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3 years ago

A neutral atom in the ground state of Sulfur has its outer most valence electron in which orbital? f d

Chemistry

1 answer:

Natasha_Volkova [10]3 years ago

3 0

<u>Answer:</u> The outermost valence electron enters the p orbital.

<u>Explanation:</u>

Valence electrons are defined as the electrons which are present in outer most orbital of an atom.

Sulfur is the 16th element of the periodic table having 16 electrons.

Electronic configuration of sulfur atom is $1s^22s^22p^63s^23p^4$

The number of valence electrons are 2 + 4 = 6

These 6 electrons enter s-orbital and p-orbital but the outermost valence electron will enter the p-orbital.

Hence, the outermost valence electron enters p orbital.

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When 28 g of nitrogen and 6 g of hydrogen react, 34 g of ammonia are produced. if 100 g of nitrogen react with 6 g of hydrogen,

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Sergeeva-Olga [200]

Explanation:

The given data is as follows.

Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml

Therefore, molarity of toulene is calculated as follows.

Molarity = $\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}$

= 4.11 M

Hence, molarity of toulene is 4.11 M.

As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

Molality = $\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}$

= 8.6 m

Hence, molality of given toulene solution is 8.6 m.

Now, calculate the number of moles of toulene as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{75.8 g}{92.13 g/mol}$

= 0.8227 mol

Now, no. of moles of benzene will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{95.6 g}{78.11 g/mol}$

= 1.2239 mol

Hence, the mole fraction of toulene is as follows.

Mole fraction = $\frac{\text{moles of toulene}}{\text{total moles}}$

= $\frac{0.8227 mol}{(0.8227 + 1.2239) mol}$

= 0.402

Hence, mole fraction of toulene is 0.402.

As density of given solution is 0.857 $g/cm^{3}$ so, we will calculate the mass of solution as follows.

Density = $\frac{mass}{volume}$

0.857 $g/cm^{3}$ = $\frac{mass}{200 ml}$ (As 1 $cm^{3}$ = 1 g)

mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

Mass % = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

= $\frac{75.8 g}{171.4 g} \times 100$

= 44.22%

Therefore, mass percent of toulene is 44.22%.

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