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Effectus [21]
3 years ago
8

CUAL ES EL PRESUPUESTO ANUAL, DE LA MUNICIPALIDAD DISTRITAL DE MARCONA.

Social Studies
1 answer:
boyakko [2]3 years ago
6 0

Answer:

Cual de Paul answeèr ang bang ďe śeiòna.

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How does the Constitution cause potential problems concerning rights?
denis-greek [22]
The Constitution leaves a lot of space for personal interpretation, which means different people can view the document in different ways and to different degrees. For example, one may say, something that isn't stated in the Constitution means the government shouldn't do it. This is called strict interpretation of the Constitution. However, someone else may say that there wasn't anything in the Constitution prohibiting it. These differences have played a role in the formation of political parties. Hope this helps! :)
6 0
3 years ago
If the inflation rate in Mexico was twice the rate in the United States, but the Mexican monetary authorities kept the peso-doll
stellarik [79]

Answer:

E. Mexican products would be more expensive, while U.S.-made products would become comparatively less expensive.

Explanation:

Imagine the following scenario:

  • MXN: mexican peso
  • USD: united states dollar
  • There is no cost in shipping or transport

if the fixed relation is 2 MXN = 1 USD. Say you want to buy a TV that costs 200MXN in Mexico or 100USD in the US. That means that buying a TV costs the same in each country.

Now suppose that the TV price in Mexico, due to inflation, changes to MXN 300.

If you had 300MXN in your wallet, you could:

  • use MXN 200 to buy 100USD and use those to buy the TV, and have MXN 100 left, to buy other stuff.
  • or
  • you could buy the TV in Mexico, you would expend all the money.

That mean that, with the same amount of MXN you can buy more things in the US than in Mexico.

Another way to think about this is.

Say you are a mexican, if there is high inflation, the price of everything will rise, if you considder the exchange rate as another price that means that everything will cost more, except the US Dollar, which will stay in the same price

6 0
3 years ago
Is a natural state of rest for the body and mind that involves the reversible loss of consciousness.
nevsk [136]

Answer:

Sleep is a natural state of rest for the body and mind that involves the reversible loss of consciousness.

8 0
2 years ago
How did the colonies react to Britain’s economic policies
Vladimir [108]

Answer:

The colonists accepted the trade laws at first, but became exceedingly unhappy with the way that Britain was taxing them over time. They became angry, resulting in the Boston Tea Party.

5 0
3 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.

8 0
3 years ago
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