CHEM 110 Exam 1 - Practice Test 1 Solutions
1D
The number of protons is always equal to the Atomic number which is the number found
on the periodic table above the symbol. The mass number is the number of neutrons
added to the number of protons (Mass=#p+ + #n0). In this problem the mass number is 80.
We know that the number of protons is equal to 35, so the number of neutrons must be 45
(80 = 35p+ + #n0). The charge on an atom is the number of protons minus the number of
electrons. (Charge = #p+- #e-). Here, we know the charge is -1 so (-1= 35 - #e-) = 36. This
makes sense because we would assume that there should be more electrons since the
overall charge is negative.
2B
s orbitals are regular circles like A. p orbitals are 8 shaped like B. D orbitals are shaped
like answer D. You will learn later in class about hybrid orbitals, which can look like
answer C. Bromine on the periodic table is found in the 3-8 group. We know that these
groups are filled with electron in the p orbitals. The last electron in the Bromine atom is
part of a p orbital so the answer is B.
3C
Keep in mind that for elements other than hydrogen, the 4s is always filled before the 3d.
The electron configuration shows the valence electrons to be in the 4 row/family of the
periodic table. We see that the s orbital is filled and the d orbital has 8 electrons in it. The
only choice is Ni. Answers A and D are tricks. They both have the same number of
electrons but different configuration. Remember that when ionizing an atom other than
Hydrogen, always take from the highest energy level, not the last filled.
4D
The l quantum number refers to the s orbital. Fe3+ has 0 valence electons in the s orbital,
but it has 6 in the core electrons. Keep in mind that electrons in the 4s are ionized before
3d.
5C
This question is based off a periodic trend. Reactivity increases as you go down a group.
Therefore this answer is based on simply going down the correct group. Group I is the
left most group.
6C
The definition of isoelectronic is to have the same electron configuration. The easiest way
to do this problem is find the base (S) on the periodic table, then find where its ion would
be. In this case, S2- would have the configuration of Ar. The next step it to figure out
which other choice has the same configuration as Ar. The best way to do this is by
following the steps above. The answer must be C. Ca2+.
7E
The quantum number l=2 represents the d orbitals. If we look at the Co element, it
normally has 2 electrons in the 4s orbitals and 7 in the 3d. If we take away 1 electron as
suggested by the charge, we end up with 1 electron in the 4s but still 7 in the 3d. There
are 5 orbitals in the d subshell so we fill them one at a time. In the end we have 4 paired
and 3 unpaired electrons.
8D
First realize the question asked, which is NOT false. So it is really asking which is true.
The key to answering this question is knowledge learned in class. They love to throw
these kinds of questions on the test because they test random knowledge. So as a
reminder, For Hydrogen atoms, the electrons fill in order of the energy level. They do not
follow the normal quantum mechanics method. A is false. B and C are false because it is
practically always possible to add or take away an electron to any element. D is true. All
the subshells in any energy level are degenerate (have the same energy). This only
applies to hydrogen. E is false because the 3s electrons are far from the nucleus, yielding
a small effective nuclear charge.
9C
This question appears much harder than it actually is. In Chem 110 the word “heat”
usually refers to energy. On the data sheet you are given (E=hv), where h is a constant.
We are not given v so we have to find another equation to use. the other equation is
(c=λv). Solve for v and plug it into the first equation and we end up with (E=hc/λ). h and
c are constants and we are given λ so we can easily find the energy of one photon of this
laser. Notice that c involves meters and λ is in nanometers, so some conversion is
necessary. 454.6nm * (1m/109)nm = 4.55*10-7 m. Now we can find E=hc/λ or
E=(6.63*10-34 J*s)(3*108 m/s)/(4.55*10-7m)= 4.37*10-19 J. This is the energy of one
photon. to find out how many photons we need, divide. 21J/(4.37*10-19 J/photon) and
you get 4.8*1019 photons. So C is the correct answer.
10D
First we must convert each mass value to the correct number of moles using the molar
mass of each element:
1mol
0.378mol
200.59 g
1mol
N : 10.6 g ×
0.757mol
14.007 g
1mol
O : 36.3g ×
2.269mol
15.999g
Hg : 75.8 g ×
The next step is to divide each value by 0.378 moles (the lowest value) and get the ratio
of atoms to each other, and thus the empirical formula HgN2O6. However, this compound
uses the N and O to make the nitrate ion and so the better way of expressing the formula
is Hg(NO3)2.
