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Ira Lisetskai [31]
3 years ago
14

Julia owns a house and property in Salt Lake City, Utah and another house in Florida. She stays in Florida for about three month

s a year and stays in Utah the other nine months. Julia is a consultant. Her employer and office are located in Utah, but she is able to work remotely when she is in Florida. What type of dwelling unit is the Utah home to Julia?
Social Studies
1 answer:
alexandr402 [8]3 years ago
7 0

The Utah house of the Julia is an alternative and permanent unit.

Explanation:

The Utah house of the Julia is an alternative and permanent unit to which she can use anytime without any restriction and permission of the second person.  Due to her active work mode she always needs another option where she can stay at anytime without delaying in her work.

This house is an alternative and permanent unit in terms of her Florida property because she only three months there. This property also saves her money because of this these properties she doesn’t need to live in hotels and resorts.  

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2 years ago
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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