<u>Answer: </u>The equation which is wrong is 
<u>Explanation:</u>
For the given reaction:

The expression for
is given by:
![K_c=\frac{1}{[O_2]^3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B1%7D%7B%5BO_2%5D%5E3%7D)
![K_p=\frac{1}{[O_2]^3}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B1%7D%7B%5BO_2%5D%5E3%7D)
The concentration of solids are taken to be 1, only concentration of gases and liquid states are taken. The pressure of only gases are taken.
Relationship between
is given by the expression:

where,
= number of moles of gaseous products - number of moles of gaseous reactants
R = gas constant
T= temperature
For the above reaction,
= number of moles of gaseous products - number of moles of gaseous reactants = 0 - 3 = -3
Hence, the expression for
is:

Therefore, the equation which is wrong is 
Answer:
1.125 moles
Explanation:
2mole of HCl produced 1mole of H2
2.25moles of HCl will produce x moles
cross multiply
2x=™1×2.25
x= 2.25÷2
x=1.125mole
<span>1. Tap water has a small concentration of H+ & OH- ions as well as water molecules, hence there would be permanent dipole-permanent dipole (p.d.-p.d.) forces of attraction between the water molecules (aka H-bonds) as well as ionic bonds between the H+ & OH- ions.
2. Distilled water does not have H+ & OH- ions, hence only H-bonds exist between the water molecules.
3. There are covalent bonds between the individual sugar molecules.
4. There are ionic bonds between the Na+ & Cl- ions in NaCl.
5. There are p.d.-p.d. forces of attraction between the Na+ ions and the O2- partial ions of the water molecules as well as between the Cl- ions and the H+ partial ions of the water molecules. There are also H-bonds between the individual water molecules and ionic bonds between the Na+ & Cl- ions (although these are in much lower abundance than in unsolvated solid NaCl).
6. There are i.d.-i.d. as well as p.d.-p.d. forces of attraction between the sugar molecules and the water molecules. There are also H-bonds between the individual water molecules and covalent bonds within the sugar molecules.</span>
Answer:
Explanation:
To find the concentration; let's first compute the average density and the average atomic weight.
For the average density
; we have:

The average atomic weight is:

So; in terms of vanadium, the Concentration of iron is:

From a unit cell volume 

where;
= number of Avogadro constant.
SO; replacing
with
;
with
;
with
and
with 
Then:
![a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2FA_%7BFe%7D%20%5D%20%2B%20%5BC_v%2FA_v%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2F%5Crho_%7BFe%7D%20%5D%20%2B%20%5BC_v%2F%5Crho_v%5D%7D%20%5CBig%29%20%20%7D)
![a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100-C_v%29A_%7Bv%7D%20%5D%20%2B%20%5BC_v%2FA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100-C_v%29%2F%5Crho_%7Bv%7D%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
![a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100A_%7Bv%7D-C_vA_%7Bv%7D%29%20%5D%20%2B%20%5BC_vA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100%5Crho_%7Bv%7D%20-%20C_v%20%5Crho_%7Bv%7D%29%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
Replacing the values; we have:



