Answer:
Final temperature of water is 83.8°C
Explanation:
In 500.0mL of a 6.00M NaOH solution, there are:
0.5000L × (6.00mol NaOH/ L) = <em>3.00moles of NaOH</em>
As the heat involved in dissolution is 44.5 kJ/mol, when the solution is dissolved:
44.5kJ/mol × 3.00moles = 133.5kJ = 133500J
Using coffee-cup calorimeter equation:
q = m×C×ΔT
<em>Where q is heat, m is mass of water (500.0g), C is specific heat of water (4.186J/g°C) and ΔT is change in temperature</em>
<em />
133500J = 500.0g×4.186J/g°C×(X-20.0°C)
63.8 = (X-20.0°C)
83.8°C = X
<em>Final temperature of water is 83.8°C</em>
It varies with container, temperature, pressure and amount of gas. But, the volume occupied by a gas at a fixed temperature and pressure remains constant which is 22.4 L for 1 mole at STP.
Reaction involved in present electrochemical cell,
At Anode: Zn → Zn^2+ + 2e^2-
At cathode: Zn^2+ + 2e^2- → Zn
Net Reaction: Zn + Zn^2+ ('x' m) → Zn^2+(0.1 m) + Zn
Number of electrons involved in present electrochemical cell = n = 2
According to Nernst equation for electrochemical cell,
Ecell = -2.303
![\frac{RT}{nT} log \frac{[Zn^2+]R}{[Zn^2+]L}](https://tex.z-dn.net/?f=%20%5Cfrac%7BRT%7D%7BnT%7D%20log%20%20%5Cfrac%7B%5BZn%5E2%2B%5DR%7D%7B%5BZn%5E2%2B%5DL%7D%20%20)
= 0.014
Given: T =

, F = 96500 C, R = gas constant = 8.314J/K.mol, [Zn^2+]R = 0.1 m , Ecell = 0.014 v
∴ 0.014 = - 2.303

∴ log

=

= -2.1117
∴ log x = log(0.1) + 2.1117
∴x = 13.09 m
I may be totally wrong but I’ll make a guess to it being thermodynamics.