Answer:
Amount of salt needed is around 2.3*10³ g
Explanation:
The salt content in sea water = 3.5 %
This implies that there is 3.5 g salt in 100 g sea water
Density of seawater = 1.03 g/ml
Volume of seawater = volume of tank = 62.5 L = 62500 ml
Therefore, the amount of seawater required is:
The amount of salt needed for the calculated amount of seawater is:
Answer:
Kc for this equilibrium is 2.30*10⁻⁶
Explanation:
Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products are held constant.
Being:
aA + bB ⇔ cC + dD
the equilibrium constant Kc is defined as:
In other words, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients. Kc is constant for a given temperature, that is to say that as the reaction temperature varies, its value varies.
In this case, being:
2 NH₃(g) ⇔ N₂(g) + 3 H₂(g)
the equilibrium constant Kc is:
Being:
and replacing:
you get:
Kc= 2.30*10⁻⁶
<u><em>Kc for this equilibrium is 2.30*10⁻⁶</em></u>