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4 years ago

How many electrons does a co atom have in its 3d subshell?

1 answer:

6 0

Co is the symbol for element cobalt. Atomic number of Co is 27.
1st shell has s orbital and can hold up to 2 electrons.
2nd shell has s and p orbital and can hold upto 8 electrons
3rd shell has s, p and d orbital and can hold upto 18 electrons
In Co electronic configuration upto p orbital in 3rd energy shell is
1s² 2s² 2p⁶ 3s² 3p⁶
After 3p orbital, electrons start filling in the s orbital in 4th energy shell as 4s orbital has lower energy than 3d therefore it fills 4s before 3d, after it fills 2 electrons in 4s, remaining electrons fill 3d orbital

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷
in the 3d subshell there are 7 electrons

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Will give medal! What is the reason the group 13 metals have a typical charge of 3+?

Art [367]

My answer:

13 group of the periodic table represented by boron, aluminum and gallium subgroup. It includes gallium, indium, thallium. Typical steper oxidation in the subset gallium 3 is explained by the presence of (n-1)d^10 E-configuration.
Aluminium oxidation degree has +3 an electronic configuration of noble gases S^2P^6

Hope this helps yah!!!

8 0

3 years ago

Three blocks are shown here pls HELPPPP:

White raven [17]

Answer: I am confident the answer is B

Explanation:

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5 0

3 years ago

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

4 years ago

17) A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with what volume? HI

galina1969 [7]

Answer: A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

Explanation:

Given: $M_{1}$ = 0.20 M, $V_{1}$ = 15.0 mL

$M_{2}$ = 0.10 M, $V_{2}$ = ?

Formula used is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula s follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.20 M ]times 15.0 mL = 0.10 M ]times V_{2}\\V_{2} = 30 mL$

Thus, we can conclude that a 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

4 0

3 years ago

How do you balance AlC3+H2O➡️Al(OH)3+CH4?

Alex17521 [72]

Info: Al(oh)3 might be an improperly capitalized: Al(OH)3
Error: Some elements or groups in the reagents are not present in the products: O
Error: equation Al4C3+H2O=Al(oh)3+CH4 is an impossible reaction
Please correct your reaction or click on one of the suggestions below:
Al4C3 + H2O = Al(OH)3 + CH4

5 0

4 years ago