Which substance is the reducing agent in this reaction? 2KMnO4+3Na2SO3+H2O→2MnO2+3Na2SO4+2KOH

Chemistry

2 answers:

boyakko [2]3 years ago

6 0

Answer:

Na2SO3

Explanation:

LuckyWell [14K]3 years ago

5 0

2KMnO4+3Na2SO3+H2O→2MnO2+3Na2SO4+2KOH
In a reaction, the reducing agent is the element or compound that donates electron or the one tht loses electrons. The oxidized species. The opposite is called the oxidizing agent. It is the one who accepts the electrons lost. For this reaction KMnO4 is reduced into MnO2.

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Draw the Lewis structure for methane (CH4) and ethane (C2H6) in the box below. Then predict which would have the higher boiling

Alinara [238K]

Answer:

Ethane would have a higher boiling point.

Explanation:

In this case, for the lewis structures, we have to keep in mind that all atoms must have 8 electrons (except hydrogen). Additionally, each carbon would have 4 valence electrons, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.

Now, the main difference between methane and ethane is an additional carbon. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have more area of interaction for ethane. If we have more area of interaction we have to give more energy to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.

I hope it helps!