Question

НУ Я ПРОШУ ПОМОГИТЕ ПОЖАЛУЙСТА Я НИКАК НЕ МОГУ РЕШИТЬ!!!!:1)H2S+Br2--->S+2HBr ОВР(Окислительно-востановительные реакции) 2)Решите полное и сокращенное ионное уравнение 1.С металлами Na+H2S--->Na2+S+H2 Потом надо самим составить уравнения и также решить полное ионное и сокращенное но с условием то что: 2.С основанием(щелочами) 3.С оксидами металла

Answer 1

Тут понятно и внятно объяснено как решать такие уравнения и состовлять тоже 

Answer 2

H2S + Br2 -> S + 2HBr (Окислительно-восстановительные реакции)

The given chemical equation represents an oxidation-reduction reaction, also known as a redox reaction. In this reaction, hydrogen sulfide (H2S) reacts with bromine (Br2) to produce sulfur (S) and hydrogen bromide (HBr).

To understand the redox nature of this reaction, we need to assign oxidation numbers to each element in the equation. The oxidation number is a positive or negative number that represents the charge an atom would have if electrons were transferred completely.

In H2S, hydrogen (H) has an oxidation number of +1, and sulfur (S) has an oxidation number of -2. In Br2, bromine (Br) has an oxidation number of 0. In S, sulfur has an oxidation number of 0, and in HBr, hydrogen has an oxidation number of +1, and bromine has an oxidation number of -1.

By comparing the oxidation numbers before and after the reaction, we can determine which elements are oxidized and which are reduced. In this case, sulfur is oxidized from -2 to 0, and bromine is reduced from 0 to -1.

The balanced equation for the reaction is as follows:

H2S + Br2 -> S + 2HBr

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. In this case, we have:

H2S + Br2 -> S + 2HBr

This equation is already balanced, as there is one sulfur atom, two hydrogen atoms, and two bromine atoms on both sides of the equation.

Now, let's examine the ionic and net ionic equations for the given reaction:

Ionic equation: 2H^+ + S^2- + Br2 -> S + 2HBr

Net ionic equation: 2H^+ + S^2- -> S + H2

In the ionic equation, we show all the ions present in the reaction, while in the net ionic equation, we only include the ions that participate in the reaction.

С металлами Na + H2S -> Na2S + H2

The given chemical equation represents the reaction between sodium (Na) and hydrogen sulfide (H2S) to produce sodium sulfide (Na2S) and hydrogen gas (H2).

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. In this case, we have:

2Na + H2S -> Na2S + H2

This equation is already balanced, as there are two sodium atoms, two hydrogen atoms, and one sulfur atom on both sides of the equation.

Now, let's examine the ionic and net ionic equations for the given reaction:

Ionic equation: 2Na^+ + S^2- + 2H^+ -> Na2S + H2

Net ionic equation: 2Na^+ + S^2- -> Na2S

In the ionic equation, we show all the ions present in the reaction, while in the net ionic equation, we only include the ions that participate in the reaction.

С основаниями (щелочами)

To solve the equation with bases (alkalis), we need a specific reaction or compound to work with. Could you please provide more information or a specific example?

С оксидами металла

To solve the equation with metal oxides, we also need a specific reaction or compound to work with. Could you please provide more information or a specific example?