Answer:
CH₃CH₂CH₂COOH > CH₃CH₂COOH > ClCH₂CH₂COOH > ClCH₂COOH
Explanation:
Electron-withdrawing groups (EWGs) increase acidity by inductive removal of electrons from the carboxyl group.
Electron-donating groups (EDGs) decrease acidity by inductive donation of electrons to the carboxyl group.
- The closer the substituent is to the carboxyl group, the greater is its effect.
- The more substituents, the greater the effect.
- The effect tails off rapidly and is almost zero after about three C-C bonds.
CH₃CH₂-CH₂COOH — EDG — weakest — pKₐ = 4.82
CH₃-CH₂COOH — reference — pKₐ = 4.75
ClCH₂-CH₂COOH — EWG on β-carbon— stronger — pKₐ = 4.00
ClCH₂COOH — EWG on α-carbon — strongest — pKₐ = 2.87
The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.
A(t) = A(o) (1/2)^(t/d)
where t is the certain period of time. Substituting the known values,
A(t) = (20 mg)(1/2)^(85.80/14.30)
Solving,
A(t) = 0.3125 mg
Hence, the answer is 0.3125 mg.
Answer:
Most viscous to least viscous: 
Explanation:
For hydrocarbons, viscosity increases with increasing molar mass. Because increasing molar mass signifies increase in number of electrons in molecules.
We know that in non-polar hydrocarbons, only van der waal intermolecular force exists. Van der waal force is proportional to number of electrons in a molecule.
Therefore with increasing molar mass, van der waal force increases. hence molecules gets more tightly bind with each other resulting increase in viscosity.
Here molar mass order :
Therefore viscosity order :
Hydrogen-1, Carbon-13, Nitrogen-15, Fluorine-19, and Phosphorus-31 are the most useful. Out of these, Hydrogen-1 and Carbon-13 in NMR are the most useful nuclei because the these atoms are the most commonly present in organic molecules.
