Answer:
Explanation:
The formula of the reaction:
KClO₂ → KCl + O₂
To assign oxidation numbers, we have to obey some rules:
- Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
- The charge on simple ions signifies their oxidation number.
- The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.
The oxidation number of K in KClO₂:
K + (-1) + 2(-2) = 0
K-5 = 0
K = +5
The oxidation number of K in KCl:
K + (-1) = 0
K = +1
The oxidation number Cl in KClO₂ is -1
For Cl in KCl, the oxidation number is -1
For O in KClO₂, the oxidation number is (2 x -2) = -4
For O in O₂, the oxidation number is 0
K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.
O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.
Answer:
c. Many of their bonds are C-C and C-H
Explanation:
The majority of bonds in carbohydrates and lipids( being an organic compound) are C-C and C-H. Like glucose, fructose or galactose ,etc.
These bonds are strong and do require a lot of energy to break. Thus, a lot of energy are required to break carbs and lipids into simpler compounds.Therefore, carbohydrates and lipids have high potential energy.
The correct answer is c.
Answer:
E) A, B, and C
Explanation:
Syn addition refers to the addition of two substituents on the same face or side of a double bond. This differed from anti addition which a occurs across opposite face of the double bond.
Hydrogenation, hydroboration and dihydroxylation all involve syn addition to the double bond, hence the answer chosen above.
Answer:
The answer is C
Explanation:
This is the answer because it is the only way that it is okay to split in half.
Answer:

Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

Best regards.