Nonmetals and negative. Answer is c
Answer:
Explanation:
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Oxidizing agents:
Oxidizing agents oxidize the other elements and itself gets reduced.
Reducing agents:
Reducing agents reduced the other element are it self gets oxidized.
Consider the following reaction:
2AgCl + Zn → 2Ag + ZnCl₂
In this reaction oxidation state of Zn on left side is 0 while on right side +2 so it gets oxidized and oxidation state of Ag on left side is +1 and on right side 0 so it get reduced.
4NH₃ + 3O₂ → 2N₂ + 6H₂O
In this reaction oxidation state of nitrogen on left side is -3 while on right side 0 so it gets oxidized and oxidation state of oxygen on left side is 0 and on right side -2 so it get reduced.
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
In this reaction oxidation state of iron on left side is +3 while on right side 0 so it gets reduced and oxidation state of Al on left side is 0 and on right side +3 so it get oxidized.
Answer:
Are transferred completely from the valence shell of an element to the other
Explanation:
Basically, to form a chemical bond, you either transfer or you share. When you share, it is a case of covalent bonding which can be in several other forms. When there is a transfer, it is a case of ionic bonding.
The basic explanation for this is that while some atoms are electronically sufficient, some are electronically deficient. This means while some atoms are having an excess number of electrons, then some are having less number of electrons.
To satisfy both parties, there must be a transfer if electrons between the two parties. While the one with the excess numbers serves as the donor, the one with insufficient number of electrons serve as the acceptor
Answer:
See the answer below
Explanation:
<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>
1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.
2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.
