Hypochlorous acid decays in the presence of ultraviolet radiation. Assume that degradation occurs accord- ing to first-order kinetics and the rate of degradation was measured to be 0.12 day−1 (at a particular sun- light intensity and temperature). Given this, how long does it take for the concentration of hypochlorous acid to reach nondetectable levels (0.05 mg · L−1) if the initial concentration were 3.65 mg · L−1?

35.75 days

Explanation:

From the given information:

For first-order kinetics, the rate law can be expressed as:

Given that:

the rate degradation constant = 0.12 / day

current concentration C = 0.05 mg/L

initial concentration C₀ = 3.65 mg/L

㏑(0.01369863014) = -(0.12) t

-4.29 = -(0.12)

t = -4.29/-0.12

t = 35.75 days

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

b. examine and describe cells