Read the following chemical equations. reaction 1: ch4 + 2o2 → 2h2o + co2 reaction 2: 2cuo + c → 2cu + co2 which of the following statements is true for both the chemical equations? carbon is reduced in both the reactions. carbon is oxidized in both the reactions. carbon is reduced in reaction 1 and oxidized in reaction 2. carbon is oxidized in reaction 1 and reduced in reaction 2.

The correct option is B) Carbon is oxidized in both the reactions.

Explanation:

Oxidation can be described as the loss of electrons, ions or atoms. Reduction can be described as the gain of electrons, ions or atoms.

In reaction 1: CH4 + 2O2 → 2H2O + CO2

Carbon is losing hydrogen atoms to become carbon dioxide. Hence, we will say that carbon is being oxidized.

In reaction 2: 2CuO + C → 2Cu + CO2

In this reaction also, carbon is oxidized into carbon dioxide by losing electrons.

Hence, in both the reactions, carbon is being oxidized.

it is important to conserve water because there is very limited amount of freshwater here on earth and if we are not smart about how we use it we will only be runnning out of our supply faster than neccessarry.

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