Test2 HSG F2006.pdf

Vanier College
Chemistry HSG
Fall 2006
Test #2
Name:
Student ID#
Mauro Di Renzo
October 26, 2006

Chemistry HSG – A’06
Test #2
Question #1 (10 marks):
• Name the following compounds: (1 mark each)
a) H 2 CO 3 (aq)
b) CuO
c) P 4
d) KMnO 4
e) N 2 O
• Give the chemical formula of each of the following compounds:
(1 mark each)
f) Boron monophosphide
g) Hydrobromic acid
h) Chromium (III) iodide
i) Sulfurous acid
j) Lithium chlorate
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Chemistry HSG – A’06
Test #2
Question #2 (15 marks):
Hydrazine (N 2 H 4 ) is used as both a rocket fuel and a pesticide. It is the result
of a reaction between sodium hydroxide, ammonia (NH 3 ) and molecular
chlorine. Sodium chloride and water are by-products of the reaction.
a) Write the balanced equation for this reaction. (2 marks)
b) What is the yield in hydrazine formed from the reaction of 120.00 g of
sodium hydroxide, 50.00 g of ammonia, and 100.00 g of chlorine
(8 marks)
c) Calculate the number of moles of each reactant that remain at the end of
the reaction. (5 marks)
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Chemistry HSG – A’06
Test #2
Question #3 (10 marks):
The most intense line in the cerium (Ce) spectrum is at 418.7 nm.
a) Determine the frequency of the radiation producing this line. (5 marks)
b) In which part of the electromagnetic spectrum does this line occur
(2 marks)
c) Is it visible to the human eye If so, what color is it If not, is this line at
a higher or lower energy than visible light (3 marks)
Question #4 (10 marks):
Calculate the wavelength, in nanometers, of light with an energy content of
1799 kJ/mol.
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Chemistry HSG – A’06
Test #2
Question #5 (10 marks):
Using quantum numbers explain why there are always three “p” orbitals on
any shell with a principal quantum number (n) value higher than 1 (In other
words: Why are there no “p” orbitals on the first shell; and why are there
only three on all the other shells)
Question #6 (10 marks):
Bohr’s equation for hydrogen-like elements (He + , Li 2+ , Be 3+ , etc…) is
presented below (where Z is the atomic number):
2
Z B
E
n
= −
2
n
For hydrogen, electronic transitions from n > 2 to n = 2 emit photons which
are in the visible range of the electromagnetic spectrum. Demonstrate
whether a photon emitted from the electronic transition n = 3 to n = 2 for the
He + ion is also in the visible range of the spectrum.
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Chemistry HSG – A’06
Test #2
Question #7 (10 marks):
a) For each of these elements, give the abbreviated electronic configuration
for each of the elements below and underline any sub-shells that contain
valence electrons.
• Y:
• At:
Question #8 (10 marks):
Which of the statements below is (are) correct for an electron with n = 4 and
m l = −2.
• The electron is in the fourth shell.
• The electron may be in a “d” orbital.
• The electron may be in a “p” orbital.
• The electron must have an m s = +½.
Question #9 (15 marks):
For the following elements: C, Mg, Na, O, Rb, Si
a) Place them in increasing order of atomic radius: (8 marks)
b) Place them in increasing order of ionization energy: (7 marks)
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Chemistry HSG – A’06
Test #2
Bonus Question (5 marks):
The first four ionization energies of the element beryllium are I 1 = 900
kJ/mol, I 2 = 1760 kJ/mol, I 3 = 14 850 kJ/mol and I 4 = 21 010 kJ/mol.
Estimate beryllium’s energy of fifth ionization (I 5 ).
Mauro’s Official Cheat Sheet
Good Luck !
Electromagnetic radiation: c = λν
Planck’s equation: E = hν
Bohr’s equation (hydrogen): E n = – B/n 2
Some constants:
Avogadro’s number = N A = 6.022 x 10 23
particles
Planck’s constant = h = 6.63 x 10 –34 J•s
B = 2.18 x 10 –18 J
1 J = 1 kg m 2 s –2
Speed of light = c = 3.00 x 10 8 m s –1
1 nm = 10 –9 m
The EM Spectrum
Wavelength (m)
visible
Violet
Visible spectrum
Blue Green Orange Red
400 450 500 550 600 650 700
Wavelength (nm)
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