Glucose and Oxygen
Hope this helped you out
3600 years.
Since 65% of the carbon-14 is remaining, we need to get the logarithm to base 2 of 0.65 to determine how many half lives have expired. So: log(0.65)/log(2) = -0.187086643/0.301029996 = -0.621488377
So we know that 0.621488377 half-lives has gone by to the bone sample. Now we just need to multiply by the half-life of carbon-14 which is 5730 years. So: 0.621488377 * 5730 = 3561.128399 years. Rounding to the nearest 100 years gives us 3600 years.
is having lowest oxidation number of sulphur in teh options given. The oxidation number is -2.
The correct option is E.
Explanation:
have oxidation number of -2, which is lowest in all the options given.
In the given examples 
will have oxidation number zero because it is in elemental form. In the rules for oxidation number it is given that any element in free state has its oxidation number zero.
All other options are in compound form as:
S
= oxidation number is +4. It is calculated as oxygen having O.N of -2,
The charge on the atom is -2, so sum of the oxidation numbers of all the elements should be equal to -2.
x+3(-2)= -2
x = 4
S
= The charge on the atom is -2, so sum of the oxidation numbers of all the elements should be equal to -2.
x +4(-2) = -2
x = +6
The charge on the atom is -2, so sum of the oxidation numbers of all the elements should be equal to -2.
2x + 3(-2) = -2
2x = 4
x = 2
= oxidation number is -2 as the charge on ion in equal to its oxidation number.
