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Arada [10]
2 years ago
10

PLEASE HELPPPPPP! I'll appreciate!

Chemistry
1 answer:
blagie [28]2 years ago
8 0

Answer:

A,Z,H,Y,T

Hope this helpsss!!!!! Best!!!!

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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
09. Where does Mitosis take place? I
mash [69]

Answer:

In plants mitosis takes place throughout life in growing regions called the meristems. Replacements as cells wear out. The cells of the skin and bone marrow are sites of active mitosis replacing skin cells and red blood cells that only have a limited life. Repair.

Explanation:

5 0
2 years ago
Whether or not the process is observed in nature, which of the following could account for the transformation of gallium-67 to z
Sonja [21]

Answer:

Option a: positron emission.

Explanation:

In the transformation we have:

⁶⁷Ga  →  ⁶⁷Zn  

The reaction is:

^{67}_{Z}X \rightarrow ^{67}_{Z -1}Y

For Ga to become Zn, the atom nucleus has to lose a proton, so in the given options, the reaction that involves the transformation of a proton is the option a, positron emission.

In a positron emission, a proton becomes into a neutron and a positron:

^{A}_{Z}X \rightarrow ^{A}_{Z-1}Y + ^{0}_{+1}e

Therefore, the correct answer is option a: positron emission.

I hope it helps you!

7 0
2 years ago
A candle is placed between a concave mirror and its focal point.
natali 33 [55]

Virtual and enlarged..............

What grade are you in btw ?

6 0
2 years ago
Read 2 more answers
A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
2 years ago
Read 2 more answers
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