Name the acids: HBrO

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$

6 0

3 years ago

G. Amount of charge required to reduce

Nady [450]

Answer:

$\boxed{\text{c) 4 F}}$

Explanation:

1. Write the skeleton equation for the half-reaction

NO₃⁻ ⟶ N₂O

2. Balance all atoms other than H and O

2NO₃⁻ ⟶ N₂O

3. Balance O by adding H₂O molecules to the deficient side.

2NO₃⁻ ⟶ N₂O + 5H₂O

4. Balance H by adding H⁺ ions to the deficient side.

2NO₃⁻ + 10H⁺ ⟶ N₂O + 5H₂O

5. Balance charge by adding electrons to the deficient side.

2NO₃⁻ + 10H⁺ + 8e⁻ ⟶ N₂O + 5H₂O

The amount of charge required to reduce 2 mol of NO₃⁻ is 8 F

$\text{The amount of charge required to reduce 1 mol of NO$_{3}^{-}$ is \boxed{\textbf{4 F}}}$

4 0

2 years ago

Read 2 more answers

Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)

Brilliant_brown [7]

The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).

The general reaction is:

2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)

We can write the above reaction in two reactions, one for oxidation and the other for reduction:

Oxidation reaction

Li⁰(s) → Li⁺(aq) + e⁻ (2)

Reduction reaction

Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)

We can see that Li⁰ is oxidizing to Li⁺ (by losing one electron) in the lithium acetate (reaction 2) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by gaining two electrons) (reaction 3).

We must remember that the reducing agent is the one that will be oxidized by reducing another element and that the oxidizing agent is the one that will be reduced by oxidizing another species.

In reaction (1), the reducing agent is Li (it is oxidizing to Li⁺), and the oxidizing agent is Fe(CH₃COO)₂ (it is reducing to Fe⁰).

Therefore, the reducing agent in reaction (1) is lithium (Li).

Learn more here: