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CHEM 1411
PRACTICE EXAM I (Chapters 1, 2, 3): 25 questions.
Q1-7: Chapter 1; Q8-13: Chapter 2; Q14-25: Chapter 3.
Multiple Choice: Select one best answer.
1. Which of the following is a physical change?
(a) color of carpet faded (by sun light)
(b) water evaporates
(c) zinc strip dissolves in vinegar (to produce hydrogen gas)
(d) wine turns sour
(e) scramble an egg
Hint: Both 9th and 10 th editions: p.p. 14-15. Physical change is the change
using the physical
properties, which are the ones you can get them back by changing the temperature
or pressure. The
procedure used by adapting physical properties is called the physical process.
Chemical properties are
the ones you cannot get them back.
2. Which of the following is a heterogeneous mixture?
(a) benzene and hexane
(b) salt, water, and sugar (c) oil and alcohol
(d) 14-K gold ring
(e) air
Hint: Both 9the and 10th editions: p. 11. Heterogeneous mixture indicates there
are two or more phases
in the mixture/solution. If the mixture is made by solid and liquid, seeing the
solid indicates it’s the
heterogeneous mixture. If the mixture is made by liquid and liquid, seeing drops
(by shaking it) or
layer indicating it’s the heterogeneous mixture.
3. Which of the following is inaccurate?
(a) silicon, Si
(b) mercury, Hg (c) silver, Ag
(d) chromium, Cr
(e) iron, Ir
Hint: Both 9th and 10th editions: Check with front cover or periodic table and
memorize them
4. What is the equivalent temperature for 98.6 Fahrenheit in Kelvin?
(a) 37
(b) 310
(c) 471.6
(d) 594.8
(e) -273.1
Hint: Both 9th and 10th editions: p.p. 19-20. Example 1.3. There is no direct
conversion between the K
and oF. So in this question, you must convert oF to oC first and then convert oC
to K by applying two
formulas: one is oC = (5/9) x (oF – 32) and the other is K = 273.15 + oC.
5. An unknown sample has a mass of 13.9 g and a volume of 17.4 mL. What is its
density (g/mL)?
(a) 0.798
(b) 1.04
(c) 3.16
(d) 4.62
(e) 5.07
Hint: Both 9th and 10th editions: p. 18. Example 1.1. Density = mass/volume
6. How many significant figures are there in the measurement of 6.124 x 10-5?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
Hint: Both 9th and 10th editions: p.p. 23-26. Memorize the rules. Example 1.4.
7.
How many significant figures are appropriate to show in the result after
carrying out the operation
below?
(223.7 + 0.27) ÷ 4.21 =?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
Hint: Both 9th and 10th editions: p. 25. Example 1.5. You need to memorize all
the rules concerning the
significant figures and the rules related to addition or subtraction and
multiplication or division. Many
students do not know the differences between addition/subtraction and
multiplication/division. See two
examples below:
Remember that the answers obtained from calculators are always WRONG when
considering the
significant figures. So when the question combining addition/subtraction and
multiplication/division
together, each follows its own rule.
1
Thus, 223.7 + 0.27 = 223.97 from calculator, which must be corrected with the
least decimal points of
the components, and thus it is 224.0. So the question turns to 224.0 ÷ 4.21 =
53.20665083 from
calculator, which must be corrected with the lease digit of the significant
figures of the components,
and thus it is 53.2.
Since this question only asks how many significant figures in the result, thus
from the components,
224.0 and 4.21, we know one has four significant figures and one has three
significant figures. Thus
the number of significant figures of the result should follow the one with the
least significant figures,
that is, three significant figures.
8. How many protons, electrons, and neutrons are there in 72Br - ?
(a) 35p, 35e, 72n
(b) 34p, 35e, 37n
(c) 35p, 36e, 72n
(d) 36p, 34e, 72n
(e) 35p, 36e, 37n
Hint: 9th edition: p. 53; 10th edition: p. 54. Also p. 50: Example 2.1 to
calculate numbers of protons,
neutrons, and electrons. For Positive charge indicates there are more protons
than electrons; while the
negative charge indicates there are more electrons than protons. This is because
neutron does not carry
charge, a proton carries one positive charge, and an electron carries one
negative charge.
The ion in this question carries one negative charge indicating that the
electron is one more than the
protons. From periodic table, the atomic number (that is, the proton number) of
Br is 35, and thus there
are 35-(-1) = 36 electrons.
9. Which of the following pairs would have similar chemical and physical
properties?
(a) Ni, Mg
(b) H, Na
(c) C, Si
(d) Cu, Ca
(e) Cr, Cl
Hint: 9th edition: p.p. 50-51; 10th edition: p. 51. The elements from the same
group (i.e. the vertical
column) will have similar physical and chemical properties.
10. Which of the following is a molecular compound?
(a) KCl
(b) CsF
(c) HCN
(d) AlBr3
(e) NaOH
Hint: Both 9th and 10th editions: p. Starting 62. Molecular compound is composed
by nonmetal reacts
with nonmetal atoms.
11. Which of the following is the empirical formula?
(a) C8H18
(b) CH3COOH
(c) C4H10O2
(d) NH4NO3
(e) B2H6
Hint: 9th edition: p.p. 55-57; 10th edition: p. 56-57. Both editions: Example
2.3. Empirical formula is
the simplest integral ratio among each atom.
12. Which of the following is not a correct match?
(a) AlCl3, aluminum chloride
(b) H2O dihydrogen monoxide
(c) CrF3, chromium (III) fluoride
(d) HNO3, nitrous acid
(e) CuSO4.5H2O, copper (II) sulfate pentahydrate
Hint: Both 9th and 10th editions: p.p. 59-68. Follow the rules of naming the
compounds. Be able to
differentiate the rules between ionic and molecular compounds.
13. Which of the following pairs is a correct match?
(a) Ba3(PO4)2, barium (II) phosphate
(b) (NH4)2SO4, diammonium sulfate
(c) Na2O2, sodium oxide
(d) Ca(NO3)2, calcium nitrate
(e) CH3COOH, carbonic acid
Hint: Both 9th and 10th editions: p.p. 59-68. Follow the rules of naming the
compounds. Be able to
differentiate the rules between ionic and molecular compounds.
14. The atomic masses of 10B and 11B are 10.0129 amu (natural abundance 19.78%)
and 11.0093 amu
(natural abundance 80.22%), respectively. What is the average atomic mass of B?
(a) 9.467
(b) 9.966
(c) 10.042
(c) 10.504
(e) 10.810
Hint: 9th edition: p.p. 78-79. Example 3.1; 10th edition: p.p. 80-81 Example
3.1. You must convert the
% into decimals first and then apply the formula and memorize it: M = M1X1 +
M2X2 =
2
10.0129x0.1978 + 11.0093x0.8022 = 10.810. Note that the average atomic mass
(with decimal) can be
found in the periodic table, while the mass number (a whole number) can not be
found.
For three isotopes: Example Exercise 5.3: 28Si (27.977 amu; 92.21% abundance),
29Si (28.976 amu;
4.70% abundance) and 30Si (29.974 amu; 3.09% abundance). The atomic mass M =
M1X1 + M2X2 +
M3X3 = 27.977x0.9221 + 28.976x0.0470 + 29.974x0.0309 = 28.09 amu by rounding to
two decimals.
15. How many moles of S are there in 64.2 g of S?
(a) 1.325
(b) 1.764
(c) 1.968
(d) 2.003
(e) 2.475
Hint: 9th edition: p. 81. Examples 3.2 and 3.6. 10th edition: p.p. 82-83:
Example 3.2.
Mole = mass (g) / molar mass (g/mole) = 64.2/32 = 2.003
16. What is the molar mass for calcium nitrate, Ca(NO3)2?
(a) 44
(b) 56
(c) 87
(d) 93
(e) 164
Hint: 9th edition: p.p. 83-84. Example 3.5. 10th edition: p.p. 85-87 Example
3.5.
