The table shows the melting point of an unknown compound Unknown Compound Melting Point (°C)
А
801
А
Kayla knows that compounds can be either ionic or covalent. She thinks this unknown compound must be ionic. Is she correct, and why or why not?
Yes, because the unknown compound has a high melting point caused by a weak ionic bond,
Yes, because the unknown compound has a high melting point caused by a strong ionic bond.
No, because the unknown compound has a high melting point caused by a weak covalent bond.
No, because the unknown compound has a high melting point caused by a strong covalent bond.
B
С
D

Idon't know hope someone you

The answer to this is the 3rd choice!

Answerit curves up and levels out.explanationwe assume that the container contains absolutely no ammonia to begin with as it states that "nitrogen gas and hydrogen gas" react to "produce ammonia."at dynamic equilibrium, the product and reactant concentrations are equal.this means that, starting from a concentration of zero, [nh₃] must curve "up" from zero and then level out to a constant concentration.

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.