grams of water at its boiling point? *
1 answer:
Energy required to vaporize : 32.3 kJ
<h3>Further explanation </h3>The heat to change the phase can be formulated :
Q = mLf (melting/freezing)
Q = mLv (vaporization/condensation)
Lf=latent heat of fusion
Lv=latent heat of vaporization
Mass of water = 14.27 g
Latent heat of vaporization ( boiling point of 100 ºC) : 2260 J/g
Energy required :
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<h2>Question:- </h2>
A solution has a pH of 5.4, the determination of [H+].
<h2>Given :- </h2>- pH:- 5.4
- pH = - log[H+]
Take negative to other side
-pH = log H+
multiple Antilog on both side
(Antilog and log cancel each other )
Antilog (-pH) = [ H+ ]
New Formula :- Antilog (-pH) = [+H]
Now put the values of pH in new formula
Antilog (-5.4) = [+H]
we can write -5.4 as (-6+0.6) just to solve Antilog
Antilog ( -6+0.6 ) = [+H]
Antilog (-6) × Antilog (0.6) = [+H]
put the value in equation