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>
They may break bonds,form new compounds, new ions etc...
Answer:
flourine(1681),carbon(1086),lithium(520) and potassium(419)
as we can see that the ionization energy of flourine is the highest than carbon than lithium and than potassium
Explanation:
i hope it will help you
Answer:
+1
Explanation:
Na₂O₂
NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.
Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:
Na₂O₂ = 0 (oxidation number of ground state compound is zero)
2Na + 2O = 0
O = –1
2Na + 2(–1) = 0
2Na – 2 = 0
Collect like terms
2Na = 0 + 2
2Na = 2
Divide both side by 2
Na = 2/2
Na = +1
Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1
Sodium-22 remain : 1.13 g
<h3>Further explanation
</h3>
The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually, radioactive elements have an unstable atomic nucleus.
General formulas used in decay:
T = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
half-life = t 1/2=2.6 years
T=15.6 years
No=72.5 g
