How many grams of Cu(OH)2 will precipitate when excess KOH solution is added to 76.0 mL of 0.764 M Cu(NO3)2 solution

5.66g of Cu(OH)2 can be produced

Explanation:

Cu(NO3)2 reacts with KOH, as follows:

Cu(NO3)2 + 2KOH → Cu(OH)2 + 2KNO3

76.0mL of 0.764M Cu(NO3)2 are, in moles:

0.0760L * (0.764mol / L) = 0.0581 moles Cu(NO3)2

As 1 mole of Cu(NO3)2 produce 1 mole of Cu(OH)2, the moles of Cu(OH)2 produced are 0.0581 moles Cu(OH)2.

Molar mass of Cu(OH)2 is 97.561g/mol. In 0.0581 moles there are:

0.0581 moles Cu(OH) * (97.561g / mol) =

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

level 4 to level 2 : ))