In the formula, Ca(NO3)2, it indicates there are 1 Ca, 1x2 = 2 N and 3x2 = 6 O.
Go to the periodic table
and locate the atomic mass for each atom. Thus the formula mass of Ca(NO 3)2 =
1x40 + 2x14 + 6x16
= 164
17. How many hydrogen atoms are there in 48.0 g of CH4?
(a) 1.81x1023
(b) 7.22x1024
(c) 6.02x1023
(d) 1.20x1025
(e) 4.70x1025
th
th
Hint: 9 edition: p.p. 85. Example 3.7. 10 edition: Example 3.7.
According to the chemical formula, one mole of CH4 contains 1 mole of C atoms
and 4 moles of
hydrogen atoms. Thus, the mole of H = 4 x {mass of CH 4/molar mass of CH4}. When
converting
moles into atoms, it needs to multiply the Avogadro’s number, that is, 6.02x10
23. Thus the number of
H atoms = moles of H atoms x Avogadro’s number = (6.02x10 23)x{4x(48.0/16.0)} =
7.22x1024
18. What is the mass percent (%) for O in SO2?
(a) 38.09
(b) 45.41
(c) 50.00
(d) 53.86
(e) 56.43
Hint: 9th edition: p. 87. Equation (3.1). 10th edition: p.p. 88-89: Example 3.8.
% S = (mass of 2 O/ mass of SO2) x 100% = {2x16/(1x32+2x16)}x100% = 50.0%
19. Chemical analysis shows the composition of a compound containing carbon,
hydrogen, chlorine,
and oxygen, to be 37.84% C, 2.12% H, 55.84% Cl, and 4.20% O. What is its
empirical formula?
(a) CHClO
(b) C2HClO4
(c) C12H8Cl6O
(d) C12H8Cl6O4
Hint: 9th edition: p. 88. Example 3.9 (Vitamin C): very important or the most
difficult for
empirical formula calculation. Figure 3.5. or p. 109 (3.50). 10th edition: p.p.
92-94: Example 3.9 (p.
90). The empirical formula is the simplest integral ratio of moles among each
atom. Here, there are
four different kinds of atoms, C, H, Cl and O. Thus mole of C = 37.84/12 = 3.15;
mole of H = 2.12/1 =
2.12; mole of Cl = 55.84/35.45 = 1.58; mole of O = 4.20/16 = 0.26.
Note that as long as one of the moles is not an integer, we have to divide the
smallest value among
them: here the smallest value is 0.26. So C : H : Cl : O = 3.15/0.26 : 2.12/0.26
: 1.58/0.26 : 0.26/0.26 =
12.1: 8.1 : 6.1: 1 = 12: 8: 6: 1, which indicates that the empirical formula
contains 12 C, 8 H, 6 Cl and
1 O. Thus, the empirical formula is written as C12H8Cl6O as 1 is usually not
written in the formula.
*****What is the empirical formula for methyl benzoate, a compound used in the
manufacture of
perfumes, contains 70.57% carbon, 5.94% hydrogen, and 23.49% oxygen? Note:
subscripts must be
integers.
A) C4H4O
B) C2H2O0.5
C) C8H8O2
D) CHO
Hint: Section 9.8. This is a very, very important question. The empirical
formula is the simplest
integral ratio of moles among each atom. Here, there are three different kinds
of atoms, C, H and O.
Thus mole of C = 70.57/12 = 5.88; mole of H = 5.94/1 = 5.94; mole of O =
23.49/16 = 1.47.
Note that as long as one of the moles is not an integer, we have to divide the
smallest value among
them: here the smallest value is 1.47. So C : H : O = 5.88/1.47 : 5.94/1.47 :
1.47/1.47 = 4 : 4.04 : 1.
