Medical experts generally consider a lead of `30 mu g Pb` per `(dL)` of blood to pose a significant health risk `(Pb = 208)`. Number of lead atoms per `cm^3` of blood is.

Persons are medically considered to have lead poisoning if they have a concentration greater than 10 micrograms of lead per decilitre of blood. What is the concentration in parts per billion?

Persons are medically considered to have lead poisoning if they have a concentration greater than 10 micrograms of lead per decilitre of blood. What is the concentration in parts per billion?

"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor") n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H^(+)//OH^(-) furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant. Example 1: (1)In acidic medium : KMnO_(4) (n=5)to Mn^(2+) (2) In neutral medium : KMnO_(4)(n=3) to Mn^(2+) (3) In basic medium : KMnO_(4)(n=1) to Mn^(6+) Example 2 : FeC_(2)O_(4)to Fe^(3+)+2CO_(2) Total number of moles e^(-) lost by 1 mole of FeC_(2)O_(4) =1+1xx2 implies 3 Consider the following reaction. H_(3)PO_(2)+NaOH to NaH_(2)PO_(2)+H_(2)O What is the equivalent mass of H_(3)PO_(2) ?(mol.Wt.is M)

"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor") n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H^(+)//OH^(-) furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant. Example 1: (1)In acidic medium : KMnO_(4) (n=5)to Mn^(2+) (2) In neutral medium : KMnO_(4)(n=3) to Mn^(2+) (3) In basic medium : KMnO_(4)(n=1) to Mn^(6+) Example 2 : FeC_(2)O_(4)to Fe^(3+)+2CO_(2) Total number of moles e^(-) lost by 1 mole of FeC_(2)O_(4) =1+1xx2 implies 3 Consider the following reaction. H_(3)PO_(2)+NaOH to NaH_(2)PO_(2)+H_(2)O What is the equivalent mass of H_(3)PO_(2) ?(mol.Wt.is M)

"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor") n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H^(+)//OH^(-) furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant. Example 1: (1)In acidic medium : KMnO_(4) (n=5)to Mn^(2+) (2) In neutral medium : KMnO_(4)(n=3) to Mn^(2+) (3) In basic medium : KMnO_(4)(n=1) to Mn^(6+) Example 2 : FeC_(2)O_(4)to Fe^(3+)+2CO_(2) Total number of moles e^(-) lost by 1 mole of FeC_(2)O_(4) =1+1xx2 implies 3 Consider the following reaction. H_(3)PO_(2)+NaOH to NaH_(2)PO_(2)+H_(2)O What is the equivalent mass of H_(3)PO_(2) ?(mol.Wt.is M)

Find the electronegativity of lead with the help of the given valus. Screening consitant (sigma) of Pb = 76.70 . Atomic number of lead = 82 and covalent radius of Pb = 5.3 Å

Find the electronegativity of lead with the help of the given valus. Screening consitant (sigma) of Pb = 76.70 . Atomic number of lead = 82 and covalent radius of Pb = 5.3 Å

An element (atomic mass = 31) crystallises in a cubic structure. The density of the metal is 5.4g*cm^(-3) . The number of unit cells is 3.1g of metal is 6.022xx10^(22) . The number of atoms per unit cell is-