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vodka [1.7K]
2 years ago
6

The image shows a representative sample of 50 atoms of a fictious element, called lemonium ( Le ). Lemonium consists of three is

otopes. The red spheres are Le – 19 , the light blue spheres are Le – 17 , and the dark blue spheres are Le – 15 . Determine the percent natural abundance of each of the three isotopes.
Chemistry
1 answer:
Keith_Richards [23]2 years ago
3 0

The percentage abundance is a measure of ratio of the number of each isotope and the total number of the element. The percentage abundance of each isotope is :

  • Le - 19 = 0.32
  • Le - 17 = 0.44
  • Le - 15 = 0.24

Number of each isotope in the sample :

  • Le - 19 = 16
  • Le - 17 = 22
  • Le - 15 = 12

Total number of sample = 50

The percentage natural abundance of the isotopes can be calculated :

Number of each isotope ÷ total number of sample

Percentage abundance of Le isotopes

  • Le - 19 = \frac{16}{50}=0.32
  • Le - 17 = \frac{22}{50}=0.44
  • Le - 15 = \frac{12}{50}=0.24

Therefore, the percentage abundance of Le - 19, Le - 17 and Le - 15 isotopes are 0.32, 0.44 and 0.24 respectively.

Learn more : brainly.com/question/16078327

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3 years ago
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As water being amphoteric in nature can react with HCl as follow,

                           HCl  +  H₂O   ⇆  H₃O⁺  +  OH⁻

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3 0
2 years ago
Read 2 more answers
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
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hodyreva [135]

Answer:

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Explanation:

From Graham's law;

Let the rate of diffusion of gas X be 1.25

Let the rate of diffusion of CO2 be 1

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1.25/1=√44/M

(1.25/1)^2 = 44/M

1.5625 = 44/M

M= 44/1.5625

M= 28.16 g/mol

4 0
3 years ago
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