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3 years ago

Please explain to me how to answer no.2 questions

Chemistry

1 answer:

bulgar [2K]3 years ago

4 0

Answer: -

IE 1 for X = 801

Here X is told to be in the third period.

So n = 3 for X.

For 1st ionization energy the expression is

IE1 = 13.6 x Z ^2 / n^2

Where Z =atomic number.

Thus Z =( n^2 x IE 1 / 13.6)^(1/2)

Z = ( 3^2 x 801 / 13.6 )^ (1/2)

= 23

Number of electrons = Z = 23

Nearest noble gas = Argon

Argon atomic number = 18

Number of extra electrons = 23 – 18 = 5

a) Electronic Configuration= [Ar] 3d34s2

We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.

So more the first ionization energy, less is the atomic radius.

X has more IE1 than Y.

b) So the atomic radius of X is lesser than that of Y.

c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.

Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.

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Read 2 more answers

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.