Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109

 

Chem 109, Dr. Paselk

Hour Exam II
 Name
Fall 1994

(100 pts)
 Lab

Key

Return to Exam II

Show Work on All Problems For Credit!

(15) 1. Briefly define/describe each of the following:

a. Isotope: an atom of an element with the same number of protons but a different number of neutron in its nucleus.

b. Y2: the probability of finding an electron at a location a given distance from a nucleus.

c. Anion: a negatively chaged ion

d. Unit cell: a small portion of a crystal lattice which, when moved in steps one edge length along its axis will reproduce the lattice

e. Hydrogen bond: a polar bond with partial covalent charecter which forms between a hydrogen attached to a highly electronegative atom (N, O, or F) and another highly electronegative atom (N, O, or F)

(10) 2. Predict the formal charge for each atom in: a) POCl3 and, b) Sodium permanganate. No credit if work is not shown!

a)

P = 5 - 4 = +1

Cl = 7 - 7 = 0

O = 6 - 7 = -1

b) This is a bit more subtle since we can't assume octets around Mn. But we can assume octets around O's, and we know the charge on the permanganate ion = -1. Thus:

O = 6 - 7 = -1 (assume normal octets)

Mn = +3 (since it must neutralize the 4 - of O's and the 1- showing on the ion)

Na = 1+ (trivial, charge on ion since total FC = charge on species)

(8) 3. Consider a lone ion of vanadium (III) and provide the requested information. Read carefully!

V3+ = [Ar] 4so3d2 (outermost, s, electons are lost first)

a. Give a set of quantum numbers for the outermost electron of this ion: n = 3, l = 2, ml= -2, -1, 0, 1, 2 (any one value OK), ±1/2

b. How many radial nodes will this electron's wave function (distribution) have? one radial node at infinity

c. How many angular nodes will this electron's wave function (distribution) have? two angular nodes

d. Should this ion be paramagnetic? Why or why not? yes, it has two unpaired electrons

 

(14) 4. Draw Lewis structures for:

a. Sodium chlorate

b. Bromine trifluoride

c. Based on your structure for (a) above, predict the molecular geometry of chlorate.

SN = 4, 3 bonded atoms, \ trigonal pyramidal

d. Based on your structure for (b) above predict the electronic geometry around bromine.

SN = 5, \ trigonal bipyramidal

e. What is the molecular shape of Bromine trifluoride?

3 bonded atoms, 2 lone pairs, \ T-Shaped

(10) 5. Find the requested concentrations in the solutions described below. Show work for credit!

a) Calculate the molarity of a solution made up by dissolving 56.2 g of calcium chloride in water to give a final volume of 500.00 mL.

FWCaCl2 = 40.08 + 2(35.45) = 110.94 g/mol

M = (56.2g / 110.94g/mol) / 0.50000 L = 1.01 M

b) Find the mole fraction (X) of sodium ions in a 0.200 molal, aqueous solution of sodium chloride.

0.200 mol NaCl in 1.0000 kg H2O

moles H2O in 1 kg = 1000g/18.016g/mol = 55.506 moles

X = 0.200mol / (55.506 + 0.200)mol = 0.00359

(8) 6. Indicate locations on the sketch of the Periodic table provided when answering the questions below.

Periodic Table of the Elements
 IA IIA IIIA IVA VA VIA VIIA VIIIA
   H  He
Li Be  B Groups  B C N O F Ne
Na Mg IIIB IVB VB VI VIIB VIIIB IB IIB  Al Si P S Cl Ar
K Ca   Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
 Rb Sr                  Ag  Cd   Sn     I  Xe
 Cs  Ba        W        Pt Au  Hg    Pb        

                           
                           

a) Draw a box around the lanthanides. Boxes in red

b) Indicate with an arrow the group with highest electron affinities. Group VIIA

c) Show with an arrow the trend, from highest to lowest, of electronegativities Draw arrow from F Cs

d) "Circle" the non-metals Boxes in purple

 

(10) 7. Twenty five grams of a non-electrolyte is dissolved in 469.0 g of water. The resultant solution was found to have a boiling point of 100.58 °C at one atmosphere of pressure. Find the apparent molecular mass of this substance. (The boiling point elevation constant for water is 0.512 °C/m.) Show Work!

DT = Kbmsolute

msolute = DT / Kb = 0.58°C / (0.512°C / m) = 1.1328 m = 1.1328 mol/kg

but 25.00g/0.4690kg = 53.305g/kg

MW = (53.305g/kg) / (1.1328 mol/kg) = 47.056 g/mol = 47.06 g/mol

(10) 8. Draw and label completely a heating curve for water going from 200 K to 400 K at one atmosphere.

(15) 9. Write net ionic equations for each of the following: (If there is no reaction write NR)

a. Solutions of silver nitrate and sodium iodide are mixed

Ag+ + I- AgI(s)

b. Copper sulfate solution is mixed with excess ammonia.

Cu2+ + 4 NH3 Cu(NH3)42+

c. Acetic acid is mixed with a solution of sodium hydroxide.

HOC2H3 + OH- C2H3O- + H2O

d. Solutions of cadmium (II) chloride and potassium sulfide are mixed.

Cd2+ + S2- CdS(s)

e. Hydrochloric acid is poured over zinc metal.

H+ + Zn Zn2+ + H2 (g)


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Last modified 4 April 2007