Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole
Moles = 0.246
Volume = 280 mL = 0.280 L
Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g
Moles = 1.98
Volume = 2.6 L
Molarity = 0.762 M
Molarity = 0.76 M
Answer: Option (C) is the correct answer.
Explanation:
Chemical formula of a secondary amide is R'-CONH-R, where R and R' can be same of different alkyl or aryl groups. Here, the hydrogen atom of amide is attached to more electronegative oxygen atom of the C=O group.
Therefore, the hydrogen atom will be more strongly held by the electronegative oxygen atom. As a result, there will be strongly hydrogen bonded in the liquid phase of secondary amide.
Whereas chemical formula of nitriles is RCN, ester is RCOOR' and acid chlorides are RCOCl. As no hydrogen bonding occurs in any of these compounds because hydrogen atom is not being attached to an electronegative atom.
Thus, we can conclude that secondary amides are strongly hydrogen bonded in the liquid phase.
1) Zn(CH₃COO)₂(s) + 2KOH(aq) = Zn(OH)₂(s) + 2CH₃COOK(aq)
Ksp{Zn(OH)₂}=1.2*10⁻¹⁷
2) Zn(CH₃COO)₂(s) + 2NaCN(aq) = Zn(CN)₂(s) + 2CH₃COONa(aq)
Ksp{Zn(CN)₂}=2.6*10⁻¹³
Ksp{Zn(OH)₂}<Ksp{Zn(CN)₂}
Zn(OH)₂ precipitates first
Explanation:
it helps to make juices.
It helps to make concentrated acid into dilute acid.
