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
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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
c.
the nitrogen atom in the compound NH
3
(3 points)? H is +1; N is -3 11.
Write a reaction to show how the oxide of the nonmetal gas SO
3
can react with water to produce an acidic solution (4 points). SO
3
(g) + H
2
O(
l
)
H
2
SO
4
(aq) H
2
SO
4
(aq)
H
+
(aq) + HSO
4
-
(aq) HSO
4
-
(aq)
H
+
(aq) + SO
4
2-
(aq) 12.
A reducing agent (4 points) a.
donates electrons and becomes reduced b.
accepts electrons and becomes oxidized c.
donates electrons and becomes oxidized d.
accepts electrons and becomes reduced e.
none of the above 13.
You mix 52 mL of 0.95 M silver nitrate with 28 mL of 1.25 M sodium chloride. What mass of silver chloride (AgCl) would form (4 points)? AgNO
3
(aq) + NaCl(aq)
NaNO
3
(aq) + AgCl(s) AgNO
3
(52 mL)(1 L / 1000 mL)0.95 moles/L) = 0.0494 moles NaCl (28 mL)(1 L / 1000 mL)(1.25 moles/L) = 0.035 moles NaCl is limiting. (0.035 moles)(1 mol AgCl / 1 mol NaCl)(143.32 g/mol) = 5.02 g SO
3
(g) + H
2
O(
l
)
H
2
SO
4
(aq) C 5.02 g -3
14.
All of the following are oxidation-reduction reactions EXCEPT a.
2 Na(s) + Br
2
(g)
2 NaBr(g) b.
Fe(s) + 2 HCl(aq)
FeCl
2
(aq) + H
2
(g) c.
C(s) + O
2
(g)
2 CO(g) d.
2 H
2
O(
l
)
2 H
2
(g) + O
2
(g) e.
CaCO
3
(s)
CaO(s) + CO
2
(g) 15.
A chemist weighs out 1.102 g of an unknown diprotic acid into a 1.0 L volumetric flask and dilutes it to the mark with distilled water. 49.2 mL of a 0.200 M NaOH solution is required to reach the equivalence point. What is the molar mass of the acid? (4 points) H
2
A + 2 NaOH
Na
2
A + 2 H
2
O (49.2 mL)(1 L / 1000 mL)(0.2 moles NaOH / L)(1 mol acid / 2 mol base) = 0.00492 mol Molar mass = g / mole = 1.102 g / 0.00492 moles = 224 g/mol 16.
What mass of O
2
(g) would fill a 5.22 L container with a pressure of 1.30 atm and a temperature of 299 K (4 points; report answer with 3 significant figures)? PV = nRT solve for n and then convert from moles to grams N = PV / RT (1.30 atm)(5.22 L) / (0.08206 L atm/mol K)(299 K) = 0.277 moles (0.277 moles of O
2
)(32 g /mol ) = 8.85 g A balloon filled with O
2
has a volume of 18.2 L at room temperature (25
C) and sea level (1 atm). a.
What would happen to the volume of the balloon if the temperature decreased to 19
C? (write increase, decrease, or no change in the box; 3 points) Decrease b.
What would happen to the volume of the balloon if the pressure inside the balloon was somehow decreased but the temperature remained constant? (write increase, decrease, or no change in the box; 3 points) Increase 224 g /mol E 8.85 g
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17.
A mass of 0.645 g of an unknown gas is introduced into an evacuated 1.50 L flask. If the pressure in the flask is 0.764 atm at 369 K, which of the following gases might be in the flask (4 points)? a.
N
2
O b.
C
2
H
2
c.
O
2
d.
HCl e.
NH
3
Density = PM / RT M = (density)RT /P = (0.645 g/1.50 L)(0.08206 L atm mol K)(369 K) / (0.764 atm) M = 17.0 g/mol Consistent with NH
3
18.
Two gases, HCl (6.08 L) and O
2
(1.82 L), were combined at standard pressure and temperature conditions to synthesize chlorine gas. What is the theoretical yield of chlorine for this reaction (4 points; report answer in L with 3 significant figures)? 4 HCl(g) + O
2
(g)
2 Cl
2
(g) + 2 H
2
O(g) Limiting reagent problem Use Avogadro’s law Theoretical ratio = 4 vol HCl/ 1 vol O
2
= 4 Actual ratio = 6.08 L HCl / 1.82 L O
2
= 3.34 Not enough HCl; HCl is limiting (6.08 L HCl)(2 L Cl
2
/4 L HCl) = 3.04 L 19.
If 6.08 atm of HCl and 1.82 atm of O
2
(same reaction as problem 23) were combined in a flask at room temperature (298 K), what would be the final pressure after the synthesis of chlorine and water assuming the temperature and volume do not change? (4 points). a.
6.38 atm b.
6.08 atm c.
7.28 atm d.
7.90 atm e.
None of the above answers are correct HCl would be consumed (it is the limiting reagent) O
2
– consume 6.08/4 atm or 1.52 atm; O
2
remaining is 1.82 – 1.52 = 0.30 atm Cl
2
– make 3.04 atm H
2
O – make 3.04 atm Total pressure = 0.30 atm O
2
+ 3.04 atm Cl
2
+ 3.04 atm H
2
O = 6.38 atm E 3.04 L A
20.
