The half-life of carbon-14 is 5730 years. therefore, 5730 years after being cut from a tree, a piece of wood will have 50.0% of the carbon-14 that it had originally. what percentage of the original carbon-14 will remain in the wood after another 5730 years?

For this problem, we use the formula for radioactive decay which is expressed as follows:

An = Aoe^-kt

where An is the amount left after time t, Ao is the initial amount and k is a constant. 

We can use the equation above to find the percent of the original substance at a given time. However, the time given here is equal to the half life of C-14. So the percent left would be 50% or half the original.

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

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).