For the purpose, we will use the equation for determining the dissociation constant from concentration and <span>percent of ionization:
Kd = c </span>× α²
α = √(Kd/c) × 100%
Kd = 6.0×10⁻⁷
c(HA) = 0.1M
α = √(6.0×10⁻⁷/0.1) × 100% = 0.23%
So, in the solution, the acid <span>percent of ionization will be just 0.23%.</span>
Answer:
Mass = 381.28 g
Explanation:
Given data:
Number of moles of HNO₃ = 16 mol
Mass of Cu needed to react with 16 mol of HNO₃ = ?
Solution:
Chemical equation:
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O + 2NO
Now we will compare the moles of Cu with HNO₃ from balance chemical equation.
HNO₃ : Cu
8 : 3
16 : 3/8×16 = 6
Mass of Cu needed:
Mass = number of moles × molar mass
Mass = 6 mol × 63.546 g/mol
Mass = 381.28 g
Its inorganic as MgCO3 is contains no carbon more hydrogen which is a crutial component of all organic compounds
Atomic mass / mass number / atomic weight
(all of which mean the same thing)
By Boyles Law (P1V1=P2V2), substituting values in and solving for V2, we find that the new volume is 3.6 L<span />
