Answer:
HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)
Explanation:
The HI donates a proton to the water, converting it to a hydronium ion
HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)
Thus, the HI is behaving like a Brønsted acid.
They go through radioactive decay because when atoms are unstable by going through it they are emitting radiation in natural process and they gain stability by losing energy.
Answer:
NaNO3 (solubility = 89.0 g/100 g H2O)
Explanation:
The solubility of a specie is the amount of solute that will dissolve in one litre of the solvent. Solubility is usually expressed in units of molarity.
Now let us calculate the molarity of the NaNO3 (solubility = 89.0 g/100 g H2O)
Molar mass of NaNO3= 23+14+3(16)= 85gmol-1
Mass of solute=89.0g
Amount of solute= mass of NaNO3/molar mass of NaNO3
Amount of solute= 89.0g/85.0 gmol-1
= 1.0moles of NaNO3
Note that 100g of water=100cm^3 of water.
If 1.0 moles of NaNO3 dissolve in 100cm^3 or water therefore,
x moles of NaNO3 will dissolve in 1000cm^3 of water
x= 1.0 × 1000/ 100
x= 10.0 moles of NaNO3
