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saul85 [17]
2 years ago
9

I. A set of sp2 orbitals can be thought of as one s orbital one-third of the time and two p orbitals two-thirds of the time. II.

A set of sp orbitals can accommodate a maximum of six electrons. III. The orbitals resulting from sp3d2 hybridization are directed toward the corners of an octahedron. IV. A set of sp3 orbitals results from the mixing of one s orbital and three p orbitals.
Chemistry
1 answer:
dybincka [34]2 years ago
5 0

Answer:

I. FALSE

II. FALSE

III. TRUE

IV. TRUE

Explanation:

<em>I. A set of sp² orbitals can be thought of as one s orbital one-third of the time and two p orbitals two-thirds of the time.</em> FALSE. 1 s orbital hybridize with 2 p orbitals to form 3 sp² permanent orbitals.

<em>II. A set of sp orbitals can accommodate a maximum of six electrons. </em>FALSE. sp orbitals, as any orbital, can accommodate a maximum of 2 electrons.

<em>III. The orbitals resulting from sp³d² hybridization are directed toward the corners of an octahedron.</em> TRUE. 1 s orbital, 3 p orbitals and 2 d orbitals hybridize to form 6 sp³d² orbitals, which are directed toward the corners of an octahedron.

<em>IV. A set of sp³ orbitals results from the mixing of one s orbital and three p orbitals.</em> TRUE. 1 s orbital hybridize with 3 p orbitals to form 4 sp³ orbitals.

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certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 40.00 gram sample of the alcohol pro
seraphim [82]

Answer:

The answer to your question is:    C₂H₆O₁   = C₂H₆O

Explanation:

Data

CxHyOz

mass = 40 g  produced 76.40 g of CO2

                                       46.96 g of H2O

Empirical formula = ?

Process

                         CxHyOz    + O2    ⇒    CO2  + H2O

                         44g of CO2 --------------------  12 g of Carbon

                         76.40 g of CO2 --------------- x

                          x = 20.84 g of Carbon

                         12 g of Carbon ---------------  1 mol

                         20.84 g of C    ---------------   x

                         x = (20.84 x 1) / 12

                         x = 1.74 mol of Carbon

                        18 g of H2O --------------------  2 g of H

                        46.96 g of H2O --------------   x

                        x = (46.96 x 2) / 18

                        x = 5.22 g of H

                        1 g of H ------------------------  1 mol of H

                        5.22 g of H -------------------   x

                        x = 5.22 mol of H

Mass of Oxygen = 40 - 20.84 - 5.22g

                           = 13.94 g

                         16 g of Oxygen ----------------  1 mol

                         13.94 g of O --------------------   x

                         x = 0.87 mol of O

Divide by the lowest number of moles

Carbon = 1.74 / 0.87 = 2

Hydrogen = 5.22 / 0.87 = 6

Oxygen = 0.87 / 0.87 = 1

                           

Empirical formula

                                 C₂H₆O₁   = C₂H₆O

                         

3 0
1 year ago
(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW
Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M  InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M  InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M  In⁺³

[F⁻] = 2.3x10⁻³M  InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M  In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

<em></em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

3 0
2 years ago
How do acids and bases affect molecules such Proteins? ​
Taya2010 [7]

Answer:

Strong acids and bases both denature proteins by severing disulphide bonds and at higher temperatures, can break proteins into peptides, or even individual amino acids.

5 0
1 year ago
You have a racemic mixture of d-2-butanol and l-2-butanol. the d isomer rotates polarized light by +13.5∘. what is the rotation
Bumek [7]
L- isomer is considered as the Enantiomer of d- isomer and since the d-isomer optical rotation is + 13.5° so the optical rotation of l-isomer will be the same degree but with opposite sign which equal to -13.5° 

So the degree of rotation of racemic mixture will equal 0° 


- This is because racemic mixture contains equal amount of both enantiomers
8 0
2 years ago
An aluminum ion has a charge of +3, and an oxide has a charge of -2. What would be the product of a reaction between these two e
kozerog [31]

The product of a reaction between these two elements is Al_{2} O_{3}.

Explanation:

The oxidation state of an ion in a compound is equal to its charge.

The aluminum having a charge of +3 because oxidation state is +3

The oxide is having charge of -2

The product of these reactants will produce a chemical compound.

The compound formed is Al_{2} O_{3}  i.e Aluminium oxide. The compound while getting formed will share the charge and cation A+ will have the charge of anion and anion will have the charge of cation. This will result in a compound as there should be a neutral charge on the compound formed.

The <em>+</em><em>3 charge of the cation Al+ will go to anion oxide O2- and the charge of anion -2 will go with cation Al+. </em>

<em />

8 0
2 years ago
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