Answer:
127.0665 amu
Explanation:
Firstly, to answer the question correctly, we need to access the percentage compositions of the iodine and the contaminant iodine. We can do this by placing their individual masses over the total and multiplying by 100%.
We do this as follows. Since the mass of the contaminant iodine is 1.00070g, the mass of the 129I in that particular sample will be 12.3849 - 1.00070 = 11.3842g
The percentage abundances is as follows:
Synthetic radioisotope % = 1.0007/12.3849 * 100% = 8.1%
Since there are only two constituents, the percentage abundance of the 129I would be 100 - 8.1 = 91.9%
Now, we can use these percentages to get the apparent atomic mass. We get this by multiplying the percentage abundance’s by the atomic masses of both and adding together.
That is :
[8.1/100 * 128.9050] + [91.9/100 * 126.9045] = 10.441305 + 116.6252355 = 127.0665 amu
Answer:
9.70 × 10^4 is equal to 97000
Answer:
a) The structure of anthracene is planar with all the pi electrons delocalized in the structure to maintain aromaticity.
b) The C-C bond length in anthracene is about 140 pm with all the bond lengths being similar to each other.
The standard C-C bond length is 154 pm while standard C=C bond is about 134 pm. Therefore the bond length in anthracene is smaller than standard C-C bond length and longer than standard C=C bond length. This can be explained from the fact that the C-C bonds in anthracene has be mixed characteristics of single and double bond because of the delocalization of pi electrons over the whole structure. As a result, they are neither fully single nor fully double bond in nature. Hence the observed bond lengths.
c) This molecule is not flat. The N-atom is sp3 hybridized here and the H-atom attached to N will remain out of plane.
Explanation:
The roots could no longer access depleted groundwater.
The topsoil in the area eroded.
Answer:
C is the element thats has been oxidized.
Explanation:
MnO₄⁻ (aq) + H₂C₂O₄ (aq) → Mn²⁺ (aq) + CO₂(g)
This is a reaction where the manganese from the permanganate, it's reduced to Mn²⁺.
In the oxalic acid, this are the oxidation states:
H: +1
C: +3
O: -2
In the product side, in CO₂ the oxidation states are:
C: +4
O: -2
Carbon from the oxalate has increased the oxidation state, so it has been oxidized.
