<u>Answer:</u> The potential of electrode is -0.79 V
<u>Explanation:</u>
When zinc is dipped in zinc sulfate solution, the electrode formed is 
Reduction reaction follows: 
To calculate the potential of electrode, we use the equation given by Nernst equation:
![E_{(Zn^{2+}/Zn)}=E^o_{(Zn^{2+}/Zn)}-\frac{0.059}{n}\log \frac{[Zn]}{[Zn^{2+}]}](https://tex.z-dn.net/?f=E_%7B%28Zn%5E%7B2%2B%7D%2FZn%29%7D%3DE%5Eo_%7B%28Zn%5E%7B2%2B%7D%2FZn%29%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BZn%5D%7D%7B%5BZn%5E%7B2%2B%7D%5D%7D)
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = -0.76 V
n = number of electrons exchanged = 2
![[Zn]=1M](https://tex.z-dn.net/?f=%5BZn%5D%3D1M)
(concentration of pure solids are taken as 1)
![[Zn^{2+}]=0.1M](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D%3D0.1M)
Putting values in above equation, we get:
Hence, the potential of electrode is -0.79 V
M₁=6.584 g
m₂=4,194 g
m(H₂O)=m₁-m₂
w(H₂O)=m(H₂O)/m₁
w(H₂O)=(m₁-m₂)/m₁
w(H₂O)=(6.584-4.194)/6.584=0.3630 (36.30%)
the percentage by mass of water in the hydrate 36.30
Answer:
<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
Explanation:
given amount of salt at time t is A(t)
initial amount of salt =300 gm =0.3kg
=>A(0)=0.3
rate of salt inflow =5*0.4= 2 kg/min
rate of salt out flow =5*A/(200)=A/40
rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow
integrating factor
integrating factor 
multiply on both sides by
integrate on both sides
b)
after long period of time means t - > ∞
<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
Answer:
Hydrogen Fluoride will dissolve glass & eat concrete; BUT mixed with water, it is very nasty - but fairly weak!
A strong acid EASILY donates a Proton (H+).
Look up dissociation of acids and the ones that give up that H+ is the strong one.
Explanation:
