All categories

3 years ago

Pluto is a planet True False

Chemistry

1 answer:

koban [17]3 years ago

6 0

Answer:

False

Explanation:

It's classified as a dwarf planet, but not technically a planet.

You might be interested in

A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f

deff fn [24]

Explanation:

Entropy means the amount of randomness present within the molecules of the body of a substance.

Relation between entropy and microstate is as follows.

S = $K_{b} \times ln \Omega$

where, S = entropy

$K_{b}$ = Boltzmann constant

$\Omega$ = number of microstates

This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.

Also, let us assume that $\Omega$ = 1, and $\Omega'$ = 0.833

Therefore, change in entropy will be calculated as follows.

$\Delta S = K_{b} \times ln \Omega' - K_{b} \times ln \Omega$

= $1.38 \times 10^{-23} \times ln(0.833) - 1.38 \times 10^{-23} \times \times ln(1)$

= $1.38 \times 10^{-23} \times (-0.182)$

= $-0.251 \times 10^{-23}$

or, = $-2.51 \times 10^{-24}$

Thus, we can conclude that the entropy change for a particle in the given system is $-2.51 \times 10^{-24}$ J/K particle.

8 0

3 years ago

From left to right, the three regions of a phase diagram represent the _____ of a substance.

timofeeve [1]

Your answer is D.

:good luck

5 0

3 years ago

Read 2 more answers

Click the "draw structure" button to launch the drawing utility. under certain reaction conditions, 2,3−dibromobutane reacts wit

REY [17]

Answer:

Explanation:

According to this. Let's analize the possible products a, b and c.

First, the problem states that we have 2 eq. of base, For this case, let's assume it's KOH. Now, As we are doing a reaction with base, means that this reaction can only take places under conditions of SN2 and E2, a fast reaction that is taking place in only 1 step.

With this in mind, let's analyze product a. This states that it has two sp hybridized carbon, in other words, a triple bond between two carbons. So the product is with no doubt, an alkyne.

Product b has only one sp hybridized carbon, which means that this carbon should cannot be an alkyne because we need two carbon atoms. The only way to have one atom of C sp hybridized, is with two double bonds, so product b would have to a alkene with two double bonds.

Product c do not have sp hybridized carbon, therefore, it only has two double bonds in two different Carbon atoms, so it's another alkene with two double bonds, but in two different atoms of carbon.

Picture attached show the product a, b and c. Hope this can help