How many grams of N2 can be produced when 6.50 g of O2
reacts?

3.79 g of N2.

Explanation:

We'll begin by writing the balanced equation for the reaction.

This is given below:

4NH3 + 3O2 → 2N2 + 6H2O

Next, we shall determine the mass of O2 that reacted and the mass of N2 produced from the balanced equation.

This is illustrated below:

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Summary:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Finally, we shall determine the mass of N2 produced by reacting 6.50 g of O2.

This can be obtained as follow:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.50 g of O2 will react to produce = (6.50 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were obtained from the reaction..

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

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.