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Harlamova29_29 [7]
2 years ago
12

usando el concepto de valencia para los elementos S, P Y Br predice las formulas de los compuestos mas simples formados por esto

s elementos al combinars, cada uno, con el hidrogeno.
Chemistry
1 answer:
hram777 [196]2 years ago
6 0

Explanation:

Teniendo en cuenta los numeros de oxidacion negativos de cada uno

P:fosforo(-3)

S:Azufre (-2)

Br:Bromo (-1)

Y el H: hidrogeno con una valencia

positiva de +1

los compuesto que se formaran son los siguientes

1 H3P= fosfuro de hidrogeno

2 H2S= sulfuro de hidrogeno

3 HBr= Acido bromhidrico

ESPERO ABERTE AYUDADO ;D

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In the following net ionic equation, identify each reactant as either a Bronsted-Lowry acid or a Bronsted-Lowry base. HCN(aq) H2
vfiekz [6]

Answer:

Explanation:

The definition of acids and bases by Arrhenius Theory  was modified and extended by  Bronsted-Lowry.

Bronsted-Lowry defined acid as a molecule or ion which donates a proton while a base is a molecule or ions that accepts the proton. This definition can be extended to include acid -base titrations in non-aqueous solutions.

In this theory, the reaction of an acid with a base constitutes a transfer of a proton from the acid to the base.

From the given information:

\mathsf{HCN _{(aq)} + H_2O_{(l)} \to CN^{-}_{(aq)} + H_3O_{(aq)}}

From above:

We will see that HCN releases an H⁺ ion, thus it is a Bronsted-Lowry acid

H_2O accepts the H⁺ ion ,thus it is a Bronsted-Lowry base.

The formula of the reactant that acts as a proton donor is <u>HCN</u>

The formula of the reactant that acts as a proton acceptor is <u>H2O</u>

8 0
3 years ago
When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
Svet_ta [14]

Answer:

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

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3 years ago
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sergij07 [2.7K]

Answer:

4121 years

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