The answer you are looking for would be a colliod. Hope this helps have a great day!!
Answer:
C. The half-life of C-14 is about 40,000 years.
Explanation:
The only false statement from the options is that the half-life of C-14 is 40,000yrs.
The half-life of an isotope is the time it takes for half of a radioactive material to decay to half of its original amount. C-14 has an half-life of 5730yrs. This implies that during every 5730yrs, C-14 will reduce to half of its initial amount.
- All living organisms contain both stable C-12 and the unstable isotope of C-14
- The lower the C-14 compared to the C-12 ratio in an organism, the older it is.
Answer:
The four resonance structures of the phenoxide ion are shown in the image attached
The conjugate base of cyclohexanol has only one resonance contributor, while
the conjugate base of phenol has four resonance contributors.
Explanation:
In organic chemistry, it is known that structures are more stable if they possess more resonance contributors. The greater the number of contributing canonical structures, the more stable the organic specie. Since the phenoxide ion has four contributing canonical structures, it is quite much more stable than cyclohexanol having only one contributing structure to its conjugate base. Hence the PKa(acid dissociation constant) of phenol is lesser than that of cyclohexanol. The conjugate base of phenol is stabilized by resonance.
<span>The choices are as follows:
h2o + 2o2 = h2o2
fe2o3 + 3h2 = 2fe + 3h2o
al + 3br2 = albr3
caco3 = </span><span>cao + co2
The correct answers would be the second and the last option. The equations that are correctly balanced are:
</span> fe2o3 + 3h2 = 2fe + 3h2o
caco3 = cao + co2
To balance, it should be that the number of atoms of each element in the reactant and the product side is equal.
