What concentration of NO−3NO3− results when 897 mL897 mL of 0.497 M NaNO30.497 M NaNO3 is mixed with 813 mL813 mL of 0.341 M Ca(NO3)2?

Explanation:

NaNO₃  = Na⁺  + NO₃⁻¹

.497 M                 .497 M

moles of NO₃⁻¹ = .897 x .497 = .4458 moles

Ca( NO₃)₂  = Ca + 2 NO₃⁻¹

.341 M                    2 x .341 M = .682 M

moles of NO₃⁻¹ = .813 x .682 = .5544 moles

Total moles =  .4458 moles  +  .5544 moles

= 1.0002 moles

volume of solution = 897 + 813 = 1710 mL

= 1.710 L

concentration of nitrate ion = 1.0002 / 1.710 M

= .585 M

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

the maximum mass of nh3 is nh3 in grams