dentify the compound that does NOT have hydrogen bonding. Identify the compound that does NOT have hydrogen bonding. HF (CH3)2N(CH2) 4 CH3 H2O CH3(CH2) 4 OH CH3(CH2) 2 NH2

(CH3)2N(CH2)

Explanation:

A hydrogen bond occurs in a molecule when hydrogen is covalently bonded to a highly electronegative atom.

That means that if we are looking for a compound in which there is no hydrogen bonding, we are actually looking for the compound in which hydrogen is not directly linked to an electronegative atom.

If we look at the options carefully, the compound that has such characteristic is only (CH3)2N(CH2) which is a tertiary amine. The nitrogen is directly linked to three alkyl groups and no hydrogen atom hence the molecule does not exhibit hydrogen bonding.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

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