Answer:
(a) O = Valance Electrons (6), Inner electrons (8)
(b) Sn = Valance Electrons (2), Inner electrons (36)
(c) Ca = Valance Electrons (2), Inner electrons (20)
(d) Fe = Valance Electrons (2), Inner electrons (26)
(e) Se = Valance Electrons (6), Inner electrons (34)
Answer:
Explanation:
There are some radioactive nuclides can be used to measure time on an archeological scale. One is the best example of this is radiocarbon dating. This process is based on the ratio of caebon-14 to carbon-12 in the atmosphere which is relatively constant.
The half time of C-14 5730 years
Carbon-14 is a radioactive nucleus. It has a half-life of 5730 years.
All living tissues like plants and animal absorbed carbon-12 along with carbon-14 with same ratio of caebon-14 to carbon-12 in the atmosphere.
Carbon-14 dating is based on the ratio of carbon-14 to carbon-12 in the atmosphere which is relatively constant
7.30 x 10^-7 km. the others have 4 significant figures.
Answer:
5.8μg
Explanation:
According to the rate or decay law:
N/N₀ = exp(-λt)------------------------------- (1)
Where N = Current quantity, μg
N₀ = Original quantity, μg
λ= Decay constant day⁻¹
t = time in days
Since the half life is 4.5 days, we can calculate the λ from (1) by substituting N/N₀ = 0.5
0.5 = exp (-4.5λ)
ln 0.5 = -4.5λ
-0.6931 = -4.5λ
λ = -0.6931 /-4.5
=0.1540 day⁻¹
Substituting into (1) we have :
N/N₀ = exp(-0.154t)----------------------------- (2)
To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:
N = 5.0 μg
N₀ = Unknown
t = 1 day
Substituting into (2) we have
[5/N₀] = exp (-0.154 x 1)
5/N₀ = 0.8572
N₀ = 5/0.8572
= 5.8329μg
≈ 5.8μg
The Chemist must order 5.8μg of 47-CaCO3
