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Citrus2011 [14]
3 years ago
15

Why do electrons enter the 4s orbital before entering the 3d orbital?

Chemistry
2 answers:
LUCKY_DIMON [66]3 years ago
8 0
C. Because as it get farther from the nucleus the amount of energy is decreasing
salantis [7]3 years ago
7 0

Answer:

C. because the 4s orbital is at a lower energy level

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Rivers, oceans, lakes, ponds
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5 0
3 years ago
If the molecular weight of a semiconductor is 27.9 grams/mole and the diamond lattice constant is 0.503 nm, what is the density
adelina 88 [10]

Explanation:

The given data is as follows.

      Mass = 27.9 g/mol

As we know that according to Avogadro's number there are 6.023 \times 10^{26} atom present in 1 mole. Therefore, weight of 1 atom will be as follows.

            1 atoms weight = \frac{38}{6.023 \times 10^{26}}    

In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.

Therefore, total weight of atoms in a unit cell will be as follows.

            = \frac{8 \times 27.9 g/mol}{6.023 \times 10^{26}}

            = 37.06 \times 10^{-26}

Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.

                   = a^{3}

                   = (0.503 \times 10^{-9})^{3}

                   = 0.127 \times 10^{-27} m^{3}

Formula to calculate density of diamond cell is as follows.

               Density = \frac{mass}{volume}

                             = \frac{37.06 \times 10^{-26}}{0.127 \times 10^{-27} m^{3}}

                            = 2918.1 g/m^{3}

or,                         = 0.0029 g/cc       (as 1 m^{3} = 10^{6} cm^{3})

Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.

4 0
3 years ago
In yeast, ethanol is produced from glucose under anaerobic conditions. A cell‑free yeast extract is placed in a solution that co
12345 [234]

Answer:

650 mmol.

Explanation:

The equation for the fermentation of one mole of glucose is:

C₆H₁₂O₆ + 2 NAD⁺ + 2 ADP + 2 P i + 2 NADH → 2 EtOH + 2 ATP + 2 NADH + 2 NAD⁺

Since NAD⁺/NADH is used and regenerated, we can eliminate it from the equation:

C₆H₁₂O₆ + 2 ADP + 2 P i  → 2 EtOH + 2 ATP

With the equation, we calculate the maximum amount of ethanol that could be obtained theoretically:

1000 mmol C₆H₁₂O₆ ------------ 2000 mmol EtOH

325 mmol C₆H₁₂O₆ ------------- x= 650 mmol EtOH

Therefore, the maximum amount of ethanol that could be produced is 650 mmol.

8 0
3 years ago
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