An experiment with 55 co takes 47.5 hours. at the end of the experiment, 1.90 ng of 55-co remains. if the half-life is 18.0 hours, how many ng of 55-co were originally present?

Explanation:

Let A₀ = the original amount of ⁵⁵Co .

The amount remaining after one half-life is ½A₀.

After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.

After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀.

The general formula for the amount remaining is:

A =A₀(½)ⁿ

where n is the number of half-lives

n = t/t_½

Data:

   A = 1.90 ng

    t = 45 h

t_½ = 18.0 h

Calculation:

(a) Calculate n

n = 45/18.0 = 2.5

(b) Calculate A

1.90 = A₀ × (½)^2.5

1.90 = A₀ × 0.178

A₀ = 1.90/0.178 = 10.7 ng

The original mass of ⁵⁵Co was .

answer: option (4) is the correct answer.

explanation:

areas which are at low altitudes and near the equator tend to receive direct sunlight from the sun. hence, heat energy or radiations from the sun make these areas warmer and they have warmest water.

whereas areas which are at high altitudes and away from the equator do no receive direct sunlight. hence, no direct contact from the sun makes these regions colder and thus, they have colder water.

thus, we can conclude that the statement areas at low latitudes near the equator have the warmest water because they receive the most direct sunlight best explains the relationship between an area's geography and the temperature of its surface water.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.