How much limestone (in kg) is required to completely neutralize a 4.5 billion liter lake with a pH of 5.3

1105.7kg of limestone must be added

Explanation:

A solution is neutralized when pH = 7.0. That is [H+] = 1x10⁻⁷M.

The initial [H+] is 10^-5.3 = 5.01x10⁻⁶M.

That means we need to neutralize:

5.01x10⁻⁶M - 1x10⁻⁷M = 4.91x10⁻⁶M

The moles of H+ in 4.5 billion liters = 4.5x10⁹L are:

4.91x10⁻⁶M * (4.5x10⁹L) = 22095 moles of H+

That reacts with limestone (CaCO3), as follows:

2H+ + CaCO3 → H2O + CO2 + Ca²⁺

Moles of limestone needed to neutralize the lake are:

22095 moles of H⁺ * (1 mole Limestone / 2 moles H⁺) = 11047.5 moles of limestone.

Molar mass of limestone is 100.09g/mol. Mass of limestone we need to add are:

11047.5 moles of limestone * (100.09g / mol) = 1.106x10⁶g of limestone. In kg:

1.106x10⁶g of limestone * (1kg / 1000g) =

1105.7kg of limestone must be added

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.