If 50.0 g of water saturated with potassium chloride at 80.0°C is slowly evaporator to dryness, how many grams of the dry salt w
ill be recovered
1 answer:
Answer:
25.0 g.
Explanation:
- From the solubility curve that is shown in the attached image, the solubility of KCl per 100.0 g of water is about 50.0 g.
- <em>So, a saturated solution of KCl in 50.0 g of water will contain about 25.0 g.</em>
- <em>Thus, the grams of the dry salt that will be recovered is 25.0 g.</em>
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![pH=-log[ H_{3}O^{+}] \\ \\ pH=-log(1.6 * 10^{-3}) \\ \\ pH=2.8](https://tex.z-dn.net/?f=pH%3D-log%5B%20H_%7B3%7DO%5E%7B%2B%7D%5D%20%5C%5C%20%20%5C%5C%20%0ApH%3D-log%281.6%20%20%2A%2010%5E%7B-3%7D%29%20%20%5C%5C%20%20%5C%5C%20%0ApH%3D2.8)
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