chemistry - help me answer questions please Write names of three typical unit ce

Question

chemistry - help me answer questions please

Write names of three typical unit cells of crystals and the number of atoms each unit cell. a. # of atoms in a cell b. # of atoms in a cell c. # of atoms in a cell 4) The freezing point depression (Deltat) of 8.2g of lauric acid (solvent) and 1.0 g of benzoic acid (solute) was 3.98degreeC. Calculate the molar mass of benzoic acid. Deltat = K_i times m, where K_i of lauric acid is 3.9degreeC-kg/mol, m is the molarity (mole of solute/kg of solvent) of the solution. 5) What is the pH of the buffer that is 0.25M lactic acid and 0.10M in sodium acetate ? The equilibrium constant K_a for lactic acid is 1.4 times 10^-4. Use the Henderson-Hasselbalch oquation to calculate the pH. pH = pKa + log [conjugate base] [weak acid] 6) The K_sp for BaCrO_4 is 1.2times10^-10. Will BaCrO_4 precipitate when 20.0mL of 1times10^-5M of Ba(NO_3)_2 is mixed with 10.0mL of 1times10^-6M K_2CrO_4? Justify your answer.

Explanation / Answer

3) The typical unit cells are

1) Simple cubic cell, with number of atoms = 1

2) Body centred cubic cell, with number of atoms = 2

3) Face centred cubic cell, with number of atoms = 4

4) We know that

Depression in freezing point = Kf X Molality

Molality = Number of moles of solute / Kg of solvents

Kf = 3.9

3.98 = Molality X 3.9

Molality = 3.98 / 3.9 = 1.02

Molality = 1.02 = Mass of solute / Molecular weight of solute X Kg of solvent

Kg of solvent = 8.2 / 1000 Kg = 0.0082

1.02 = 1 / Molecular weight of benzoic acid X 0.0082

Molecular weight = 1 / 0.0082 X 1.02 = 119.56 g / mole

5) Ka = 1.4 X 10^-4

pKa = -logKa = 3.85

pH = ?

[Lactic acid] = 0.25 M

[sodium acetate] = 0.1

pH = pKa + log [salt] / [acid]

pH = 3.85 + log [0.1] / [0.25]

pH = 3.45

6) We know that

Ksp = [Ba+2] [ CrO4-2]

Moles of [Ba+2] = Molarity of Ba(NO3)2 X volume = 20 X 10^-5 millimoles

Moles of [CrO4-2] = 10^-6 X 10 millimoles

Total volume =30mL

[Ba+2] = 20 X 10^-5 millimoles / 30 = 0.667 X 10^-5 M

[CrO4-2] = 10^-6 X 10 millimoles / 30 = 0.333 X 10^-6 M

[Ba+2] [ CrO4-2] = 0.667 X 10^-5 M X 0.333 X 10^-6 M = 0.222 X 10^-11 = 2.22 X 10^-10

So ionic product > Ksp

So precipitate will be formed

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