Chem119-Exam4-practice-answers
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FALL 2023 BARONDEAU CHEMISTRY 119 EXAM 4 practice Name (Print): _________________________________ Name (Signature):_____________________________ Student ID:_____________________________ Instructions: You can remove the formula page/periodic table on the last page from the exam. The exam must be completed during the class period and the answers written neatly in the boxes provided next to the problems for grading.
1.
How many molecules of oxygen must be included in the balanced chemical equation for the complete combustion reaction of heptane, C
7
H
16
(hint: the complete combustion of a hydrocarbon is the reaction with molecular oxygen to form carbon dioxide and water) (4 points)? C
7
H
16
+ 11 O
2
7 CO
2
+ 8 H
2
O 2.
Mixing MgCl
2
and Na
3
PO
4
results in the precipitation of Mg
3
(PO
4
)
2
. What mass of magnesium phosphate can be produced from 55.2 g of MgCl
2
and excess Na
3
PO
4
(give answer with 3 significant figures; 4 points)? 3 MgCl
2
(aq) + 2 Na
3
PO
4
(aq)
6 NaCl(aq) + Mg
3
(PO
4
)
2
(s) (55.2 g MgCl
2
)(1 mol / 95.21 g)(1 mol Mg
3
(PO
4
)
2
/ 3 mol MgCl
2
) = 0.193 mol Mg
3
(PO
4
)
2
(0.193 mol Mg
3
(PO
4
)
2
)(262.86 g/ 1 mol) = 50.8 g 3.
How many moles of Al
2
O
3
can be produced from the reaction of 105. g of Al and 108. g of O
2
(give answer with 3 significant figures; 4 points)? 4 Al(s) + 3O
2
(g)
2 Al
2
O
3
(s) Al 105 g (1 mol / 26.98 g) = 3.89 mol / 4 = 0.973 O
2 108 g (1 mol / 32.00 g) = 3.38 mol / 3 = 1.125 Al is limiting reagent (3.89 mol Al)( 2 mol Al
2
O
3
/ 4 mol Al) = 1.95 mol 11 50.8 g 1.95 mol
4.
Nitric oxide is produced by reacting ammonia with oxygen gas. What is the percent yield if 28.7 g of NO (molar mass 30.01 g/mol) is isolated from the reaction of excess NH
3
with 45.2 g of O
2
? Report the answer with 3 significant figures (4 points). 4 NH
3
(g) + 5 O
2
(g)
4 NO(g) + 6 H
2
O(g) (45.2 g O
2
)(1 mol / 32 g)(4 mol NO / 5 mol O
2
)(30.01 g NO / 1 mol) = 33.9 g theoretical yield Percent yield = (28.7 / 33.9)(100 %) = 84.6% 5.
What is the concentration of the acid in a HI solution with a measured pH of 2.25 (4 points)? Report the answer with 3 significant figures. pH = -log (H
+
) HI is strong acid – completely dissociates antilog (-2.25) = H
+
= 5.62 10
-3
M 6.
What is the pH of a 7.58
10
-4
M Sr(OH)
2
solution (4 points)? Report the answer with 3 significant figures. Strong base 2 OH
-
per Sr(OH)
2
[OH
-
] = (7.58 10
-4
M)(2) = 1.516 10
-3
M pOH = -log[1.516 10
-3
] = 2.82 pH = 14 – pOH = 11.2 84.6% 5.62 10
-3
M 11.2
7.
According to the Arrhenius definition, an acid (4 points) a.
increases the H
3
O
+
concentration in an aqueous solution b.
increases the OH
-
concentration in an aqueous solution c.
is a strong electrolyte d.
is a proton acceptor e.
increases the pH of a solution f.
is a proton donor g.
is an electron-pair acceptor 8.