What volume of oxygen will react with 42 mL of acetylene, assuming the gases are present at the same temperature and pressure (4 points)? C
2
H
2
(g) + 5/2 O
2
(g)
2 CO
2
(g) + H
2
O(g) a.
42 mL b.
84 mL c.
105 mL d.
16.8 mL e.
None of the above answers are correct (42 mL C
2
H
2
)(5/2 mol O
2
/ 1 mL C
2
H
2
) = 105 mL Answer the following questions for the gas molecules He, N
2
, Cl
2
, and F
2
a.
Which of the four gas molecules would have the largest density (3 points; if they are all the same write the answer “all the same”) Cl
2
b.
Which of the four gas molecules would have the largest average kinetic energy at room temperature (3 points; if they are all the same write the answer “all the same”) All the same c.
Which of the four gas molecules would effuse at the fastest rate? (3 points; if they are all the same write the answer “all the same”) He C
Equations and constants that might be useful Avogadro’s number = N
A
= 6.022
10
23
mol
-1
c =
K =
C + 273.15 E = h
J = 1 kg m
2
s
-2 E= mc
2
R = 8.314 J/K mol E
n
= - Rhc / n
2
R = 0.08206 L atm/mol K
E = -Rhc (1/n
2
final
– 1/n
2
initial
) 1 atm = 760 mm Hg
E = -N
A
Rhc (1/n
2
final
– 1/n
2
initial
) P
1
V
1
= P
2
V
2 V
1
/T
1
= V
2
/T
2
= h / mv (de Broglie equation) P
1
V
1
/T
1
= P
2
V
2
/T
2 c = 2.998
10
8
m/s V
n (at constant T and P) Plank’s constant = h = 6.626
10
-34
J s PV = nRT d = m/v = PM/RT R (Rydberg constant) = 1.097
10
7
m
-1
X
A
= n
A / n
total
pH = -log [H
+
] = -log [H
3
O
+
] P
A
= X
A
P
total pOH = -log [OH
-
] pH + pOH = 14 (at room temp)
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- History stry Bookmarks X O8 https://app.101edu.co FEB 13 Tools Window 80 Aktiv Chemistry F3 Balance the following chemical equation (if necessary): Si₂H3(s) + O₂(g) → SiO₂(g) + H₂O(g) 1 $ + Reset Help SIO₂ O F4 X 04 03- 3 0₂ 03 ) + F5 Question 40 of 47 0²- 4 0 ↑ H₂O < 0- 5 05 □+ ALL F6 6 0 1 MacBook Air 2 2+ & 7 Si₂H3 07 (s) (1) (g) (aq) 4 F7 8 9 0 08 09 Do • x H₂O 4+ ♫ A * 0₂ FB Delete tv w O O DD F9arrow_forward← X g 單 0 Priva Tern C My Account - MySam - Sam Hou X Content X Q mathway calculator - Search https://shsu.blackboard.com/ultra/organizations/_202946_1/cl/outline LI Lets start with the medical components. We need to determine how many units of the syringe connections can be made out of one metric ton of PEI, or around 2205 lbs of the plastic. X M Mathway | Basic Math Problem S X + 1 metric ton = 2205 lbs Starting material - PEI 1 metric ton ≈ 2205 lbs Calculation for number of units of each component: 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 This calculation assumes all starting material is used during the production of the components. Medical components Products Stopcock with Luer Connection, 1-way, male lock unit weight 2.72 g 3/500 lbs Female Luer Lock 3/16" unit weight 0.63 g 7/5000 lbs Units = If necessary round to the…arrow_forwardRefer to the pictures:arrow_forward
- Chemical & Physical Properties Reviewarrow_forwardView History Bookmarks Develop Window Help A east.cengagenow.com [References) 1 pt EXERCISE Unit Conversion 1 pt Drag conversion units onto the boxes in the equation to make conversions. Some boxes can be left empty. Click on a unit to remove it from its position. 1 pt 1 cm³ Cu 9 g Cu 9.5 x 1021 atoms Cu 1 g Cu 1 pt 1 kg 1000 g 1 cm = 1 mL 1 pt 1 L 1000 cm3 1 pt 1 pt 520 L x 1 pt 1 pt 1 pt A piece of copper has a volume 520 L. What is the mass of the sample, in units of grams? In the boxes above, enter the correct setup that would be used to solve this problem. 1 pt 1 pt Check Next (2 of 3) 1 pt 1 pt Submit Answer Try Another Version 2 item attempts remaining 1 ptarrow_forwardWhich two options describe behaviors of particles that are related to the chemical properties of the materials?arrow_forward
- Question 20arrow_forwardE 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. C.The unknown…arrow_forwardQuestion 19arrow_forward
- please help mearrow_forwardmaining Time: 21 minutes, 27 seconds. estion Completion Status: aLY OAC'K. QUESTION 25 Show all your work on a sheet of paper to solve the following significant figures. You can earn partial credit for this problem What is the mass in grams of 4.75 x 1044atoms of Ne? PE 1. IA 2 1.0079 IIA 4 Be Li 2.arrow_forwardQuestion 21arrow_forward
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