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.
Answer:
Glucose
Explanation:
Glucose is broken down into water and carbon dioxide.
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Answer:
The oxidizing agent is the MnO₄⁻
Explanation:
This is the redox reaction:
10 I⁻ (aq) + 2 MnO₄⁻ (aq) + 16 H⁺ (aq) → 5 I₂ (s) + 2 Mn²⁺ (aq) + 8 H2O (l)
Let's determine the oxidation and the reduction.
I⁻ acts with -1 in oxidation state and changes to 0, at I₂.
All elements in ground state has 0 as oxidation state.
As the oxidation state has increased, this is the oxidation, so the iodide is the reducing agent.
In the permanganate (MnO₄⁻), Mn acts with +7 in oxidation state and decreased to Mn²⁺. As the oxidation state is lower, we talk about the reduction. Therefore, the permanganate is the oxidizing agent because it oxidizes iodide to iodine
The false statement from the above is that: Temporary charge imbalances in the molecules lead to London dispersion forces.
<h3>What are the factors that affect London dispersion forces?</h3>
Generally, the factors which affects the London dispersion forces a dispersion force are as follows:
- Shape of the molecules
- Distance between molecules
- Polarizability of the molecules
However, London dispersion forces simply refers to a sort of temporary attractive force formed when electrons in two adjacent atoms occupy positions that make the atoms form dipoles.
So therefore, temporary charge imbalances in the molecules lead to London dispersion forces is a false statement
Learn more about London dispersion forces:
brainly.com/question/1454795