11B
Solve using dimensional analysis:
6.85 g C2H2 x 1mol C2H2 x 2 mol C x 6.02 x 1023 C atoms = 3.17 x 1023 C
1
26 g
1 mol
1 mol
12D
Since they tell you it’s a hydrocarbon, you know there won’t be any oxygen
in the molecule. This makes it a somewhat easier problem than the typical
combustion analysis questions. The general formula of all combustion
reactions is an organic compound reacting with O2 to form CO2 and H2O.
The empirical formula of this hydrocarbon can be found if we know the
moles of C and H present in this reaction. Both the 36 g of CO2 and the
19.64 g of H2O can be used to find the moles of C and H, respectively,
because they are the only products in a combustion reaction that have those
atoms.
36 g CO2 x 1 mol CO2 x 1mol C = 0.818 mol C
1
44 g
1 mol
19.64 g H20 x 1 mol H2O x 2 mol H = 2.18 mol H
1
18 g
1 mol
If you divide the moles of C by itself and the moles of H by the moles of C,
the values of 1 for C and 2.67 for H result. In order to make these whole
numbers both must be multiplied by 3, so the empirical formula is C3H8.
13C
The electron configuration of the Zn2+ ion is 1s22s22p63s23p63d10. The two electrons are
removed from the s orbital because orbital prefer to be completely full, empty, or half
full. The l = 0 subshell corresponds to the s orbital, so there are 6 total electrons in the s
orbitals of this ion.
14D
The electron configuration of ground state silicon is 1s22s22p63s23p2. Only two electrons
in the p orbitals of each energy level can have an ml value of -1, so there would be a
maximum of 2 electrons out of the 2p6 orbitals and 1 electron out of the 3p2 orbitals.
15C
The molecular formula of sodium nitrite, which is also its empirical formula
because it cannot be further reduced, is NaNO2. This has a molar mass
(formula weight) of 69g/mol. In one mole, there would be 2 x (6.02 x 1023
molecules of oxygen, so the first statement is false.
16E
Electronegativity increases going up and to the right of the periodic table; fluorine is the
most electronegative element. The bond with the greatest difference in electronegativity
between both atoms will be the most polar. B and F are separated by the most space
compared to the other bonds, so that will be the most polar bond.
17D
18A
Ammonium, NH4+, has an overall charge of +1 and oxygen has a charge of -2, so answer
choice A gives the neutral molecule.
19C
If an Al and X containing molecule contains 3 moles of X in one mole of the molecule,
then each molecule contains 3 X atoms. Aluminum has a charge of +3, so the X atom
must have a charge of -2 to make a neutral molecule. Therefore, S2- is the only
possibility.
20C
For Ionic compounds, melting point is directly related to lattice energy. The weaker the
lattice energy, the lower the melting/boiling point. The first thing to look for when
determining lattice energy is the charges on the ions. Smaller charges have weaker lattice
energies, so KCl and NaCl are the weakest of the group, with KCl being the weakest of
the two because K is larger than Na. Of the remaining compounds, both have +2 and -2
charges and so that is why they have the highest melting points, with CaSe having the
highest melting point of them all since Se is smaller than Te.
21B
Density is mass/volume, and we already have the mass, so we are basically just looking
for volume. Since the side-length is 10.0 cm, the volume of the cube is (10.0 cm)3 =
1000 cm3. Remember, the question asked for the density with units g/cm3, which means
the mass has to be converted from kg to g. Density = (2210 g) / (1000 cm3) = 2.21 g/cm3.
22C
Each of the atoms has its mass number given. Using the periodic table we can look up the
atomic number (number of protons) of each element and then subtract that value from the
mass number to get the number of neutrons. The correct answer is 14N, since 14 – 7 = 7 is
the lowest number of all the available choices.
23C
Remember people, the only way two atoms can be isotopes of one another is if they are
the same element, or more importantly, if they have the same atomic number. Thus C is
the correct answer since they both have 22 protons.
24C
The equation that relates frequency with energy, is E = hv. However, remember that this
equation only works if the energy has the units of Joules, J, (not kJ) and also if it is for
just one photon, not a mole of photons as is the case here. So first we must convert from
kJ/mol to J/photon:
189kJ 1000 J
1mol
×
×
= 3.138 × 10 −19 J / photon
23
1mol
1kJ
6.022 × 10 photons
Then we solve for frequency:
E 3.138 × 10 −19 J
ν= =
= 4.74 × 1014 s −1
−34
h 6.626 × 10 Js
Remember, h is a constant and never changes.
25D
We need an equation that relates changes in energy level (n) and wavelength (λ).
