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frozen [14]
3 years ago
8

The oxygen atom of a ketone (such as cyclohexanone) contains 2 lone pairs of electrons. These pairs of electrons most likely res

ide in what type of orbital
Chemistry
1 answer:
skelet666 [1.2K]3 years ago
5 0

Answer: The given pairs of electrons most likely reside in sp^{2} type of orbital.

Explanation:

As it is given that two lone pair of electrons are present on the oxygen atom of ketone (such as cyclohexanone).

Also, there will be one bond pair between carbon and oxygen atom.

Hence, total electrons present in the domain are as follows.

2 lone pairs + 1 bon pair of electron = 3 electron domains

This means that there will be sp^{2} type of orbital present.

Thus, we can conclude that given pairs of electrons most likely reside in sp^{2} type of orbital.

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If 0.600mol of chloride gas reacted with 0.500mol of aluminium metal to produce aluminium chloride,which reactant is in excess?h
Rus_ich [418]

Answer:

The balanced chemical equation: 2 Al + 3Cl2→ 2 AlCl3

Mole-mole relationship: 2 moles Al + 3 moles Cl2→ 2 moles AlCl3

Given: 0.600 moleCl2; 0.500 mole Al

Required: Excess reactant___; Number of moles of AlCl3 produced__

Solution: Use dimensional analysis using the mole-mole rel

0.600 mole Cl2 * 2 moles Al/3 moles Cl2 = 0.4 mole Al

0.5 mole Al* 3 moles Cl2/2 moles Al = 0.75 mole Cl2

Based on the given:

0.6mole Cl2 + 0.4 mole Al ( this is possible based on the given)

0.5mole Al + 0.75 mole Cl2 (this is not possible because the given is only 0.600 mole of Cl 2)

Answer: Excess reactant is Al; Limiting reactant is Cl2

The amount of AlCl3 produced = 0.6 mole Cl2 + 0.4 mole Al = 1.0 mole AlCl3

4 0
3 years ago
How many grams are in 10.25 moles of zinc chromate?
Kobotan [32]

Answer:

1859.4 g of ZnCrO₄ in 10.25 moles

Explanation:

First of all, we determine the molecular formula of the compound:

Zinc → Zn²⁺  (cation)

Chromate → CrO₄⁻²  (anion)

Zinc chromate → ZnCrO₄

Molar mass for the compound is:

Molar mass of Zn + Molar mass of Cr + (Molar mass of O) . 4 = 181.41 g/mol

65.41 g/mol + 52 g/mol + 16 g/mol . 4 = 181.41 g/mol

Let's apply this conversion factor: 10.25 mol . 181.41 g/mol = 1859.4 g

5 0
3 years ago
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