The procedure for testing your unknown solution in this week's lab is identical to the procedure which you conducted in Week 1. The only difference is, of course, your Unknown Solution may or may not contain all of the ions which you tested for in Week 1. With that being said, please consider the following scenario:
You enter the lab and obtain an Unknown Solution from the Stockroom. You begin testing the solution through the steps outlined in the flowchart on p. 9 of the Exp 22 document. You first add HCl, and centrifuge your mixture. You observe the formation of a white precipitate in the bottom of the test tube. After pouring off the supernatant liquid, you add hot water to the white precipitate. Upon addition of the hot water, you still have some white precipitate in the bottom of the test tube. You add ammonia, NH3, to the test tube and observe the formation of a gray-black precipitate. Which of the following is the best conclusion to draw at this point? (Choose only one answer.)
a. The unknown solution definitely has Ag+ present.
b. The unknown solution could have Agt present, or Hg22+ present, or BOTH.
c. The unknown solution definitely has Hg22+ present.
d. The Unknown Solution definitely has Pbb present.

c. The unknown solution definitely has Hg22+ present.

Explanation:

In the analysis of group 1 metal cation, the unknown solution is treated with sufficient quantity of 6 M HCl solution and if group 1 metal cations are present they will selectively precipitates as white precipitate of chlorides. These cations are respectevely: Ag+,Pb2+,Hg22+ . If we consider the whole periodic table, the only elements whose chlorides are insoluble are those of silver, lead (II) and mercury (I), while chlorides of the other elements are soluble.

However,  the precipitate of PbCl2 is soluble in hot water but the other two remains insoluble after treating with hot water. The precipitate of AgCl disappears upon treatment of NH3 solution but Hg2Cl2 becomes gray--black in the reaction with NH3. The black Colour appears due to the formation of metallic Hg.

Balanced chemical equation of the reaction is :

Hg2Cl2 + 2NH3  > HgNH2Cl (white ppt.) + Hg (black ppt.) + NH4Cl

Therefore, the unknown solution definitely has Hg22+ present.

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this