Afirst-order reaction has a half-life of 29.2 s . how long does it take for the concentration of the reactant in the reaction to fall to one-sixteenth of its initial value?

Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value

The half-life for the chemical reaction is 29,2 s and is independent of initial concentration.
c₀ - initial concentration the reactant.

c - concentration of the reactant remaining at time.

t = 29,2 s.
First calculate the rate constant k:
k = 0,693 ÷ t = 0,693 ÷ 29,2 s = 0,0237 1/s.
ln(c/c₀) = -k·t₁.
ln(1/16 ÷ 1) = -0,0237 1/s · t₁.

t₁ = 116,8 s.

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.