The reaction involved in present case is:
Net Reaction: Ca + 1/2 O2 → CaO. ..................(1)
In terms of oxidation and reduction, the reaction can be shown at
Oxidation: Ca → Ca2+ + 2e- .................(2)
Reduction: 1/2O2 + 2e- → O2-...................(3)
From, reaction 1 it can be seen that 1 mol of Ca reacts with 1/2 mol of O2 to form 1 mol of CaO.
From, reaction 2 it can be seen that 1 mol of Ca, generates 2 mol of e-.
Thus, when 1/2 mol of Ca is used in reaction, it will lose 1 mol of electrons.
Answer:
Answer is letter B
Explanation:
The first one is wrong because acids release H+, not bases.
The third one is wrong because the pH is exactly 7, not greater.
The last one is wrong because it is vague and does not fit a neutralization reaction.
Answer:
yo try and search it on Google
From the fact that oxygen is in group 16 and carbon is in group 14, the structure of CO2 must be O=C=O. In methane, there is no bond between any of the hydrogen atoms. The structure of H2O2 is H–O–O–H.
Carbon is in group 14 hence it has four valence electrons and oxygen is in group 16 hence it has six valence electrons. This implies that each oxygen atom will share four electrons with carbon in a covalent bond to form the structure O=C=O.
In CH4, we know that carbon is tetravalent so it forms for bonds. Therefore, there is no bond between hydrogen atoms so it bonds with each hydrogen atom; hydrogen only forms one bond.
In H2O2, there is the peroxide ion that has the structure O-O. Hence, the correct structure of H2O2 is H–O–O–H.
Learn more: brainly.com/question/24775418
Answer:
I guess you just answered a lot of questions
Explanation:
Thanks for the points btw :)