A student has two compounds in two separate bottles but with no labels on either one. One is an unbranched alkane, octane, C8H18
1 answer:
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Answer:
All three are present
Explanation:
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: .
- Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
- Secondly, addition of liquid ammonia would form a precipitate with silver:
; Silver hydroxide at higher temperatures decomposes into black silver oxide:
.
- Thirdly, we also know we have
in the mixture, since addition of potassium chromate produces a yellow precipitate:
. The latter precipitate is yellow.