Following are the possible isomers of secondary alcohol and ketones for six carbon molecules. In order to distinguish between sec. alcohol and ketone we can simply treat the unknown compound with acidified Potassium Dichromate (VI) in the presence of acid. If with treatment with unknown compound the colour of K2Cr2O7 (potassium dichromate VI) changes from orange to green then it is confirmed that the unknown compound is sec. alcohol, or if no change in colour is detected then ketone is confirmed. This is because ketone can not be further oxidized while, sec. alcohol can be oxidized to ketones as shown below,
<span>6s²4f¹⁴5d¹⁰6p²
6 shows that the element is in the 6 period,
6p² shows that the element is in the 14th group. (1 and 2 groups have s -electrons as last ones, 13 group has s²p¹, and 14 group has s²p²)
The element is Pb.
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Answer:
the atomic packing factor of Sn is 0.24
Explanation:
a = b = 5.83A and c = 3.18A.
Volume of unit cell = a²c
= (5.83)² * 3.18 * 10⁻²⁴ cm³
= 1.08 * 10⁻²²cm³
Volume of atoms =

(∴ BCC, effective number of atom is 2)
Volume of atoms =

= 2.55*10⁻²³cm³


<h3>therefore, the atomic packing factor of Sn is 0.24</h3>
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