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
C. pouring honey on a plate so the density of thickness and stickyness would
be the highest
Answer:
He was the first scientist to observe and describe bacteria and protozoa by looking at a drop of water from a pound under a microscope. He also was the one to build the first compound microscope.
Hope this helps :)
Answer:
Actual yield = 86.5g
Explanation:
Percent yield = 82.38%
Theoretical yield = 105g
Actual yield = x
Equation of reaction,
CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O
Percentage yield = (actual yield / theoretical yield) * 100
82.38% = actual yield / theoretical yield
82.38 / 100 = x / 105
Cross multiply and make x the subject of formula
X = (105 * 82.38) / 100
X = 86.499g
X = 86.5g
Actual yield of CaCl₂ is 86.5g
The SI unit for amount of substance is the mole. It has the unit symbol mol. <span>The mole is defined as the amount of substance that contains an equal number of elementary entities as there are atoms in 12g of the </span>isotope carbon-12<span>.</span>
