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Natali5045456 [20]
3 years ago
7

For the oxidation–reduction reaction equation 2 Rb + Br 2 ⟶ 2 RbBr 2Rb+Br2⟶2RbBr indicate how many electrons are transferred in

the formation of one formula unit of product.
Chemistry
1 answer:
Vladimir79 [104]3 years ago
8 0

Answer:

2e⁻ are transferred in the formation of one formula unit of product

Explanation:

The reaction is:

2Rb + Br₂ → 2RbBr

In ground state, both elements have 0 as oxidation state.

In the RbBr, Rb acts with+1 and Br with -1.

Rb increased the oxidation state → Oxidation

Br decreased the oxidation state → Reduction

Rb → Rb⁺ + 1e⁻

Br₂ + 2e⁻ → 2Br⁻

Bromine had to gained 2 electrons, to decreased the oxidation state, so

(Rb → Rb⁺ + 1e⁻ ) ₓ2 = 2Rb → 2Rb⁺ + 2e⁻

(Br₂ + 2e⁻ → 2Br⁻) ₓ1 =  Br₂ + 2e⁻ → 2Br⁻

2Rb + Br₂ + 2e⁻ → 2Br⁻ + 2Rb⁺ + 2e⁻

If we cancel the electrons, we reach the balanced reaction. In conclussion, 2 e⁻

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I believe "h" stands for hecta-, which means "one hundred."

I will convert from g to hg (divide by 100), then from mL to L (multiply by 1000).

\frac{4.2g}{mL}*\frac{1hg}{100g}*\frac{1000mL}{1L}=42\frac{hg}{L}

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alexandr402 [8]

First, let's state the chemical reaction:

2Al+3Cl_2\to2AlCl_3\text{.}

We can find the number of moles of Cl2 required to produce 0.0923 moles of AlCl3, doing a rule of three: 3 moles of Cl2 reacted produces 2 moles of AlCl3:

\begin{gathered} 3molesCl_2\to2molesAlCl_3 \\ \text{?moles Cl}_2\to0.0923\text{ moles }AlCl_3\text{.} \end{gathered}

The calculation would be:

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The nuclear reaction here resembles a beta-plus decay. The mass numbers of the two nuclei are equal. However, the atomic number of the product nucleus is lower than that of the reactant nucleus by 1.

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