3
Since 4.04 is very close to 4.00 and thus we can round it to 4.00. So C : H : O
= 4 : 4 : 1, which
indicates that the empirical formula contains four C, four H and one O. Thus,
the empirical formula is
written as C4H4O as 1 is usually not written in the formula.
20. Chemical analysis shows the composition of a compound containing carbon and
hydrogen, to be
80.00% carbon and 20% hydrogen and the molar mass is 30 g. What is its molecular
formula?
(a) CH
(b) C2H4
(c) C2H6
(d) C6H12
(e) C10H22
Hint: 9th edition: p.p. 91-92. Example 3.11 or p. 109 (3.54). 10th edition: p.
93: Example 3.11. The
molecular formula is an integral multiple of empirical formula. That is, the
molar mass = empirical
molar mass x integer. From C: H = 80.00/12 : 20.00/1 = 6.66: 20 = 1: 3. So the
empirical formula is
CH3 and the empirical molar mass of CH3 = 12x1+1x3 =15. So the integer = 30/15 =
2. Thus there are
two empirical formulas in a molecular formula. Therefore, the molecular formula
is C2H6.
21. What are the coefficients respectively when the equation
__PH3 + __O2  __P2O5 + __H2O is balanced?
(a) 2, 2, 1, 3
(b) 2, 1, 3, 4
(c) 2, 3, 1, 2
(d) 2, 4, 1, 3
(e) 1, 1, 3, 3
Hint: 9th edition: p.p. 94-97. Example 3.12 or p. 109 (3.60). 10th edition: p.p.
94-99: Example 3.12.
Balancing an equation is a process of trial an error. Sometimes it requires more
than one trial. Usually
start visual examination and select the most bulky species, that is, the one
with the most different
kinds of atoms and number of atoms as the reference and set its coefficient as
one.
In this question, P2O5 is the most bulky one, we put 1 in front of it to remind
us we have done
examining P2O5. Now the equation is updated to __PH3 + __O2  1 P2O5 + __H2O
Since P2O5 contains 2 phosphorus atoms, so we need two phosphorus atoms at the
left side, which
leads us to put 2 (called coefficient) in front of the PH3.
Now the equation is updated to 2 PH3 + __O2  1 P2O5 + __H2O
As there are 6 hydrogen atoms in 2 PH3, thus we need to balance the hydrogen
atoms at the right side,
which lead us to put 3 in front of the H2O.
Now the equation is updated to 2 PH3 + __O2  1 P2O5 + 3 H2O
Now we need to balance the oxygen atoms. Since there are 1x5+3x1 = 8 oxygen
atoms at the right
side, and thus the left side must have the same number of oxygen atoms. That is
to say, __x2 = 8. So
__ = 4. Now the equation is updated to 2 PH3 + 4 O2  1 P2O5 + 3 H2O
Since all the atoms of each type has the same amount, this equation is balanced.
22. In the reaction of Al(OH)3 with H2SO4, how many moles of water can be
produced If the reaction
is begun with 5.500 mole of Al(OH)3?
2Al(OH)3 + 3H2SO4  Al2(SO4)3 + 6H2O
(a) 2.50
(b) 4.75
(c) 6.32
(d) 7.58
(e) 16.50
Hint: 9th edition: p.p. 97-101. Examples 3.13. and 3.14. Figure 3.8. 10th
edition: p.p.99-103: Examples
3.13. &3.14. Stoichiometrey. Method One (ratio approach): From the equation, the
involving species, 2
Al(OH)3 and 6 H2O with coefficients 2 and 6 respectively, tell us that for 2
moles of Al(OH)3 it
produces 6 moles of H2O. According to this proportion or ratio, 5.500 mole of
Al(OH)3 requires 5.500
x (6/2) = 16.50 moles of H2O.
Method Two (road map approach): Apply the road map or say the dimensional
analysis. The road map
is
grams of substance A  moles of substance A moles of substance B  grams of
substance B.
4
Note:
(1) substance A is the one with given (or known) information of mass (or mole);
substance B is the one needed to be calculated.
(2) grams of substance A  molar mass of substance A = moles of substance A.
