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12 December, 11:19

A geochemist in the field takes a small sample of the crystals of mineral compound X from a rock pool lined with more crystals of X. He notes the temperature of the pool, 26.° C, and caps the sample carefully. Back in the lab, the geochemist dissolves the crystals in 3.00 L of distilled water. He then filters this solution and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.36 kg1) Using only the information above can you calculate the solubility of X in water at 26 degrees Celsius? yes or no2) If yes calculate the solubility. Round answer to 2 signifacnt digits

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  1. 12 December, 13:28
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    The solubility is 0.13 g/mL

    Explanation:

    Step 1: Data given

    Temperature = 26.0 °C = 299 K

    Volume = 3.00 L

    The mass of the crystals, after washing and drying = 0.36 kg = 360 grams

    step 2: Calculate the solubility

    3.00 L of water contains 360 grams of crystals

    For 1.00L of water we'll have 360 / 3 = 130 grams of crystals

    This means we have 130 grams of crystals in 1 L, this gives us a solubility of 130g/L

    In 1000 mL we have 130 grams crystal

    in 1 mL we have 130/1000 = 0.130 grams of crystals

    The solubility is 0.13 g/mL
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