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Serggg [28]
1 year ago
13

A substance decays so that the amount a of the substance left at time t is given by: a = a0 ∙ (0.8)t where a0 is the original am

ount of the substance. what is the half-life (the amount of time that it takes to decay to half the original amount) of this substance rounded to the nearest tenth of a year?
Chemistry
1 answer:
Zolol [24]1 year ago
4 0

The half-life of the substance is 3.106 years.

<h3>What is the formula for exponential decay?</h3>
  • The exponential decline, which is a rapid reduction over time, can be calculated with the use of the exponential decay formula.
  • The exponential decay formula is used to determine population decay, half-life, radioactivity decay, and other phenomena.
  • The general form is F(x) = a.

Here,

a = the initial amount of substance

1-r is the decay rate

x = time span

The equation is given in its correct form as follows:

a = a_{0}×(0.8)^{t}

As this is an exponential decay of a first order reaction, t is an exponent of 0.8.

Now let's figure out the half life. Since the amount left is half of the initial amount at time t, that is when:

a = 0.5 a0

<h3>Substituting this into the equation:</h3>

0.5a_{0} = a_{0}×(0.8)^{t}

0.5 = (0.8)^{t}

taking log on both sides

t log 0.8 = log 0.5

t = log 0.5/log 0.8

t = 3.106 years

The half-life of the substance is 3.106 years.

To learn more about exponential decay formula visit:

brainly.com/question/28172854

#SPJ4

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Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.
Vsevolod [243]
Missing in your question Ka2 =6.3x10^-8
From this reaction:
 H2SO3 + H2O ↔ H3O+  + HSO3-
by using the ICE table :
                H2SO3     ↔    H3O     +    HSO3- 
intial         0.6                     0                  0
change     -X                      +X                +X
Equ         (0.6-X)                  X                   X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
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[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
                 HSO3-     ↔   H3O     +     SO3-
initial        0.088              0.088              0
change    -X                      +X                   +X
Equ         (0.088-X)          (0.088+X)          X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] =  0.088 as the value of Ka2 is very small
 6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
               = 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
       = -㏒ 0.088 = 1.06 
5 0
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Answer:

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Explanation:

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P₁V₁ = P₂V₂

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