Answer:
Specific heat of water = 33.89 KJ
Explanation:
Given:
mass of water = 81 gram
Initial temperature = 0°C
Final temperature = 100°C
Specific heat of water = 4.184
Find:
Required heat Q
Computation:
Q = Mass x Specific heat of water x (Final temperature - Initial temperature)
Q = (81)(4.184)(100-0)
Q = 33,890.4
Specific heat of water = 33.89 KJ
Answer:
Value of
for the given redox reaction is ![1.0\times 10^{-8}](https://tex.z-dn.net/?f=1.0%5Ctimes%2010%5E%7B-8%7D)
Explanation:
Redox reaction with states of species:
![14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)](https://tex.z-dn.net/?f=14H%5E%7B%2B%7D%28aq.%29%2BCr_%7B2%7DO_%7B7%7D%5E%7B2-%7D%28aq.%29%2B6Cl%5E%7B-%7D%28aq.%29%5Crightarrow%202Cr%5E%7B3%2B%7D%28aq.%29%2B3Cl_%7B2%7D%28g%29%2B7H_%7B2%7DO%28l%29)
Reaction quotient for this redox reaction:
![Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}](https://tex.z-dn.net/?f=Q_%7Bp%7D%3D%5Cfrac%7B%5BCr%5E%7B3%2B%7D%5D%5E%7B2%7D.P_%7BCl_%7B2%7D%7D%5E%7B3%7D%7D%7B%5BH%5E%7B%2B%7D%5D%5E%7B14%7D.%5BCr_%7B2%7DO_%7B7%7D%5E%7B2-%7D%5D.%5BCl%5E%7B-%7D%5D%5E%7B6%7D%7D)
Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of
is taken as 1 due to the fact that
is a pure liquid.
![pH=-log[H^{+}]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%7B%2B%7D%5D)
So, ![[H^{+}]=10^{-pH}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D10%5E%7B-pH%7D)
Plug in all the given values in the equation of
:
![Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}](https://tex.z-dn.net/?f=Q_%7Bp%7D%3D%5Cfrac%7B%280.10%29%5E%7B2%7D%5Ctimes%20%280.010%29%5E%7B3%7D%7D%7B%2810%5E%7B-0.0%7D%29%5E%7B14%7D%5Ctimes%20%281.0%29%5Ctimes%20%281.0%29%5E%7B6%7D%7D%3D1.0%5Ctimes%2010%5E%7B-8%7D)