Calculate the amount of co2 produced (in kg) per mole of acetate oxidized during aerobic oxidation. if on the other hand, acetate is oxidized to co2 by denitrification, determine the amount of co2 produced (in kg) per mole of acetate oxidized. the half reaction for denitrification is:
the balanced equation for the formation of water is as follows
2h₂ + o₂ > 2h₂o
molar ratio of oxygen to water is 1: 2
mass of water produced is - 46 g
therefore the number of moles of water produced - 46 g / 18 g/mol = 2.56 mol
1 mol of oxygen reacts to give 2 mol of water assuming oxygen is the limiting reactant
for 2 mol of water to be produced 1 mol of oxygen has to react
therefore for 2.56 mol of water to be produced - 1/2 x 2.56 mol = 1.28 mol of oxygen should react
mass of oxygen reacted is - 1.28 mol x 32 g/mol = 41.0 g
mass of oxygen that should react is 41 g
hmmmmmm let me think
4. 1.11111 x 10⁻⁹ m, 1.61616 x 10⁻⁷ m.
explanation:the main idea here to solve this problem is that to precipitate a substance, the ionic products of the ions that form this substance should be ≥ its solubility product (ksp).to precipitate baso₄, the ionic product of baso₄ ([ba²⁺][so₄²⁻]) should be ≥ ksp of baso₄ (ksp = 1.1 x 10⁻¹⁰).
baso₄ ↔ ba²⁺ + so₄²⁻
ksp = [ba²⁺][so₄²⁻][so₄²⁻] = ksp / [ba²⁺] = (1.1 x 10⁻¹⁰) / (0.099) = 1.11111 x 10⁻⁹ m.
by the same way; the other precipitate:to precipitate pbso₄, the ionic product of pbso₄ ([pb²⁺][so₄²⁻]) should be ≥ ksp of pbso₄ (ksp = 1.6 x 10⁻⁸).
pbso₄ ↔ pb²⁺ + so₄²⁻
ksp = [pb²⁺][so₄²⁻][so₄²⁻] = ksp / [pb²⁺] = (1.6 x 10⁻⁸) / (0.099) = 1.61616 x 10⁻⁷ m.
so, the right answer is 4. 1.11111 x 10⁻⁹ m, 1.61616 x 10⁻⁷ m.