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2 years ago

13

Among the following, which element has the lowest ionization energy? Question 2 options: Cs Na Cl I

2 answers:

marissa [1.9K]2 years ago

7 0

The element with the lowest ionization energy is CESIUM, CS.
Ionization energy is the energy required to remove the most loosely bound electron in an atom of an element. The higher the number of shells in an atom of an element, the lower the ionization energy that will be required to remove the valence electron.

motikmotik2 years ago

5 0

CS Caesium atomic number 55 i took the test

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To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of oxygen gas

$C_2\text{ and }p_2$ are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used: 1 atm = 760 torr

$C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm$

Putting values in above equation, we get:

$\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M$

Hence, the solubility of oxygen gas at 628 torr is $4.58\times 10^{-3}M$

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