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inessss [21]
3 years ago
5

PLEASE HELP. Give the orbital configuration of the phosphorus (p) atom.

Chemistry
2 answers:
Sonja [21]3 years ago
8 0

Answer:

Explanation:

The Electronic Configuration of the elements is the arrangement of all electrons of an element in energy levels and sub-levels (orbitals).  

There are 7 energy levels, numbered from 1 to 7, and in which electrons are distributed, logically in order according to their energy level. Each level is divided into sub-levels. These sub-levels into which each level is divided can be up to 4, called: s, p, d, f. In the sub-level s there can only be a maximum of 2 electrons, in p there can be a maximum of 6 electrons, in the sub-level d 10 electrons and finally in the sub-level f there can be a maximum of 14 electrons.

In level 1 there is only one sub-level, which will be the s. In level 2 there are 2 sub-levels, the s and the p. At level 3 there are 3 sub-levels s, p and d. And at level 4 there are 4 sub-levels, the s, the p, the d and the f.

Aufbau's principle is a principle of atomic physics, which explains how electrons are accommodated in their orbits around the nucleus of the atom. This principle states that the filling of these orbitals occurs in increasing order of energy, that is, from the orbitals of lower energy to those of higher energy.

The simplest of the sub-levels is 1s. If you fill the 1s you can start filling the 2s. After the 2s sub-level is satisfied, the 2p can begin to fill and so on.  

The atomic number Z of the element represents the number of protons of the element, and in a neutral atom it also indicates the number of electrons.  

So, given that the atomic number of the phosphorus is 15, and everything that has been said before, it is possible to say that the electrons that it has in the orbitals will be placed as follows:

<u><em>1s² 2s² 2p⁶ 3s² 3p³</em></u>

Finally, the orbital configuration of the phosphorus atom (P) will be as the image shows.

Kipish [7]3 years ago
4 0

Answer: 1s^22s^22p^63s^23p^3

Explanation:

Assuming that orbital configuration is the same as electron configuration this is the answer.

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A student is asked to determine the molarity of a strong base by titrating it with 0.250 M solution of H2SO4. The students is in
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Answer:

The molarity of the strong base is 0.625 M

Which procedural error will result in a strong base molarity that is too high?

⇒ Using a buret with a tip filled with air rather than the H2SO4 solution

Explanation:

<u>Step 1:</u> Data given

Molarity of H2SO4 = 0.250 M

The initial buret reading is 5.00 mL

The final reading is 30.00 mL

<u />

<u>Step 2:</u> Calculate volume of H2SO4 used

30.00 mL - 5.00 mL = 25.00 mL

<u>Step 3:</u> Calculate moles of H2SO4

0.250 M = 0.250 mol/L

Since there are 2 H+ ions per H2SO4

0.250 mol/L  * 2 = 0.500 mol/L

The number of moles H2SO4 = 0.500 mol/L * 0.025 L

Number of moles H2SO4 = 0.0125 mol

<u>Step 4</u>: Calculate moles of OH-

For 1 mol H2SO4, we need 1 mol of OH-

For 0.0125 mol of H2SO4, we have 0.0125 mol of OH-

<u>Step 5</u>: Calculate the molarity of the strong base

Molarity = moles / volume

Molarity OH- = 0.0125 mol / 0.02 L

Molarity OH - = 0.625 M

Which procedural error will result in a strong base molarity that is too high?

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Answer:

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What is the electric force on a proton 2.5 fmfm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point ch
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Explanation:

It is known that charge on xenon nucleus is q_{1} equal to +54e. And, charge on the proton is q_{2} equal to +e. So, radius of the nucleus is as follows.

            r = \frac{6.0}{2}

              = 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

              d = r + 2.5

                 = (3.0 + 2.5) fm

                 = 5.5 fm

                 = 5.5 \times 10^{-15} m     (as 1 fm = 10^{-15})

Therefore, electrostatic repulsive force on proton is calculated as follows.

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Putting the given values into the above formula as follows.

           F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

              = (9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}

              = (9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}

              = 411.2 N

or,           = 4.1 \times 10^{2} N

Thus, we ca conclude that 4.1 \times 10^{2} N is the electric force on a proton 2.5 fm from the surface of the nucleus.

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