A3.35 gram sample of an unknown gas is found to occupy a volume of 1.64 l at a pressure of 706 mmhg and a temperature of 59 °c. assume ideal behavior. the molar mass of the unknown gas is g/mol.

 The molar mass of the unknown gas is 59.8 g/mol

Explanation:

According to the ideal gas equation:-

P= Pressure of the gas = 706 mmHg = 0.93 atm    (760mmHg=1atm)

V= Volume of the gas = 1.64 L

T= Temperature of the gas = 59°C=(59+273)K=332 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

To calculate the moles, we use the equation:

Thus the molar mass of the unknown gas is 59.8 g/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

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.