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Given :
Volume of NaCl solution 2.5 L .
Molarity of NaCl solution is 0.070 M .
To Find :
How many moles are present in the solution.
Solution :
Let, n be the number of moles.
We know, molarity is given by :
So,
Therefore, number of moles of NaCl is 0.175 moles.
Answer:
The correct option is: c. phospholipid
Explanation:
Phospholipids, also known as glycerophospholipids, are the derivatives of fatty acids which is a major structural component of the cell membranes.
Phospholipid is the class of lipids that is composed of a <u>glycerol molecule that forms ester bonds with the two long-chain fatty acids and one phosphate group.</u>
<u>Therefore, Molecule A is a </u><u>phospholipid.</u>
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
