Calculate the mass of ammonia that contains a trillion hydrogen atoms.

              9.4 × 10⁻¹² g of NH₃

Solution:

Step 1: Calculate Number of NH₃ Molecules containing 1 Trillion Hydrogen Atoms:

As,

                  3 Hydrogen Atoms are present in  =  1 Molecule of NH₃

So,

    1 × 10¹² Hydrogen Atoms will be present in  =  X Molecules of NH₃

Solving for X,

                      X  =  (1 × 10¹² × 1) ÷ 3

                      X =  3.33 × 10¹¹ Molecules of NH₃

Step 2: Calculate Moles of NH₃:

As,

                          Moles  =  Number of Molecules ÷ 6.022 × 10²³

Putting value,

                          Moles  =  3.33 × 10¹¹ Molecules ÷ 6.022 × 10²³

                          Moles  = 5.53 × 10⁻¹³ mol

Step 3: Calculate Mass of NH₃:

As,

                          Moles  =  Mass ÷ M.Mass

Solving for Mass,

                          Mass  =  Moles ×  M.Mass

Putting values,

                          Mass  =  5.53 × 10⁻¹³ mol × 17 g.mol⁻¹

                          Mass =  9.4 × 10⁻¹² g of NH₃

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

am i too late?

explanation: