An example of a fixed cell is a fat cell.
O2 is the limited reactant
D. To answer this question refer to the periodic table and think logically about it. Adding one proton increases the atomic number so you go along the row to Nitrogen. If you lose one neutron then the atomic mass decreases by 1 so 14-1 is 13.
Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
Answer:
It means the chemical entity is a radical
Explanation:
When we talk of unsaturation, we are referring to the number of pi-bonds in a chemical entity. The alkane, alkene and alkyne organic family are used to as common examples to explain the term unsaturation.
While alkynes have 3 bonds, it must be understood that they have 2 pi bonds only and as such their degree of saturation is two.
In the case of an alkene, there is only one single pi bond and as such the degree of unsaturation is 1.
Now in this case, we have a fractional 0.5 degree of unsaturation alongside the 3 to make a total of 3.5. So what’s the issue here?
The fractional part shows that the chemical entity we are dealing with here is a radical. While the integer 3 shows that there are 3 pi-bonds, the half pi bond remaining tells us that there is a missing electron on one of the atoms involved in the chemical bonding and as such, the 1/2 extra degree of unsaturation tends to tell us this.
Kindly recall that a radical is a chemical entity within which we have at the least an unpaired electron.
