Download Practice Exam for Exam 4 Key - Ars

Practice Exam for Exam 4
1. A wooden object is claimed to have been found in an Egyptian pyramid and is offered for
sale to an art museum. Analysis of the object determines that it contains 35.4 mg of
Carbon-14. A fresh sample of the same kind of wood contains 62.1 mg of Carbon-14. Is this
object likely to have come from a pyramid from 3500 years ago?
λ=
ln
ln2
ln2
ln2
⇒ k=
=
= 0.0001291 y −1
λ
k
5730 y
Nt
= −kt
No
N
35.4
ln t
ln
No
62.1 = 4354 y
=
t =
−k
− 0.0001291 y −1
(
)
Because the measured age is older than 3500 years, the item is likely to be authentic.
2. How much energy in MJ is release when 1.00 mol of Lithium-6 undergoes the following
reaction:
6
3Li
+ 10n  42He + 31H
Use the data from your textbook.
∆m =
−0.005135 g
( 4.00260 g + 3.01605 g ) − ( 6.01512 g + 1.008665 g ) =
=
−5.135 × 10 −6 kg
(
)(
E=
∆mc 2 =
−5.135 × 10 −6 kg 2.998 × 10 8 m s−1
)
2
=
−4.615 × 1011 J
=
−4.615 × 105 MJ
3. Complete or write the following nuclear reactions
a.
59
b.
235
92U
c.
Positron emission of Phosphorus-30
27Co
+ 10n  5625Mn + 42He
+ 10n  9940Zr + 13552Te + 2 10n
30
15P
 01e + 3014Si
4. Which of the following complexes are tetrahedral and which are square planar based on the
data provided?
a. Ni(en)22+
(diamagnetic)
because it is diamagnetic and Nickel is a d8 ion
2
the hybridization must be dsp and it is square planar
(green color)
because we see green it must absorb red
b. FeCl4therefore  is small, therefore the complex is tetrahedral
(low spin)
With Ag+ it’s always low spin so the only way to
c. Ag(CO)4+
tell is by the fact that CO is a strong ligand so  must be large so the complex is square
planar
5. Name the following transition metal complexes or complex ions
a. [CrBr2(NH3)4]+
Tetraamminedibromidochromium(III) ion
b. [AlF6]-
Hexafluoridoaluminate(V) ion
c. [Fe(CN)5(SCN)]4Pentacyanatothiocyanato-S-ferrate(II) ion
d. [Ag(S2O3)2]3dithiosulfatoargentate(I) ion
e. [Co(NH3)5(SO4)]+
Pentaamminesulfatocobalt(III) ion
6. Write formulas from the following names
a. dicyanidoargentate(I) ion
[Ag(CN)2]b. pentaaquahydroxidoaluminum(III) chloride
[Al(H2O)5(OH)]Cl2
c. (ethylenediamine)dithiocyanato-S-copper(II)
[Cu (NH2CH2CH2NH2)(SCN)2]
7. Cr(en)32+ is low spin, what sort of hybridization does it have?
Low spin, maximum pairing of electrons d orbital available it is d2sp3.
8. Fe(NO2)2(en)2 is polar, does it have optical isomers?
If it’s polar the NO2 ligands are cis and it has optical isomers.
9. MoCl2(H2O)2+ is non-polar, does it have geometric isomers?
It’s non polar so it’s square planar, so it does have geometric isomers.
10. A compound with the overall formula CoCl6  3H2O is dissolved in water. Electrical
conductivity measurements indicate that there are 3 ions per formula unit. Gentle heating
of the compound results in a slight mass loss. X-ray crystallography measurements indicate
that the cobalt has octahedral geometry. Write a formula for this compound indicating
what the ligands are, what the anions are and if how many waters of hydration there are.
(i.e. [ML1nL2m]Ay  xH2O)
3 ions per formula unit indicates that 2 of the chlorides are anions to the complex ion.
Gentle heating resulting in a slight mass loss indicates that at least one of the water molecules is
a hydrate water.
Because the complex ion is octahedral we have to have six ligands. 2 of the chlorides are anions
so the other 4 are ligands which means that 2 of the waters also have to be ligands. The
remaining water is the water of hydration. The formula of the compound is:
[CoCl4(H2O)2]Cl2  H2O
11. Predict the type of decay for the following isotopes and write the nuclear equation.
a.
29
14Si
N/Z = 15/14 > 1.0 so the likely decay type is beta decay:
29
14Si
 0-1e + 2915P
b.
N/Z = 67/51<1.5 (because Z >20) so the likely decay type is positron emission or
electron capture
118
0
118
51Sb  +1e +
50Sn
c.
232
90Th
d.
11
118
51Sb
5B
Z>61 so the likely decay type is alpha decay
232
4
228
90Th  2He +
88Ra
N/Z = 6/5 > 1 so the likely decay type is beta decay:
11
0
11
5B  -1e + 6C