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
Answer: 462 g
Explanation:molar mass is M= 63.55 +2·(12.01+14.01)= 115.59 g/mol.
Mass m= n·M = 4.0 mol·115.59 g/mol= 462.36 g
First calculate for the molar mass of the given formula unit, CaCO₃. This can be done by adding up the product when the number of atom is multiplied to its individual molar mass as shown below.
molar mass of CaCO₃ = (1 mol Ca)(40 g Ca/mol Ca) + (1 mol C)(12 g of C/1 mol of C) + (3 mols of O)(16 g O/1 mol O) = 100 g/mol of CaCO₃
Then, divide the given amount of substance by the calculated molar mass.
number of moles = (20 g)(1 mol of CaCO₃/100 g)
number of moles = 0.2 moles of CaCO₃
<em>Answer: 0.2 moles</em>
Answer:
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