It should be potential energy!!!
Chromatography is used in purification. Drugs analysts may use the technique to separate the active molecule in a drug molecule, for efficacy or toxicity analysis, from the other drug components.
Explanation:
Chromatography is used to separate a mixture of different components based on the size of their molecules. In liquid chromatography, the mixture is dissolved in a solvent that acts as the mobile phase and then passed along a stationary phase with different kinds of pores, As the mixture passes through the pores, their different components are separated because they take different times to pass through the stationary phase because of their different rates in passing through the pores.
In gas chromatography, a gas is used as a mobile phase while a liquid is used as the stationary phase.
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Answer:
The order of solubility is AgBr < Ag₂CO₃ < AgCl
Explanation:
The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:
Ksp = (A) (B) where A and B are the molar solubilities = s² (for compounds with 1:1 ratio).
It follows then that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:
Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.
Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹² with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of the ratio of ions 2:1 in Ag2CO3, so the answer is not obvious. But since we know that
Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋
Ksp Ag2CO3 = 2s x s = 2 s² = 8.0 x 10-12
s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶
And for AgCl
AgCl ⇄ Ag⁺ + Cl⁻
Ksp = s² = 1.8 x 10⁻¹⁰ ∴ s = √ 1.8 x 10⁻¹⁰ = 1.3 x 10⁻⁵
Therefore, AgCl is more soluble than Ag₂CO₃
The order of solubility is AgBr < Ag₂CO₃ < AgCl
The density of the stone is 5 because the formula is mass/volume and the volume is 5 and the mass is 25
