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
The number of hydrogen atoms that are in 4.40 mol of ammonium sulfide is 2.12 x10^25 atoms
calculation
find the number of moles of Hydrogen in ammonium sulfide (NH4)2S
that is 4.40 x number of hydrogen atoms in (NH4)2S ( 4x2= 8 atoms)
moles is therefore= 4.40 x8= 35.2 moles
by use of Avogadro's law constant
that is 1mole = 6.02 x10^23 atoms
35.2 moles=?
by cross multiplication
{35.2 moles x 6.02 x10^23} /1 mole = 2.12 x10^25 atoms
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