The correct answer is letter a, whereas, carrying capacity
in captivity increases population size. It is because the carrying captivity is
the one responsible for having a maximum population size in which will help in
sustaining the necessities that the species need in the environment in which
makes it responsible for the population size to increase depending on its
capacity. The correct answer is letter a.
Answer:
A single compound is simultaneously oxidized and reduced.
Explanation:
In chemistry, disproportionation is a simultaneous oxidation and reduction of a single chemical specie.
What this means is that; in a disproportionation reaction, only one compound is both oxidized and reduced. This implies that two products are formed during disproportionation. One is the oxidized product while the other is the reduced product.
Consider the disproportionation of CuCl shown below;
2CuCl -----> CuCl2 + Cu
Here, CuCl2 is the oxidized product while Cu is the reduced product.
Answer:
<u>When small organic molecules bind together, they form larger molecules called biological macromolecules.</u>Biological macromolecules are important cellular components and perform a wide array of functions necessary for the survival and growth of living organisms. The four major classes of biological macromolecules are carbohydrates, lipids, proteins, and nucleic acids.
(i hope this helps)
Answer:
+1
Explanation:
Na₂O₂
NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.
Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:
Na₂O₂ = 0 (oxidation number of ground state compound is zero)
2Na + 2O = 0
O = –1
2Na + 2(–1) = 0
2Na – 2 = 0
Collect like terms
2Na = 0 + 2
2Na = 2
Divide both side by 2
Na = 2/2
Na = +1
Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1
Answer:
Solid metal
Explanation:
The reduced form of metal ions is the metal in elemental state (simple substance). So, if you have a solution with metal ions and they are reduced, you probably will see the deposition of the metal. For example: if you have a solution with sodium ions (Na⁺), and the ions are then reduced, you will see the aparition of a solid phase of metallic sodium (Na(s)), according to the following half-reaction:
Na⁺ + e- → Na(s)
