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3 years ago

An atom holds 7 electrons. use orbital notation to model the probable location of its electrons.

Chemistry

1 answer:

zheka24 [161]3 years ago

7 0

Answers:

(a) 1s² 2s²2p³; (b) 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²; (c) 1s² 2s²2p⁶ 3s²3p⁵

Step-by-step explanation:

One way to solve this problem is to add electrons to the orbitals one-by-one until you have added the required amount.

Fill the subshells in the order listed in the diagram below. Remember that an s subshell can hold two electrons, while a p subshell can hold six, and a d subshell can hold ten.

(a) Seven electrons

1s² 2s²2p³

There are two electrons in the 2s subshell and three in the 2p subshell. The remaining two electrons are in the inner 1s subshell.

(b) 22 electrons

1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²

There are two electrons in the 4s subshell and two in the 2p subshell. The remaining 18 electrons are in the inner subshells.

(c) 17 electrons

1s² 2s²2p⁶ 3s²3p⁵

There are two electrons in the 3s subshell and five in the 2p subshell. The remaining 10 electrons are in the inner subshells.

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What orbits from a pi bond

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3 years ago

Determine whether or not each ion contributes to water hardness.

Virty [35]

Answer: The ion that contribute to water hardness are:

--> a. Ca2+

--> b. (HCO)3^- and

--> c. Mg2+

While K+ DOES NOT contribute to water hardness.

Explanation:

WATER in chemistry is known as a universal solvent. This is so because it is polar in nature and dissolves most inorganic solutes and some polar organic solutes to form aqueous solutions. It is composed of elements such as hydrogen and oxygen in the combined ratio of 2:1.

Water is said to be HARD if it does not lather readily with soap. There are two types of water hardness:

--> Permanent hardness: This is mainly due to the presence of CALCIUM and MAGNESIUM ions in the form of soluble tetraoxosulphate(VI) and chlorides. These ions are removed by adding washing soda or caustic soda.

--> Temporary hardness: This is due to the presence of calcium HYDROGENTRIOXOCARBONATES. It can be removed by boiling and using slaked lime.

Therefore from the above given ions, Ca2+,(HCO)3^- and Mg2+ contributes to water hardness.

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2 years ago

Quick help I really need help

Sauron [17]

Answer:

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3 years ago

Please help me !!!!!!!!! i'll give brainlist !

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Answer:

A-B

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3 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

3 years ago