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Butoxors [25]
3 years ago
12

Name the acids: HBrO

Chemistry
1 answer:
solong [7]3 years ago
6 0

Answer:

Hypobromous acid

Explanation:

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I NEED HELP PLEASE, THANKS! :)
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 <u>two reactions</u>, 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 <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).  

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

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

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

 

Learn more here:

  • brainly.com/question/10547418?referrer=searchResults
  • brainly.com/question/14096111?referrer=searchResults

I hope it helps you!

3 0
2 years ago
The carbon cycle does not really involve matter. True False
stiks02 [169]
False everything involves matter

8 0
2 years ago
Read 2 more answers
Please help check over my work, sorry for spamming
steposvetlana [31]

Answer:

reflection you're right

Explanation:

7 0
3 years ago
Read 2 more answers
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