1) given the balance equation below. calculate how much na3po4 in grams you
would need to create 100 grams of naoh.
1 na3po4 + 3 koh → 3 naoh + 1 k3po4

136.67 g of Na3PO4 is required to create 100 gram of NaOH.

Explanation:

The balanced equation:

1 mole Na3PO4 = 164 g/mole (Molar mass)

1 mole NaOH = 40 g/mole (Molar mass)

Now,

1 mole of Na3PO4 produce = 3 mole of NaOH

164 g/mol of Na3PO4 produce = 3(40) g/mol of NaOH

or

120 g/mol of NaOH is produced from = 164 g/mol of Na3PO4

1 g/mol of NaOH is produced from =

100 grams of NaOH is produced from =

gram of Na3PO4

calculate,

= 136.67 g

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

your parents i think this could be it