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bekas [8.4K]
3 years ago
15

why do we have to decide what to do in a day? this is for my econ class i’m doing scarcity and opportunity cost

Social Studies
1 answer:
Readme [11.4K]3 years ago
3 0

Answer:

deciding what to do in a day is very important for many reasons. if you're having a bit of a shortage with money for instance, you'd want to decide what you'd want to do that day with your money, get food? or get things you don't need? save it? making a plan is always good before you do something, because that way you'll rarely experience any mistakes.

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Derive expressions for the half-life of zero-, first-, and second-order reactions using the integrated rate law for each order.
olganol [36]

Answer:

A very useful concept in the study of the kinetics of a reaction is the half-life. This concept is used in many other fields. Perhaps the best known is the study of radioactive decay processes. These processes are, from our point of view, first-order reactions, since the rate of decay of a radionuclide only depends on the amount of this present in the sample.

We will define the average life time of a reagent as the time necessary for half of its initial concentration to react. It is usually represented as

t

1/2

.

In the case of a first-order reaction, whose velocity equation we know is:

v = k⋅ [A]

So, according to the definition of

t

1/2

we have to:

t =

t

1/2

⟹ [A] =

[A

]

0 0

two

Using the logarithmic formula for first order reactions:

ln [A] = ln [A

]

0 0

−k⋅t

we will have to:

In

[A

]

0 0

two

= ln [A

]

0 0

−k⋅

t

1/2

so that:

ln [A

]

0 0

−ln2 - ln [A

]

0 0

= - k⋅

t

1/2

and we have that:

t

1/2

=

- ln2

- k

=

ln2

k

From this expression, we see that the average life time is, in this case, independent of the initial concentration. On the other hand, it is evident that the specific velocity constant has dimensions of

(weather

)

−1

and it is also independent of the initial concentration.

In the same way, we can apply the concept of half-life to the case of a second order reaction with respect to a reagent. For these types of reactions, we have to:

1

[A]

-

1

[A

]

0 0

= k⋅t

then, substituting the values ​​of the definition of half-life, we will have:

1

[A

]

0 0

two

-

1

[A

]

0 0

= k⋅

t

1/2

that is to say:

two

[A

]

0 0

-

1

[A

]

0 0

= k⋅

t

1/2

now we simplify

1

[A

]

0 0

= k⋅

t

1/2

and we cleared:

t

1/2

=

1

k⋅ [A

]

0 0

As we can see, in the case of second-order reactions, the half-life does depend on the initial concentration of the reference reagent.

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Answer:

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Explanation:

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