Fortunately, there’s one equation that does just that:
1
( 1
1 %
1
1
= R&& 2 − 2 ##, which we solve for λ =
=
= 9.50 × 10 −8 m
1
1
λ
n
n
(
%
(
%
1
1
7
2 $
' 1
R&& 2 − 2 ## 1.097 × 10 & 2 − 2 #
'1 5 $
' n1 n2 $
Don’t be surprised if you don’t see the answer given exactly as that. Often wavelength is
given in nm, so the correct answer is 95.0 nm.
Remember, R is a constant and never changes.
26A
When a photon is absorbed, the electron always moves to a higher energy level. So
answer choices B,C and E are not even potential candidates as they all represent photon
emissions, not absorptions. Next we need an equation that relates frequency with changes
in energy levels, n. Unfortunately, there is no equation, but don’t panic we just have to
use two equations. First we can figure out the changes in energy for these electron
transitions using the equation:
$1 1 '
ΔE = 2.18 ×10 −18 J && − ))
% ni n f (
Then we take their ΔE values and calculate frequency (v) using ΔE = hv.
The lowest frequency will come from transition n = 3 to n = 16.
Hint: You don’t actually have to make any calculations here if you know what to look
for. Using ΔE = hv, we know that E and v are proportional, so a smaller E gives a
smaller v. Furthermore, any electron transition that doesn’t involve n = 1 will always be
of lower energy than one which does involve n = 1. So from that we can deduce that A
has to be the right answer.
27C
To answer this question we need the equation Ek = hv – Eb, which we rearrange to solve
for the binding energy, Eb = hv – Ek. We know h is a constant, and we have frequency, so
all we need is the kinetic energy, Ek = ½mv2. The velocity we already have and the mass
of an electron is something you will be given on the front of your test. Remember to
convert the mass to kg first before you solve for kinetic energy:
Ek =
1
2
(9.109 × 10
− 28
)
g ×
1kg
× 1.5 × 10 5 m / s
1000 g
(
2
)
= 1.025 × 10 − 20 J
Now we can solve for binding energy:
(
)
Eb = 6.626 × 10 −34 Js (4.85 × 1014 s −1 ) − (1.025 × 10 −20 J ) = 3.11 × 10 −19 J
28E
Since L = 1 refers to the p-orbital, the question can be better interpreted by asking how
many electrons are in the p-orbitals for sulfur.
Since the electron configuration of S is1s22s22p63s23p4, we can see that there are 10
electrons that reside in the p-orbitals.
CHEM 110 Exam 1 - Practice Test 2 Solutions
1C
If n = 4, L can be 0, 1, 2 and 3.
When L = 0, mL = 0
When L = 1, mL = -1, 0, +1
When L = 2, mL = -2, -1, 0, +1, +2
When L = 3, mL = -3, -2, -1, 0, +1, +2, +3
So for all the available mL orbitals, there are 3 with mL = -1.
2E
Remember that the greatest effective nuclear charge is experienced by the electron closest
to the nucleus, with the lowest energy level, and thus 1s is the best choice.
Note: the electron that experiences the weakest Zeff would be farthest from the nucleus; in
this case 4s.
3C
Chromium is one of the exceptions that has a unique electron configuration: [Ar]4s13d5.
After removing 3 electrons the new config. will be [Ar]3d3. If you draw an orbital
diagram for this, you will find exactly 3 unpaired electrons (all up-arrows).
4E
An ionic compound requires two oppositely charged ions, usually a metal cation and a
non-metal anion. Only K3P meets this requirement.
5C
The periodic trend for atomic radius is that size increases as you approach the bottom-left
of the periodic table. Thus the correct answer is K > Na > Ar > Ne.
6B
The periodic trend for ionization energy is that it increases as you approach the top-right
of the periodic table. Thus the correct answer is Cs < Ba < Cl < F.
7A
This is a slightly more complex question than problem 13 since we have ions instead. The
first thing to do is determine how many electrons each ion has and then to ‘pretend’ each
ion is the element with the corresponding number of electrons. In other words, Sr2+ has
36 electrons, so we pretend that it’s Kr (which also has 36 electrons). We do the same for
the other ions and end up with four options that are all the same element, Kr. When this
happens (and this almost always does on chem. tests) the smallest radius goes to the ion
with the most protons. In this case, Sr2+ is the lucky winner, and thus the answer.
Note: the largest radius would go to the ion with the fewest protons (Se2-).
8C
Polarity is a measure of differing electronegativity values for the two elements that form
the bond. Another way to look at this is just to measure how far apart the elements are
from each other on the periodic table.
