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How many half-lives will it take for the concentration of the n2o to reach 30 % of its original concentration?

Respuesta :

TheSandman1337
We have that after 1 half-life, the reactant reaches half its concentration [tex] (\frac{1}{2})^1 [/tex]. After 2 half-lives it reaches [tex] (\frac{1}{2})^2 [/tex] of its concentration, namely one-fourth. After n half lives we have that it reaches [tex] (\frac{1}{2})^n [/tex] of its initial concentration. Thus, we have to solve the equation [tex] (\frac{1}{2})^n=0,30 [/tex]. Taking the log base 2 of each part of the equation we get from logarithm properties: [tex]log_{2}[(1/2)^n]=log_{2}(2^{-n})=-n[/tex]. The other hand yields: [tex]-n=log_{2}(0,3)[/tex]. Solving for n we get that n=1.74. Thus, after around 1.74 half lives the concentration becomes 30% of the initial one. Note how this is consistent with our previous analysis; if we let it a little bit more, 2 half-lives, the concentration will become 25% of the initial.

Answer:

Number of lives required to reach 30% of the its original concentration is 2.

Explanation:

Number of half lives are calculated by using formula :

[tex]a=\frac{a_o}{2^n}[/tex]        ............(1)

where,

a = amount of reactant left after n-half lives

[tex]a_o[/tex] = Initial amount of the reactant

n = number of half lives

Here we have:

Initial concentration of nitrogen dioxide =[tex]a_o=x[/tex]

Concentration of nitrogen dioxide after n number of half lives = a

a = 30% of x :

[tex]\frac{30}{100}\times x= 0.3x[/tex]

[tex]a=\frac{a_o}{2^n}[/tex]

[tex]0.3x=\frac{x}{2^n}[/tex]

[tex]2^n=\frac{1}{0.3}[/tex]

n = 1.736 ≈ 2

Number of lives required to reach 30% of the its original concentration is 2.

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