Asample has a c614 activity of 0.0015 bq per gram of carbon.
(a) find the age of the sample, assuming that the activity per gram of carbon in a living organism has been constant at a value of 0.23 bq.
(b) evidence suggests that the value of 0.23 bq might have been as much as 45% larger. repeat part (a), taking into account this 45% increase.

Explanation:

Two important equations in radioactive decay are

We use them for carbon dating.

(a) Initial activity = 0.23 Bq

(i) Calculate the rate constant

The half-life of ¹⁴C is 5730 yr.

(ii) Calculate the age of the sample

(b) Initial activity =  45 % larger

N₀ = 1.45 × 0.230 Bq = 0.334 Bq

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

thats a little if u want the answer just look it up on google thats what i do haha : )