Ameteorite contains 0.17 g of nickel-59, a radioisotope that decays to form cobalt-59. the meteorite also contains 5.27 g of cobalt-59. how many nickel-59 half-lives have passed since the meteorite formed? 1 5 5.1 5.44

Answer : The correct option is, 5

Solution : Given,

Mass of nickel-59 = 0.17 g

Mass of cobalt-59 = 5.27 g

Equation for the radioactive decay of nickel-59 is :

Now, we have to calculate the initial amount of nickel-59, we are using the stoichiometry of the reaction and moles of the reactant and product.

Formula used :

Moles of  

Moles of

By stoichiometry of the reaction,

1 mole of is produced by 1 mole

So, 0.089 moles of will be produced by =

Amount of  decomposed will be = 0.089 moles

Initial amount of  will be = Amount decomposed + Amount left = (0.00288 + 0.089)moles = 0.09188 moles

Now, to calculate the number of half lives, we use the formula :

where,

a = amount of reactant left after n-half lives = 0.00288 moles

= Initial amount of the reactant = 0.09188 moles

n = number of half lives

Now put all the given values in above equation, we get

Taking log on both sides, we get

Therefore, '5' number of half-lives have passed since the meteorite formed .

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