Answer:
METHOD 1: (surface area of a solid reactant) METHOD 2: (concentration or pressure of a reactant)
Explanation:
METHOD 1: (surface area of a solid reactant) Increasing the surface area of a solid reactant exposes more of its particles to attack. This results in an increased chance of collisions between reactant particles, so there are more collisions in any given time and the rate of reaction increases.
METHOD 2: (concentration or pressure of a reactant) Increasing the concentration means that we have more particles in the same volume of solution. This increases the chance of collisions between reactant particles, resulting in more collisions in any given time and a faster reaction. As we increase the pressure of reacting gases, we increase the rate of reaction.
Chemical reaction (dissociation) 1: C₂O₄H₂(aq) ⇄ C₂O₄H⁻(aq) + H⁺(aq).
Chemical reaction (dissociation) 2: C₂O₄H⁻(aq) ⇄ C₂O₄²⁻(aq) + H⁺(aq).
c(C₂O₄H⁻) = c(H⁺) = x.
c(C₂O₄H₂) = 0.0269 M.
pKa₁ = 1.23.
Ka₁ = 10∧(-1.23) = 0.059.
Ka₁ = c(C₂O₄H⁻) · c(H⁺) / c(C₂O₄H₂).
0.059 = x² / (0.0269 M - x).
Solve quadratic eqaution: x = c(H⁺) = 0.02 M.
pH = -log(0.02 M) = 1.7.
<span>the simplest organic molecules, consisting of only carbon and hydrogen and with only single bonds between carbon atoms.
Ex. Methane, Ethan, Propane, Butane
</span>
You forgot to post 'the following' .
