Answer: The equilibrium constant for the overall reaction is 
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
a) 
![K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
b) 
![K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
For overall reaction on adding a and b we get c
c) 
![K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5E%5Cfrac%7B5%7D%7B2%7D%7D)
![K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_c%3DK_a%5Ctimes%20K_b%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%5Ctimes%20%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
The equilibrium constant for the overall reaction is 
When battery discharge / delivering current the lead at the anode is oxidized
that is ;
pb---->pb+ 2e-
since the lead ions are in presence of aquous sulfate in insoluble lead sulfate precipitate onto the electrode
the overall reaction at the anode is therefore
Pb + SO4^2- ---> PbSO4 + 2e-
HCl is the formula for hydrocloric acid
the oxidation reaction between oxygen and sodium produces sodium oxide. In many cases, an element may form more than one oxide.
Explanation:
Answer:
0.0002 M
Explanation:
<em>The molarity of the HCl required would be 0.0002 M.</em>
First, let us consider the balanced equation of the reaction:

<em>Stoichiometrically, 1 mole of </em>
<em> reacts with 2 moles of </em>
<em> for a complete neutralization reaction.</em>
Recall that: mole = 
Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole
<em>If 1 mole </em>
<em> requires 2 moles HCl, then 0.0041 mole will require</em>:
0.0041 x 2 = 0.0082 mole HCl
Volume of the HCl = 40.95 L
Molarity = mole/volume
Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M