Calculate the volume in liters occupied by 39.6 grams of NO at 5.54 atm and 72°C

6.75 L

Explanation:

The following data were obtained from the question:

Mass of NO = 39.6 g

Pressure (P) = 5.54 atm

Temperature (T) = 72°C

Volume (V) =...?

Next, we shall determine the number of mole in 39.6 g of NO. This is illustrated below:

Mass of NO = 39.6 g

Molar mass of NO = 14 + 16 = 30 g/mol

Mole of NO =..?

Mole = mass /Molar mass

Mole of NO = 39.6 /30

Mole of NO = 1.32 moles

Next, we shall convert 72°C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Temperature (T) = 72°C

Temperature (T) = 72°C + 273

Temperature (T) = 345 K

Finally, we shall determine the Volume occupied by NO as follow:

Pressure (P) = 5.54 atm

Temperature (T) = 345 K

Number of mole (n) = 1.32 moles

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =...?

PV =nRT

5.54 × V = 1.32 × 0.0821 × 345

5.54 × V = 37.38834

Divide both side by 5.54

V = 37.38834 / 5.54

V = 6.75 L

Therefore, the volume occupied by 39.6 g of NO is 6.75 L

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