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finlep [7]
3 years ago
14

What orbitals are being filled in the transition metals

Chemistry
2 answers:
Kruka [31]3 years ago
8 0

There are four principle orbitals (s, p, d, and f) which are filled according to the energy level and valence electrons of the element. All four orbitals can hold different number of electrons. The s-orbital can hold 2 electrons, and the other three orbitals can hold up to 6, 10, and 14 electrons, respectively.

Artyom0805 [142]3 years ago
6 0
The d orbitals are being filled
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2

Explanation:

55.85 grams per mole.

times 2

111.7

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6 0
3 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
4 years ago
n a semiconductor, the bonding molecular orbitals that contain electrons are referred to as the ________, while the antibonding
Kruka [31]

Answer:

In a semiconductor, the bonding molecular orbitals that contain electrons are referred to as the valence band, while the antibonding orbitals that are completely empty are referred to as the conduction band.

The conduction band occupies a higher energy level than the valence band. The band gap is what separates the two orbitals.

3 0
2 years ago
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