Answer:
1.
643.21g 1 mol 6.022^23
262.87 g 1 mol
= 1.4735E24 [Mg3(PO4)2]
2.
4.061x10^24 1mol 22.4 (L)
6.022^23 1mol
= 151 liters H2O2
3.
479.3g 1 mol 6.022^23
18.02g 1mol
= 1.60E25 H20 atoms
4.
80.34L 1mol 164.1
22.4L 1mol
588.6g Ca(NO3)2
5.
893.7g 1mol 22.4
44.01g 1mol
= 427 L CO2 or 427.4
6.
5.39 x 10^25 1mol 78.01
6.022^23 1mol
= 6980g Al(OH)3
hope this helps!! :)
Answer
7665 years
Procedure
Let N₀ be the amount of carbon-14 present in a living organism. According to the radioactive decay law, the number of carbon-14 atoms, N, left in a dead tissue sample after a certain time, t, is given by the exponential equation:
N = N₀e^(-λt)
where λ is the decay constant which is related to half-life (T1/2) by the equation:

Here, ln(2) is the natural logarithm of 2.
The percent of carbon-14 remaining after time t is given by N/N₀.
Using the first equation, we can determine λt.
The half-life of carbon-14 is 5,720 years, thus, we can calculate λ using the second equation, and then find t.

Solving the second equation for t, and using the λ we have just calculated we will have
t= 7665 years
I think it would seeing as you typicaly have to use a match to light a gas burner