A45.9 g sample of a metal is heated to 95.2°c and then placed in a calorimeter containing 120.0 g of water (c = 4.18 j/g°c) at 21.6°c. the final temperature of the water is 24.5°c. which metal was used?

Iron

Explanation:

Heat released by the metal sample will be equivalent to the heat absorbed by  water.

But heat = mass × specific heat capacity × temperature change

Thus;

Heat released by the metal;

= 45.9 g × c ×(95.2 -24.5) , where c is the specific heat capacity of the metal

= 3245.13c joules

Heat absorbed by water;

= 120 g × 4.18 J/g°C × (24.5-21.6)

= 1454.64 joules

Therefore;

3245.13c joules = 1454.64 joules

c = 1454.64/3245.13

  = 0.448 J/g°C

The specific heat capacity of the  metal sample is 0.448 J/g°C. The metal use is most likely, Iron.

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

wher is the rest of the question