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andriy [413]
3 years ago
5

1 2 3 4 5 6 7 8 9 10TIME REMAINING52:51While developing his periodic table, Dmitri Mendeleev broke his own rule and placed the e

lement tellurium before iodine in the group. He did this even though tellurium's atomic mass was known to be greater than iodine's. In his notes, Mendeleev stated that he expected tellurium to have a lower atomic mass than iodine. However, all measurements continue to show that tellurium has a greater atomic mass than iodine even though tellurium acts like it has a lower atomic mass
What is the most likely reason that Mendeleev placed tellurium before iodine?
Chemistry
1 answer:
zheka24 [161]3 years ago
6 0
Since, Mendeleev noticed that Tellurium has chemical properties like the other elements included in the same group; he did not know that neutrons can affect atomic mass. Therefore, Mendeleev's Tellurium and Iodine aberration happened because the Tellurium that he got has a prevalence of a heavier Tellurium isotope making it a greater weight than Iodine.
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Answer:

Name

Atomic Number

Atomic Symbol

Atomic Weight

Phase

Color

Classification

Group in Periodic Table

Group Name

Period in Periodic Table

Block in Periodic Table

Electronic Configuration

Melting Point

Boiling Point

Electronic Shell Structure

CAS Number

Neighborhood Elements

Lithium

3

Li

6.941

Solid

Silver

alkali metal

1

lithium family  

2

s

[He] 2s1

180.54 °C

1342 °C

2, 1

CAS7439-93-2

Neighborhood Elements of Lithium

Cesium

55

Cs

132.90545

Solid

Silver

alkali metal

1

lithium family  

6

s

[Xe] 6s1

28.44 °C

671 °C

2, 8, 18, 18, 8, 1

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Neighborhood Elements of Cesium

Explanation:

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4157 J work is done by the gas , If 2 moles of helium undergo a temperature increase of 100 K at constant pressure.

<h3>What are Noble Gases ?</h3>

The noble gases are helium, argon, krypton, xenon, and radon, in order of their mass.

They are called noble gases because they are so majestic that they do not react with anything in general.

The work done can be calculated by the formula

Q=n Cp ΔT

Cp= ( 1+ f/2)R

R= 8.3144598 J. mol-1.

Cp = (1+3/2) * 8.314

Cp = 5*8.314/2

Q= 2 * 5 *8.314 *100/2

Q = 4157 J

Therefore 4157 J work is done by the gas , If 2 moles of helium undergo a temperature increase of 100 K at constant pressure.

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