Using this method, we get Mg – F > N – F > Br – Br.
Note: Bonds involving the same element (like Br – Br) have zero polarity and are called
non-polar.
9D
10E
11C
Start with the elemental electron configuration. Cu = [Ar] 4s2 3d9. Next Remember that it
is better to have half-filled subshells rather than randomly filled so a 4s electron will
jump into the d. This exception to the rule occurs for all the group 6B and 1B elements.
The new electron configuration is Cu=[Ar] 4s1 3d10. Next step is to ionize an electron to
make this a cation. The ionized electron must come from the highest energy level, which
would be 4s. The final electron configuration is Cu+=[Ar] 3d10. From here, you can see
that there are no unpaired electrons, so this ion is diamagnetic. Since the ion has a
positive charge, it is a cation, and this element is obviously a transition metal.
12A
Start by relating energy to wavelength. We know E=hv and 𝑐 = 𝜆 ∗ 𝑣. We can rearrange
ℎ𝑐
ℎ𝑐
these to 𝐸 = 𝜆 . Solve this for 𝜆 and we get 𝜆 = 𝐸 . The energy in this problem has to do
!
!
𝑖
𝑓
with electrons jumping levels so 𝐸 = 𝑅ℎ (𝑛! − 𝑛! ). We can simplify all of this into one
equation to answer this problem or we can break it into steps. Breaking it into steps is
better to avoid calculator errors. For these problems, use lots of parenthesis to ensure you
!
!
aren’t imputing the data incorrectly. 𝐸 = !2.18 ∗ 10!!" 𝐽 !! − !! = ! −4.84 ∗ 10!!" 𝐽.
The negative sign infers that the energy is flowing out of the system, so we can get rid of
!.!"∗!"!!" !∗!"!
it now. 𝜆 =
Violet range.
!.!"∗!"!!"
= !!410.5 ∗ 10!! 𝑚!𝑜𝑟!411!𝑛𝑚. This wavelength is in the
13D
For these problems count the number of possible emission lines. Remember photons are
emitted when electrons jump from a high level to a lower level. Below is a representation
of every possible jump.
14C
𝑘𝑄 𝑄
To answer this problem refer to Columbs law, it’s on the data sheet. 𝐸 = 𝑙 𝑑! ! . An
electron has a (-1) charge and a proton has a (+1) charge. Since 𝑘𝑙 is a constant, and
we’re not looking for exact values, it can be neglected.
!! !!
!
I)
would be 𝐸 =
= − !.
!
!! !!
!
II)
𝐸=
III)
𝐸=
= − !.
!
The sign on energy values only define direction, so according to magnitudes,
II>I>III.
!.!
!! !!
= !.
!
15B
For this problem you must remember atomic trends and the names of the groups:
Lanthanides (Not real, maybe “Lanthanoid” which is an f series.), Halogens (Group 7A),
Actinides (Not real, maybe Actinoid series), Alkali Metals (Group 1A), Chalcogens
(Group 6A). Remember that electronegativity values increase as you go to the right and
up on the table. The atomic radii trend is the exact opposite. Radii increases as you
progress towards the bottom left of the table. Therefore the highest electro negativities
and smallest radii would be the furthest right group, which would be Halogens.
16B
The key to this question is balancing charges on each side and choosing the correct
starting point. The question asked for the second ionization energy, so the reactant must
be a first-ionized Mg, which is Mg+. After this Mg+ is ionized, the products should be a
double ionized Mg and an electron, so Mg2+ and e-. (Mg+  Mg2+ + e- ) Answer choice
(c) is not correct because only 1 electron is being ionized at a time and the charges do not
work.
17C
Set up a system of equations and solve them. Let’s set the abundance of 81Br to be (a) and
the abundance of 77Br to be equal to (b). Now we know:
𝐴𝑡𝑜𝑚𝑖𝑐!𝑀𝑎𝑠𝑠 = Σ 𝑚𝑎𝑠𝑠 ∗ 𝑎𝑏𝑢𝑛𝑑𝑎𝑛𝑐𝑒 .
Atomic mass of Br is 79.904 (from the periodic table). So [79.904=81a + 77b].
Also, since (a) and (b) are the only abundances, they must add to 100%. [a+b=1].
Here is the system:
1) [79.904=81a + 77b]
2) [1= a + b]
You are free to use any method you choose to solve the system. Here is one way:
Multiply equation 2 by (-77) then add it to equation 1.
(-77) *[1= a+b]  [-77=-77a -77b]
[-77=-77a-77b] + [79.904=81a + 77b] = [2.904=4a]
a=0.726= 72.6% This is the abundance of 81Br.
