The ionization energy of
is
.
<h3>What is ionization energy?</h3>
It is the energy needed to remove one electron from a neutral atom, which results in the formation of an ion.
The measurement is based on an isolated atom in its gaseous phase and is often expressed in kJ/mol.
b) Atom is ionized when the electron is completely removed from its electron cloud (so it being moved from first,
to the infinity's shell and
), and now the equation can be written as

So the ionization energy is affected by the charge of the nucleus and the general formula can be represented as:
and as 

The
has an atomic number Z=5, therefore using the formula when
=1, we get

The ionization energy of
is
.
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Answer:
This question is incomplete
Explanation:
This question is incomplete because of the absence of options. However, the compound C₆H₁₄ is hexane. Hexane is a member of saturated hydrocarbons (homologous series) called alkanes (with the general formula CₙH₂ₙ₊₂). The structure for an hexane is shown below
H H H H H H
I I I I I I
H - C - C - C - C - C - C - H
I I I I I I
H H H H H H
which can also be written as
CH₃CH₂CH₂CH₂CH₂CH₃
IT will be Refraction of light
0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions.
The balanced neutralization equation is:
NaOH(aq) + KHC₈H₄O₄(aq) ⇒ KNaC₈H₄O₄(aq) + H₂O(l)
- Step 1: Calculate the reacting moles of KHP.
0.212 g of KHP react. The molar mass of KHP is 204.22 g/mol.
0.212 g × 1 mol/204.22 g = 1.04 × 10⁻³ mol
- Step 2: Determine the reacting moles of NaOH.
The molar ratio of NaOH to KHP is 1:1.
1.04 × 10⁻³ mol KHP × 1 mol NaOH/1 mol KHP = 1.04 × 10⁻³ mol NaOH
- Step 3: Calculate the molarity of NaOH.
1.04 × 10⁻³ moles of NaOH are in 35.00 mL of solution.
[NaOH] = 1.04 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.0297 M
0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
Learn more about titration here: brainly.com/question/4225093