What is the net ionic equation for the reaction of aqueous calcium chloride, CaCl
2
, and aqueous lead nitrate, Pb(NO
3
)
2
(4 points)? CaCl
2
(aq) + Pb(NO
3
)
2
(aq)
Ca(NO
3
)
2
(aq) + PbCl
2
(s) Ca
2+
(aq) + 2 Cl
-
(aq) + Pb
2+
(aq) + 2 NO
3
-
(aq)
Ca
2+
(aq) + 2 NO
3
-
(aq) + PbCl
2
(s) 2 Cl
-
(aq) + Pb
2+
(aq)
PbCl
2
(s) 9.
What is/are the spectator ion(s) in the reaction between aqueous phosphoric acid (H
3
PO
4
) and aqueous potassium hydroxide (KOH) (4 points)? Phosphoric acid is a weak acid and will not produce spectator ions KOH is a strong base. K
+
will be a spectator ion. 10.
What is the oxidation number of a.
the chlorine atom in the compound HClO
4
(3 points)? H is +1, Cl is -2, Cl is +7 b.
the sulfur atom in the compound SO
3
2-
(3 points)? O is -2; charge is -2; S is +4 A K
+
2 Cl
-
(aq) + Pb
2+
(aq)
PbCl
2
(s) +7 +4
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Lets start with the medical components. We need to determine how many units of the syringe
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X M Mathway | Basic Math Problem S X +
1 metric ton = 2205 lbs
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total weight of starting material PEI
X
Products
Medical components
Stopcock with Luer
Connection, 1-way,
male lock
unit weight
2.72 g = 3/500 lbs
1 unit of medical component
1 weight of unit of component
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Medical components
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520 L x
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E Copy of Fall 2020 Fina X
hHp8TPKsYapxmOBQpq19sYHEQfKQudwLFoPO2ojmMGw/edit
d-ons Help
Last edit was seconds ago
BIUA
c 田回。三=== 三ニ三▼E
三E X
Arial
14.5
2
3
4
6
A student needs to identify an unknown substance found in the lab
based on its physical properties. The substance has a mass of 130.5 g
and a volume of 36.73 ml and tarnishes when it comes in contact with
water. (AKS 2b, DOK2).
Density
e/cm3)
0.42
0.95
0.36
3.71
0.30
1.12
0.20
0.69
Name
Cherry Alloy
Cuppric Alloy
Elastica Aloy
Galvanized Alloy
Holy Alloy
Marblic Alloy
Porous Alloy
Silica Dioxica Alloy
Stuper Alloy
Trisphereica Alloy
Vulcan Alloy
Zink Alloy
0.62
246
1.88
3.55
Question options:
A.The unknown substance is Cherry Alloy. The student was able to
use the reactivity to solve the problem-because tarnishing is a
physical property of the Cherry Alloy.
B. The unknown substance is Vulcan Alloy. The student was able to
use density to solve the problem because density is a physical
property of Vulcan Alloy.
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Show all your work on a sheet of paper to solve the following
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4) __ H2 +
O2 0 H20
HCI O
Pb(OH)2 +
AIBR3 +
Al2(SO4)3
5)
H20 +
PbCl2
6).
K2SO4 O
KBr +
7)
CH4 +
O2 0
CO2 +
H2O
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_ C3H8 +
02
O2 0
СО2 +
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CO2 +
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1-27. Noble gases (Group 18 in the periodic table) have the follow-
ing volume concentrations in dry air: He, 5.24 ppm; Ne, 18.2 ppm;
Ar, 0.934%; Kr, 1.14 ppm; Xe, 87 ppb.
(a) A concentration of 5.24 ppm He means 5.24 µL of He per liter
of air. Using the ideal gas law in Problem 1-16, find how many
moles of He are contained in 5.24 µL at 25.00°C (298.15 K) and
1.000 bar. This number is the molarity of He in the air.
(b) Find the molar concentrations of Ar, Kr, and Xe in air at 25°C
and 1 bar.
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How much boiling water would you need to raise the bath to body temperature (about 37 "C)? Assume that no
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