The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation:

H2SO4 + 2KOH → K2SO4 + 2H2O

Suppose 50 mL of KOH with unknown concentration is placed in a flask with bromthymol blue indicator. A solution of 0.10 M H2SO4 is dripped into the KOH solution. After exactly 33.4 mL of H2SO4 is added, the indicator changes from blue to yellow. What is the concentration of the KOH?

A. 0.067 M
B. 0.13 M
C. 0.15 M
D. 0.26 M

heat or electricity in the atmosphere- c.

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

answer; the numbers of valence electrons in the acetic acid molecule (ch3co2h) is 24;