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.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.