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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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Kay [80]
At the first reaction when 2HBr(g) ⇄ H2(g) + Br2(g)
So Kc = [H2] [Br2] / [HBr]^2
7.04X10^-2 = [H2][Br] / [HBr]^2

at the second reaction when 1/2 H2(g) + 1/2 Br2 (g) ⇄ HBr
Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2
we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
and by taking the square root: 
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∴ Kc for the second reaction = √(1/7.04X10^-2) = 3.769 
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Natasha2012 [34]
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Chromium (III) forms a complex with diphenylcarbazide whose molar absorptivity is 4.17x104 at 540 nm.
son4ous [18]

Answer:

3.00 cm

Explanation:

The absorbance can be expressed using <em>Beer-Lambert's law</em>:

A = ε*b*c

Where ε is a constant for each compound, b is the optical path, and c is the molar concentration of the compound.

Now we <u>match the absorbance values for both solutions</u>, because we want the absorbance value to be the same for both solutions:

A = ε * 1.00 cm * 7.68x10⁻⁶M = ε * b * 2.56x10⁻⁶ M

And <u>solve for b:</u>

 ε * 1.00 cm * 7.68x10⁻⁶M = ε * b * 2.56x10⁻⁶ M

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5 0
2 years ago
How many grams of NH3 are needed to provide the same number of molecules as in 0.35 g of SF6 ?
Strike441 [17]

Answer:

<u />

  • <u>There are 0.041 g of NH₃ in the same number of molecules as in 0.35 g of SF₆.</u>

<u />

Explanation:

Using the molar mass of the chemical formula SF₆ you can find the number of moles of molecules in 0.35 g of such substance. Then, using the molar mass of NH₃, you can find mass in grams corresponding to the same number of molecules.

<u>1. Find the molar mass of SF₆:</u>

Atom   atomic mass          number of atoms   total mass in 1 mole

S           32.065 g/mol                     1                                       32.065 g

F            18.998 g/mol                     6                 6 × 18.998 = 113.988 g

                                                                               =====================

                                                          molar mass of SF₆ =    146.053 g/mol

<u>2. Find the number of moles in 0.35 g of SF₆:</u>

  • number of moles = mass in grams / molar mass
  • number of moles = 0.35 g / 146.053 g / mol = 0.0024 mol

<u>3. Find the molar mass of NH₃:</u>

Atom   atomic mass          number of atoms   total mass 1 mole                

N           14.007 g/mol                     1                                       14.007 g

H             1.008 g/mol                    3               3 × 1.008 g = 113.988 g

                                                                               =====================

                                                         molar mass of NH₃ =    17.031 g/mol

<u />

<u>4. Find the mass in 0.0024 mol of NH₃:</u>

  • mass in grams = number of moles × molar mass

  • mass = 0.0024 mol × 17.031 g/mol ≈ 0.041 grams

<u>5. Conclusion: </u>

There are 0.041 g of NH₃ in the same number of molecules as in 0.35 g of SF₆.

6 0
2 years ago
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