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AleksandrR [38]
2 years ago
6

Which element has a greater electronegativity? fluorine (9) or radium (88)

Chemistry
2 answers:
emmainna [20.7K]2 years ago
7 0

Answer:

Fluorine

General Formulas and Concepts:

<u>Chemistry</u>

  • Reading a Periodic Table
  • Periodic Trends
  • Electronegativity - the tendency for an element to attract an electron to itself
  • Z-effective and Coulomb's Law, Forces of Attraction

Explanation:

The Periodic Trend for Electronegativity is up and to the right of the Periodic Table.

Fluorine is Element 9 and has 9 protons. Radium is Element 88 and has 88 protons. Therefore, Radium has a bigger Zeff than Flourine.

However, since Radium is in Period 7 while Fluorine is in Period 2, Radium has more core e⁻ than Fluorine does. This will create a much larger shielding effect, causing Radium's outermost e⁻ to have less FOA between them. Fluorine, since it has less core e⁻, the FOA between the nucleus and outershell e⁻ will be much stronger.

Therefore, Fluorine would attract an electron more than Radium, thus bringing us to the conclusion that Fluorine has a higher electronegativity.

Lapatulllka [165]2 years ago
6 0

Answer:

Fluorine

Explanation:

Electronegativity increases as you go from left to right across the periodic table and decreases as you go from top to bottom of the periodic table.

Fluorine is in period 3, group 17

Radium is in period 7, group 2

Radium is in period 7 and we know that electronegativity decreases as you move from top to bottom.

Explanation: As you move from top to bottom, you are in higher energy level, which means that your distance from the nucleus is further away.

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Calculate the volume of one mole of a gas at 1.00 atm pressure and 0 °C.
N76 [4]

Answer:

Solution:-

The gas is in the standard temperature and pressure condition i.e. at S.T.P

Therefore,

V

i

​

=22.4dm

3

V

f

​

=?

As given that the expansion is isothermal and reversible

∴ΔU=0

Now from first law of thermodynamics,

ΔU=q+w

∵ΔU=0

∴q=–w

Given that the heat is absorbed.

∴q=1000cal

⇒w=−q=−1000cal

Now,

Work done in a reversible isothermal expansion is given by-

w=−nRTln(

V

i

​

V

f

​

​

)

Given:-

T=0℃=273K

n=1 mol

∴1000=−nRTln(

V

i

​

V

f

​

​

)

⇒1000=−1×2.303×2×273×log(

22.4

V

f

​

​

)

Explanation:

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