A geochemist 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?
a. yes
b. no
If you said yes, calculate it.
Be sure your answer has a unit symbol and the right number of significant digits.

The correct answer is 1.89130 × 10⁻² g per ml.

Explanation:

Based on the given information, the volume of the water sample is 46 ml, the temperature given is 21 degree C. Weight of the compound mineral X is 0.87 grams obtained post evaporating, washing, and drying of the sample. Yes, on the basis of the given information, one can find the solubility of compound X in water at 21 degree C.  

As 46 ml of water comprise 0.87 grams of the mineral compound X. Therefore, 1 ml of the water sample will comprise,  

= 0.87/46 g of X

= 1.89130 × 10⁻² grams

Hence, the solubility of the compound X in the sample of water is 1.89130 × 10⁻² gram per ml.  

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).