Answer: 999851 Joules of energy is needed to evaporate the sweat that is produced
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 gram of liquid water to gaseous form at its boiling point.
Given: The heat of vaporization for water is 2257 J/g.
Thus if for 1 g , the heat required is = 2257 J
For 443 g , heat required is =
Thus 999851 Joules of energy is needed to evaporate the sweat that is produced
explanation:
[NaOH] = 2M
NaOH(aq) <======> Na+(aq) + OH-(aq)
[OH-]=[NaOH]
[OH-] = 2M
<em>From</em><em> </em><em>Kw</em><em> </em><em>=</em><em> </em><em>[OH-]</em><em>[</em><em>H</em><em>+</em><em>]</em>
<em>But</em><em> </em><em>Kw</em><em> </em><em>=</em><em> </em><em>1</em><em>0</em><em>^</em><em>-</em><em>1</em><em>4</em><em> </em><em>mol^2dm^-9</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>[H+] = Kw / [OH-]
[H+] = 10^-14 ÷ 2M
[H+] = 5 × 10^-15
pH = -log[H+]
pH = -log 5 × 10^-15
pH = 14.3