Ageochemist in the field takes a 46.0 ml sample of water from a rock pool lined with crystals of a certain mineral compound x. he notes the temperature of the pool, 21·c, and caps the sample carefully. back in the lab, the geochemist filters the sample and then evaporates all the water under vacuum. crystals of x are left behind. the researcher washes, dries and weighs the crystals. they weigh 0.87 g. yes x10 1 using only the information above, can you calculate the solubility of x in water at 21.° c?

The solubility of the mineral compound X in the water sample is 0.0189 g/mL.

Explanation:

Step 1: Given data

The volume of water sample = 46.0 mL.

The weight of the mineral compound X after evaporation, drying, and washing = 0.87 g.

Step 2: Calculate the solubility of X in water

46.00 mL of water sample contains 0.87 g of the mineral compound X.

To calulate how many grams of the mineral compound  1.0 mL  of water sample contains:

0.87 g/46.0 mL = 0.0189 g.

This means the solubility of the mineral compound X in the water sample is 0.0189 g/mL.

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a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.