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3 years ago

Based on molecular orbital theory, there is/are ________ unpaired electron(s) in the of+ ion.

Chemistry

1 answer:

Ede4ka [16]3 years ago

5 0

Atomic number of O= 8

Atomic number of F = 9

Electron configurations:

⁸O = 1s² 2s² 2p⁴

⁹F = 1s² 2s² 2p⁵

Total electrons: 8+ 9 = 17

For OF the molecular electron configuration would be:

OF = (1σ)²(1σ*)²(2σ)²(2σ*)²(2pσ)²(2pπ)⁴(2pπ*)³

OF⁺ suggests a loss of one electron from the 2pπ* orbital leaving the configuration as:

OF⁺ = (1σ)²(1σ*)²(2σ)²(2σ*)²(2pσ)²(2pπ)⁴(2pπ*)²

Thus, based on MO theory there are two unpaired electrons in OF⁺ ion

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When the pressure that a gas exerts

Ronch [10]

Answer:

The pressure changes from 2.13 atm to 1.80 atm.

Explanation:

Given data:

Initial pressure = ?

Final pressure = 1.80 atm

Initial temperature = 86.0°C (86.0 + 273 = 359 K)

Final temperature = 30.0°C (30+273 =303 K)

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

P₁ = P₂T₁ /T₂

P₁ = 1.80 atm × 359 K / 303 K

P₁ = 646.2 atm. K /303 K

P₁ = 2.13 atm

The pressure changes from 2.13 atm to 1.80 atm.

5 0

3 years ago

Read 2 more answers

Compared to the nonmetals in Period 2, the metals in Period 2 generally have larger

konstantin123 [22]

The answer is atomic radii; the size or radii of an atom increases from left to right, versus the ionization energies and electronegativities of atoms which increase from right to left.

6 0

3 years ago

Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w

Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619= [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0

3 years ago

2. To what volume will you have to dilute 20.0 ml of a 8 M HCl solution to make a 0.35 M

GrogVix [38]

Answer:

You need to dilute 8 M HCl to 0.35 M to make 457 mL.

Explanation:

Use the equation M₁V₁=M₂V₂

Convert the mL to L : 20 mL= 0.02 L

8 M x 0.02 L = 0.35 M x V₂

V₂ = 0.4571429 L = 457 mL

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3 years ago

(ii) Electrons are supplied to the external circuit. How and where is this done?<br> [2]

pentagon [3]

Answer:

The electrons are supplied by the species getting oxidized. They move from anode to the cathode in the external circuit. The external battery supplies the electrons. They enter through the cathode and come out through the anode

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3 years ago