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3 years ago

The bond between metal ions and the sea of electrons

1 answer:

6 0

Answer: Metallic Bonding. ... In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.

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(a) Given that Ka for acetic acid is 1.8 X 10^-5 and that for hypochlorous acid is 3.0 X 10^-8, which is the stronger acid? (b)

Gala2k [10]

Answer:

HOAc is stronger acid than HClO

ClO⁻ is stronger conjugate base than OAc⁻

Kb(OAc⁻) = 5.5 x 10⁻¹⁰

Kb(ClO⁻) = 3.3 x 10⁻⁷

Explanation:

Assume 0.10M HOAc => H⁺ + OAc⁻ with Ka = 1.8 x 10⁻⁵

=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺

Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸

=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺

HOAc delivers more H⁺ than HClO and is more acidic.

Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴

Kb(OAc⁻) = 5.5 x 10⁻¹⁰

Kb(ClO⁻) = 3.3 x 10⁻⁷

4 0

3 years ago

How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut

san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = 8 = Concentration in Ethyl acetate / Concentration in water

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

Where X is the amount of material in grams that comes to the organic phase.

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

In the second extraction, you will extract 0.50g of material

Thus, after the two extraction you will lost:

4.44g + 0.50g = 4.94g of material

6 0

3 years ago

The following reaction has an activation energy of 262 kJ/mol.

Hitman42 [59]

Answer: i really dont know srry

Explanation:

4 0

3 years ago

Please help Explain why a molecule that has polar bonds can be a polar molecule.

Zepler [3.9K]

Answer:

Explanation:

Well, obviously a molecule with polar bonds can be polar in itself. It's like saying I am an atheltic person who can just reach the basketball rim with my head and also I can dunk.

But if the question is how can a molecule that in non-polar have polar bonds, well, its because the polar bonds' dipole cancels each other out. It's like a tight rope. If a person pulls in one direction, it intuitively, the rope would go in that direction. However, if a person pulls in the other direction with the same amount of force, the rope stays still. This is the same case. Although molecules can have different electronegativities, the pull of electrons in one direction is cancelled out by a pull in the opposite direction, making the net dipole 0.

This is common for main VSERP shaped molecules like linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral.

8 0

3 years ago

A 1.67-g sample of solid silver reacted in excess chlorine gas to give a2.21-g sample of pure solid Agcl.The heat given off in t

kotegsom [21]

Given:

Mass of Ag = 1.67 g

Mass of Cl = 2.21 g

Heat evolved = 1.96 kJ

To determine:

The enthalpy of formation of AgCl(s)

Explanation:

The reaction is:

2Ag(s) + Cl2(g) → 2AgCl(s)

Calculate the moles of Ag and Cl from the given masses

Atomic mass of Ag = 108 g/mol

# moles of Ag = 1.67/108 = 0.0155 moles

Atomic mass of Cl = 35 g/mol

# moles of Cl = 2.21/35 = 0.0631 moles

Since moles of Ag << moles of Cl, silver is the limiting reagent.

Based on reaction stoichiometry: # moles of AgCl formed = 0.0155 moles

Enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol

Ans: Formation enthalpy = 126.5 kJ/mol

6 0

3 years ago

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