If you add a surfactant in water-oil mixture, surfactant molecules would form micelles wherein the hydrophobic end of the surfactant faces inside the micelle while the hydrophilic end faces the water molecules. Inside these micelles oil particles are attached.
<span>M(HCl) * </span><span>V(HCl) </span>= <span>M(NaOH) * </span><span>V(<span>NaO<span>H)
</span></span></span>
M(HCl) = 0.35
<span>V(HCl) = 45mL
</span>M(NaOH)= 0.35
now, solne for V(NaOH) by putting these values in the above equation.
M(HCl) * <span>V(HCl) </span>= <span>M(NaOH) * </span><span>V(NaOH)</span>
<span>0.35 * 45 = 0.35 * V(NaOH)</span>
<span>V(NaOH) = 45 mL</span>
Answer:
Kc for this reaction is 0.43
Explanation:
This is the equilibrium:
N₂(g) + 2H₂O(g) → 2NO(g) +2H₂(g)
And we have all the concentration at equilibrium:
N₂: 0.25M
H₂ : 1.3M
NO: 0.33M
H₂: 1.2M
They are ok, because they are in MOLARITY. (mol/L)
Let's make the expression for Kc
Kc = ( [NO]² . [H₂]² ) / ([N₂] . [H₂O]²)
Kc = (0.33² . 1.2²) / (0.25 . 1.2²)
Kc = 0.4356
In two significant digits. 0.43
<span>Hydroxy group..
... :):) ....</span>