18B
First convert Watts to Joules.
𝐽
5!𝑊𝑎𝑡𝑡𝑠 ∗
1𝑠
∗ 60𝑠 = 300!𝐽
1!𝑊
This is work being done by the system, so the sign for this will be negative. (-300J)
The problem also states that 500 J of heat is entering the system, the sign convention for
this would be (+500J).
Finally:
Δ𝐸 = 𝑞 + 𝑤 = (+500) + −300 = 200!𝐽
19C
Electron affinity is the energy change when an electron is forced onto an atom: when
anions are formed.
I like to compare it to the price the element would pay to get an electron. Fluorine has a
very high electronegativity so it really wants an electron, so it is willing to pay more for it
hence the electron affinity value of fluorine is very high. Some elements don’t want
electrons (low electronegativities) so it is going to cost you money (energy) to make that
element an anion.
In summary, electron affinity values can be positive or negative depending on the
electronegativities of the particular element.
20C
The Heisenberg Uncertainty Principle states that it is impossible to measure position and
velocity at the same time. This principle also states that the Bohr model is bologna
because it would be inappropriate to imagine electrons moving in orderly patterns.
21B
The answer is MgBr because it should be MgBr2. It’s subtle. Remember when naming
ionic compounds, name the cation first then the anion with an –ide suffix.
22D
This type of molecule is called a hydrate molecule. For every molecule of KAl(SO4)2
there will be 12 water molecules attached. We could also say that for every 1 Al, we have
20 Oxygen molecules (8 from (SO4)2 and 12 from H2O). So in 2 grams of alum we would
have 20*(# of Aluminum atoms in 2 grams) or 20 ∗ 2.6 ∗ 10!" = 5.2 ∗ 10!! !Oxygen
atoms. Six grams of alum would just be 3 times greater than the current number.
#!𝑂𝑥𝑦𝑔𝑒𝑛!𝑎𝑡𝑜𝑚𝑠!𝑖𝑛!6!𝑔𝑟𝑎𝑚𝑠 = 3 ∗ 5.2 ∗ 10!! = 1.56 ∗ 10!" !𝑎𝑡𝑜𝑚𝑠!𝑜𝑓!𝑂
23C
Electron affinity is the energy change when an electron is forced onto an atom: when
anions are formed. The reactants must be a neutral element and an electron and the
products must be an anion. The only answer choice that satisfies this is (c).
24D
To answer this problem, first measure the wavelength. In the diagram, two complete
wavelengths are measured, so the wavelength would be half of this dimension.
𝜆=
!.!∗!"!!
!
= 335 ∗ 10!! 𝑚. Next we have to convert wavelength to energy with this
ℎ𝑐
!.!"∗!"!!" !∗!"!
equation !𝐸 = ! =
= 560.28 ∗ 10!!" 𝐽.! We calculated the energy
!""∗!"!!
associated with 1 wavelength, but each of the answer choices have units of J/mole. It is
necessary to convert to the energy associated with a mole of wavelengths to match units.
560.28 ∗ 10!!" 𝐽 6.022! ∗ 10!" Wavelengths
𝐽
𝐸=
∗
= 3.92 ∗ 10! !
.
1!𝑊𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ
1!Mole
𝑚𝑜𝑙𝑒
25E
First start with the electron configuration of Cu. Keep in mind copper follows on of the d
orbital exception rules.
𝐶𝑢 = 1𝑠! 2𝑠! 2𝑝! 3𝑠! 3𝑝! 4𝑠! 3𝑑!"
Next, determine where 𝑚𝑙 orbitals can be located. We know 𝑚𝑙 = −𝑙!𝑡𝑜 + 𝑙 so 𝑙 must be
greater than 0. So we will have a 𝑚𝑙 ! of +1 when 𝑙 = 1,2,3. In this electron configuration
we see 𝑙 = 1,2 with the p and d orbitals. Here are the subshells with +1 orbitals.
2𝑝! !3𝑝! !3𝑑!" !
Below is an electron diagram of these electrons. Simply count the electrons in +1 orbitals.
26E
Analyze the molecule to determine which bond is the most polar. Is it the C-Cl, C-I, CBr, or C-F? Since C is the same in each of these bonds, we just need to compare
electronegativity values. Whichever bond has the most electronegative element in it will
be the most polar. Since F is the most electronegative element on the periodic table, the
F-C bond is the most polar. This means in the overall molecule, F will attract the most
electrons and have the highest partial negative charge.