A neutral atom of which of the four element has the smallest radius

Chemistry

2 answers:

arsen [322]3 years ago

6 0

Answer:
Element 1

Explanation:
The ionization energy is defined as the energy required to remove electrons from the atoms.
We know that the nucleus of the atom attracts the electrons, thus, bound these electrons to the atom.
This means that as the radius decreases, the force of attraction between the nucleus and the electron will increase, therefore, the energy required to remove the electron would increase (and vice-versa).

Based on the above, the atom with the smallest radius would be the atom with the largest first ionization energy.

Hope this help :)

Dennis_Churaev [7]3 years ago

4 0

I would probably say that element 1 has the lowest atomic radius since it has the highest 1st ionization energy.

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The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

Answer: The concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

Explanation:

To calculate the molarity of solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L$