What is the unknown metal if the temperature of a beaker of 100ml of water was raised 17c to 19 c when 21 grams of the metal at 122.5c was placed into the beaker ?

The metal has a heat capacity of 0.385 J/g°C

This metal is copper.

Explanation:

Step 1: Data given

Mass of the metal = 21 grams

Volume of water = 100 mL

 ⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams

Initial temperature of metal = 122.5 °C

Initial temperature of water = 17°C

Final temperature of water and the metal = 19 °C

Heat capacity of water = 4.184 J/g°C

Step 2: Calculate the specific heat capacity

Heat lost by the metal = heat won by water

Qmetal = -Qwater

Q = m*c*ΔT

m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)

21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)

-2173.5 *c(metal) = -836.8

c(metal) = 0.385 J/g°C

The metal has a heat capacity of 0.385 J/g°C

This metal is copper.

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

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%