The half reactions as they occur at each electrode
is as follows
at the anode Sn(s) =sn^2+(aq) + 2e -
at the cathode 2 ag^+(aq) + 2e - = 2Ag (s)
net cell reaction = Sn (s) + 2Ag^+(aq) = sn^2+ (aq) + 2 Ag (s)
Na⁺¹₃P⁺⁵O⁻²₄
+1*3 + (+5) + (-2*4) = 0
I’m pretty sure the immediate answer is 2. Using the equation 60N=(30kg)x2(whatever unit of measure) also it will change because if you change the mass or the Net Force, your going to have to redo the equation based on the new information.
The first one is D hope it helps!
Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of
= 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration
.


The given equilibrium reaction is,

Initially c 0
At equilibrium

The expression of
will be,
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)

where,
= degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:



Therefore, the value of equilibrium constant for this reaction is, 1.1