(3) moles of substance A  coefficient of substance B  coefficient of substance
A =
moles of substance B.
(4) grams of substance B = moles of substance B  molar mass of substance B.
Since this question starts at moles and thus we only apply (2):
5.500 moles Al(OH)3 x {6 H2O/2 Al(OH)3} = 16.50 moles H2O
23. How many grams of H2O could be formed by the reaction of 16.0 g of CH 4 with
48.0 g of O2?
CH4 + 2O2  CO2 + 2H2O
(a) 27.0
(b) 37.3
(c) 46.8
(d) 54.1
(e) 58.7
Hint: 9th edition: p.p. 97-101. Examples 3.13. and 3.14. Figure 3.8. 10th
edition: p.p. 99-103. Examples
3.13 & 3.14. Stoichiometrey. This question provides two known quantities of
substance A and thus
this is the limiting reagent question. So we need to apply two times of allfour-step-road map to figure
out what is the true mass of water. Note that the limiting reagent limits the
(maximum) of product(s)
that can be produced. In the road map application, the one produces the smallest
amount of the
product is the true limiting reagent. The one with the left over is the excess
reagent.
Assume CH4 (i. e. substance A # 1) is the limiting reagent:
÷ 16 g/mol
x 2H2O/1CH4
x 18 g/mol
16 g CH4 ------------- 1 mole CH4 ---------------- 2 mole H2O ------------- 36
g H2O
Assume O2 is the limiting reagent:
÷ 32 g/mol
x 2H2O/2O2
x 18 g/mol
48 g O2 ------------- 1.5 mole O2 ---------------- 1.5 mole H2O-------------
27 g H2O
Because O2 produces the least amount of H2O (can compare by using mole, 1.5 < 2,
or grams, 27 <
36), it is the true limiting reagent and the maximum amount of water produced is
27 grams. The CH 4
is the excess reagent.
24. What is the excess reagent in the above reaction (Q.23)of CH4 and O2? How
many grams of the
excess reagent were consumed?
(a) CH4, 12 grams
(b) O2, 12 grams
(c) CH4, 4 grams
(d) O2, 4 grams
(e) CH4, 8 grams
Hint: 9th edition: p.p. 101-103. Example 3.15. 10th edition: p.p. 103-106:
Example 3.15.
Now the substance A is O2 and CH4 is the substance B because we need to
calculate its amount
consumed stoichiometrically. Apply the road map:
÷ 32 g/mol
x 1CH4/2O2
x 16 g/mol
48 g O2 ------------- 1.5 mole O2 ---------------- 0.75 mole CH4-------------
12 g CH4
Thus, there are (16 – 12) = 4 grams of CH4 left.
This is because # grams excess = # grams original – # grams used.
Or say # moles excess = # moles original – # moles used.
25. A 15.6 grams of benzene (C6H6) is mixed with excess HNO3 to prepare
nitrobenzene (C6H5NO2).
After the reaction there are 15.6 grams of nitrobenzene produced. What is the
percent yield of
nitrobenzene?
5
C6H6 + HNO3  C6H5NO2 + H2O
(a) 34.3%
(b) 47.6%
(c) 58.9%
(d) 63.4%
(e) 71.2%
Hint: 9th edition: p.p. 103-106. Example 3.16. 10th edition: p.p. 106-107:
Example 3.16.
The amount of product calculated according to the road map is the theoretical
yield and the one
obtained by weighing is the actual yield.
% yield = (actual yield/ theoretical yield) x 100%
Now from the road map, we calculate the theoretical yield (note that since the
actual yield given is in
unit of grams, so we need to complete the entire road map).
÷ 78 g/mol
x 1C6H5NO2 /1C6H6
x 123g/mol
15.6 gC6H6------------ 0.2mole C6H6 ----------------------- 0.2mole C6H5NO2 -------- 24.6g C6H5NO2
Thus the percent yield = {15.6 g / 24.6 g} x 100% = 63.41%
6
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