Answer:
O H C
Moles in 100g 3.33 6.65 3.33
Ratio 1.00 2.00 1.00
Possible empirical formula = 
D. With the same number of protons and different number of neutrons.
Answer:
the range should be 2.2 to 4.3
Explanation:
I think so because the numbers at the left side of the scale from 1 are more acidic so as it increases it's still acidic but lesser so 1 is more acidic than 2 so I used 2.2 as the beginning of the range because it's less acidic than A even though its a greater number and 4.3 is lesser than 4.4 but its still greater on the scale. frankly speaking I don't feel so correct because it's in decimal so try and compare facts thank you
Answer:
17202.6 years
Explanation:
Activity of the living sample (Ao) = 160 counts per minute
Activity of the wood sample (A) = 20 counts per minute
Half life of carbon-14 = 5730 years
t= age of the artifact
From;
0.693/t1/2= 2.303/t log Ao/A
Then;
0.693/ 5730= 2.303/t log Ao/A
Substituting values;
0.693/5730= 2.303/t log (160/20)
Then we obtain;
1.209×10^-4 = 2.0798/t
t= 2.0798/1.209×10^-4
Thus;
t= 17202.6 years
Therefore the artifact is 17202.6 years old.
