RCOOH + NaOH → RCOONa + H₂O (salt and water)
RCOOH + OH⁻ → RCOO⁻ + H₂O
Answer:
Considering the half-life of 10,000 years, after 20,000 years we will have a fourth of the remaining amount.
Explanation:
The half-time is the time a radioisotope takes to decay and lose half of its mass. Therefore, we can make the following scheme to know the amount remaining after a period of time:
Time_________________ Amount
t=0_____________________x
t=10,000 years____________x/2
t=20,000 years___________x/4
During the first 10,000 years the radioisotope lost half of its mass. After 10,000 years more (which means 2 half-lives), the remaining amount also lost half of its mass. Therefore, after 20,000 years, the we will have a fourth of the initial amount.
Answer:
(a) H₃O⁺(aq) + H₂PO₄⁻(aq) ⟶ H₃PO₄(aq) + H₂O(ℓ)
(b) OH⁻(aq) + H₃O⁺(aq) ⟶ 2H₂O(ℓ)
Explanation:
The equation for your buffer equilibrium is:
H₃PO₄(aq) + H₂O(ℓ) ⇌ H₃O⁺(aq)+ H₂PO₄⁻(aq)
(a) Adding H₃O⁺
The hydronium ions react with the basic dihydrogen phosphate ions.
H₃O⁺(aq) + H₂PO₄⁻(aq) ⟶ H₃PO₄(aq) + H₂O(ℓ)
(b) Adding OH⁻
The OH⁻ ions react with the more acidic hydronium ions.
OH⁻(aq) + H₃O⁺(aq) ⟶ 2H₂O(ℓ)
Beta decay converts a neutron to a proton and emits a high-energy electron, producing a daughter nucleus with the same mass number as the parent and an atomic number that is higher by 1.
