Chemistry, 09.12.2019 22:31 Isaac0213

Cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture. the initial reactant in the production of cobalt-60 is . 59co b. 56fe c. 58fe d. 61co e. 60fe

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Answer from: angelcat9137

Answer : The correct option is, (C) $^{58}\textrm{Fe}$

Explanation :

Neutron capture : In this decay process, an atomic nucleus and one or more number of neutrons collide and combine to form a heavier nucleus. The mass number changes in this process.

The neutron capture equation is represented as,

$_Z^A\textrm{X}+_{0}^1\textrm{n}\rightarrow _{Z}^{A+1}\textrm{X}+\gamma$

(A is the atomic mass number and Z is the atomic number)

Beta emission or beta minus decay : It is a type of decay process, in which a neutrons gets converted to proton, an electron and anti-neutrino. In this the atomic mass number remains same.

The beta minus decay equation is represented as,

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0e$

(A is the atomic mass number and Z is the atomic number)

As per question, the cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture.

Process 1 : Neutron capture.

$_{26}^{58}\textrm{Fe}+_{0}^1\textrm{n}\rightarrow _{26}^{59}\textrm{Fe}+\gamma$

Process 2 : Beta emission.

$_{26}^{59}\textrm{Fe}\rightarrow _{27}^{59}\textrm{Co}+_{-1}^0e$

Process 3 : Neutron capture.

$_{27}^{59}\textrm{Co}+_{0}^1\textrm{n}\rightarrow _{26}^{60}\textrm{Co}+\gamma$

From this we conclude that, the initial reactant in the production of cobalt-60 is $_{26}^{58}\textrm{Fe}$

Hence, the correct option is, (C) $^{58}\textrm{Fe}$

Answer from: Quest

Answerhave membrane-bound organelles;

Answer from: Quest

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

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