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3 years ago

Consider the elements: na , mg , al , si , p . part a which element has the highest second ionization energy?

Chemistry

2 answers:

oksian1 [2.3K]3 years ago

6 0

Explanation:

Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.

With increase in atomic size of the atom, there will be less force of attraction between the nucleus and the valence electrons of the atom. Hence, with lesser amount of energy the valence electrons can be removed.

Na, Mg, Al, Si, and P are all period 3 elements. Hence, there will be decrease in size from sodium to phosphorous. So, the amount of enenrgy required to remove an electron from phosphorous atom will be high as it is smaller is size.

And, when one electron is removed from it then it further decreases in size. So, when second electron is removed then we need to provide much more heat energy to overcome the strong force of attraction between the nucleus and valence electrons.

Thus, we can conclude that out of the given options P element has the highest second ionization energy.

Monica [59]3 years ago

4 0

I'm not sure but i would say Sodium because it has 496Kj/Mol after Magnesium
but still not quite sure if it's the second highest ionization charge or not, try searching it up :)

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3 years ago

During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.

Travka [436]

<u>Answer:</u> The mass of HCl present in 500 mL of acid solution is 36.5 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL$

Putting values in above equation, we get:

$1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

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

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

$2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g$

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

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