Dissociation of Pb₃(PO₄)₂ is;
Pb₃(PO₄)₂(s) ⇆ 3Pb²⁺(aq) + 2PO₄³⁻(aq)
initial - -
change -X +3X +2X
Equilibrium 3X 2X
Ksp = [Pb²⁺(aq)]³ [PO₄³⁻(aq)]²
1.0 x 10⁻⁵⁴ = (3X)³ (2X)²
1.0 x 10⁻⁵⁴ = 108X⁵
X = 6.21 x 10⁻¹² M
Hence the molar solubility of Pb₃(PO₄)₂ is 6.21 x 10⁻¹² M.
Answer:
The answer to your question is an acid base reaction
Explanation:
A single replacement reaction is a reaction in which one metal replaces the cation of a compound. The reaction of this problem is not of this type because here the reactants are compounds no single elements.
A decomposition reaction is a reaction in which one compound decomposes into two or more products. This is not the answer to this question because in this reaction there are two reactants not only one.
A synthesis reaction is a reaction in which two reactants form only one product. The reaction of this problem is not of this type because there are two products not only one.
An acid-base reaction is a kind of double replacement reaction. In some acid-base reactions, there is an interchange of cations and anions like is shown in this reaction.
Cells will not able to produce proteins by translation as it's happens in ribosomes
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
