Some acid-base reactions form gases. these occur when you form an acid with carbonates and sulfites. in these reactions, the species that is formed is unstable and immediately decomposes into water and a gas. for carbonates, h2co3 is formed which decomposes into water and co2 gas. for sulfites, h2so3 is formed which decompsoes into water and so2 gas. one example is the reaction between vinegar and baking soda. this reaction is often used in "volcano" demonstrations. vinegar contains acetic acid and baking soda contains sodium hydrogen carbonate. what is the net ionic reaction when vinegar and baking soda react?

CH₃COOH(aq) + HCO₃⁻(aq) ⇄ CH₃COO⁻(aq) + H₂O(l) + CO₂(g)

Explanation:

Acetic acid reacts with sodium hydrogen carbonate to produce sodium acetate, water, and carbon dioxide. The complete molecular equation is:

CH₃COOH(aq) + NaHCO₃(aq) ⇄ CH₃COONa(aq) + H₂O(l) + CO₂(g)

Electrolytes are species that dissociate in water and form ions. This allows us to write a complete ionic equation. Weak electrolytes (CH₃COOH) exist mostly in the molecular form while strong electrolytes (NaHCO₃, CH₃COONa) are mostly dissociated in aqueous solutions. The complete ionic equation would be:

CH₃COOH(aq) + Na⁺(aq) + HCO₃⁻(aq) ⇄ CH₃COO⁻(aq) + Na⁺(aq) + H₂O(l) + CO₂(g)

Some species do not change during the reaction and are known as spectator ions (Na⁺). We can omit them in the net ionic equation:

CH₃COOH(aq) + HCO₃⁻(aq) ⇄ CH₃COO⁻(aq) + H₂O(l) + CO₂(g)

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

it is the most commonly accepted one