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.
The solution needed is prepared as below
by use of the M1V1 =M2 V2 formula where
M1 = 2.25 L
v2 = 1.0M
M2 = 9.0 M
V2 =? l
make V2 the subject of the formula V2 =M1V1/M2
= 2.25 L x 1.0M/9.0 M = 0. 25 L
therefore the solution need 0.25 L of 9.0M H3PO4 and dilute it a final volume of 2.25 l
Tempratures
1. -60 C = 213.15 K
2. 250 C= 523.15 K
3. 365 K = 91.85 C
4. 205 K = -68.15 C
Pressures
1. 22.0 inches Hg = 558.8 mm Hg
2. 3.4 atm = 2584 mm Hg
3. 680 mm Hg = 0.894737 atm
4. 785 mm Hg = 104.658 kPa
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Although It showes I'm a begginer, I was An Ace, till my account got deleted.
My name was Sorry14.
Answer:
this is not and reaction of acid and base.(CaOCL2)
Explanation:
Ca2, Cl, are base and O is a neutral.
so how can that be a acid and base reaction