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The lower case letter represents the recessive allele.
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
Answer:
c. ΔH° is positive and ΔS° is positive.
Explanation:
Hello,
In this case, as the Gibbs free energy for a reaction is defined in terms of the change in the enthalpy and entropy as shown below:
Thus, as the reaction becomes spontaneous (ΔG°<0) at temperatures above 1100K (high temperatures), it necessary that c. ΔH° is positive and ΔS° is positive as the entropy will drive the spontaneousness as it becomes smaller than TΔS°.
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The answer to this question is: TRUE
