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Jlenok [28]
3 years ago
8

The reaction H2(g)+F2(g)?2HF(g) is spontaneous at all temperatures. A classmate calculates the entropy change for this reaction

and obtains a large negative value for ?S?.
Part A

Did your classmate make a mistake in the calculation?

Did your classmate make a mistake in the calculation?

There is mistake in the calculation. The second law states that in any spontaneous process there is an increase in the entropy of the universe.
There is mistake in the calculation. The second law states that in any spontaneous process the increase in the entropy of the system equals zero. Also there is a decrease in entropy of the system, as in the present case, this decrease is canceled by an increase in entropy of the surroundings.
There is no mistake in the calculation. The second law states that in any spontaneous process there is an decrease in the entropy of the system. Also there is a decrease in entropy of the system, as in the present case.
There is no mistake in the calculation. The second law states that in any spontaneous process the decrease in the entropy of the system equals zero. Also there might be a decrease in entropy of the system, as in the present case, this decrease is canceled by an increase in entropy of the surroundings.
Chemistry
1 answer:
Pachacha [2.7K]3 years ago
6 0

Answer:

The correct answer is there is a mistake in the calculation. The second law of thermodynamics state that in any spontaneous process there is an increase in the entropy of the universe.

Explanation:

According to the second law a reaction will occur in a system spontaneously if the total entropy of both system and surrounding increases during the reaction.That means in case of spontaneous reaction entropy change is always positive.

     But according to the question the reaction H2+F2=2HF is spontaneous in all temperature.So according to the second law of thermodynamics i can say that my classmate made a mistake in calculation that"s why his result for entropy change comes negative.

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Answer:

Enriched uranium-

Explanation:

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Identify whether the following example is qualitative or quantitative data.
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After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of
SIZIF [17.4K]

Answer:

b) The dehydrated sample absorbed moisture after heating

Explanation:

a) Strong initial heating caused some of the hydrate sample to splatter out.

This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).

b) The dehydrated sample absorbed moisture after heating.

Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.

c) The amount of the hydrate sample used was too small.

It will create some errors but they do not create a difference of 13% difference as stated in the problem.

d) The crucible was not heated to constant mass before use.

Here the error is small.

e) Excess heating caused the dehydrated sample to decompose.

Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.

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3 years ago
A particular reactant decomposes with a half-life of 113 s when its initial concentration is 0.372 M. The same reactant decompos
AysviL [449]

Answer:

Rate constant =  0.0237 M-1 s-1, Order = Second order

Explanation:

In this problem, it can be observed that as the concentration decreases, the half life increases. This means the concentration of the reactant is inversely proportional to the half life.

The order of reaction that exhibit this relationship is the second order of reaction.

In the second order of reaction, the relationship between rate constant and half life is given as;

t1/2 = 1 / k[A]o

Where;

k = rate constant

[A]o = Initial concentration

k = 1 / t1/2 [A]

Uisng the following values;

k = ?

t1/2 = 113

[A]o = 0.372M

k = 1 / (113)(0.372)

k = 1 / 42.036 = 0.0237 M-1 s-1

8 0
3 years ago
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