Answer:
![[Pb^{2+}]=3.9 \times 10^{-2}M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3D3.9%20%5Ctimes%2010%5E%7B-2%7DM)
this is the concentration required to initiate precipitation
Explanation:
⇄ 
Precipitation starts when ionic product is greater than solubility product.
Ip>Ksp
Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.
This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.
![Ip=[Pb^{2}][2Cl^-]^2=Ksp](https://tex.z-dn.net/?f=Ip%3D%5BPb%5E%7B2%7D%5D%5B2Cl%5E-%5D%5E2%3DKsp)
lets solubility=S
![[Pb^{2+}] = S](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%20%3D%20S)
![[Cl^-]=2S](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D2S)
![Ksp=[Pb^{2+}]\times [Cl^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5Ctimes%20%5BCl%5E-%5D%5E2)
![S=\sqrt[3]{\frac{Ksp}{4} }](https://tex.z-dn.net/?f=S%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BKsp%7D%7B4%7D%20%7D)
this is the concentration required to initiate precipitation
Answer:
The molecular formula of glucose is C₆H₁₂O₆
Explanation:
Empirical formula:
It is the simplest formula gives the ratio of smallest whole number of atoms.
Molecular formula:
It gives the total number of atoms in a molecule of compound.
The molecular formula and empirical formula can be related as follow:
Molecular formula = n × empirical formula
Given data:
Empirical formula = CH₂O
Molecular formula = ?
It is stated in given problem that molecular formula is the 6 times of the empirical formula.
Molecular formula = n × empirical formula
Molecular formula = 6 × CH₂O
Molecular formula = C₆H₁₂O₆
The molecular formula of glucose is C₆H₁₂O₆.
The answer is: the distance between two nuclei is 2.35×10⁻¹⁰ m.
r(Na⁺) = 1.16×10⁻¹⁰ m; radius of sodium cation.
r(F⁻) = 1.9×10⁻¹⁰ m; radius of fluoride anion.
d(NaF) = r(Na⁺) + r(F⁻).
d(NaF) = 1.16×10⁻¹⁰ m + 1.9×10⁻¹⁰ m.
d(NaF) = 2.35×10⁻¹⁰ m; distance between two nuclei.
The sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.
