Chemistry I Final Exam Review Key
pdf
keyboard_arrow_up
School
University of Illinois, Urbana Champaign *
*We aren’t endorsed by this school
Course
102
Subject
Chemistry
Date
Jan 9, 2024
Type
Pages
39
Uploaded by Human8
Chemistry I Final Exam Review John M. Van Wazer Dr. Tina Huang Merit FQF December 7, 2023 —1— Contents Exam I 2 i.
Significant Figures
ii.
Dimensional Analysis 3 iii.
Isotopes 4 iv.
Light Energy 5 Exam II 6 i.
Quantum Numbers
ii.
Polarity & Atomic Radii 7 iii.
Electron Configuration 9 iv.
Compound Nomenclature 10 v.
Lewis Structure & VSEPR Theory 11 Exam III 13 i.
Hybridization
ii.
Stoichiometry 17 iii.
Electrolytes 19 iv.
Intermolecular Forces 21 v.
Dilutions 22 Exam IV 23 i.
Solubility & Acid-Base Reactions
ii.
Gases, Gas Kinetics, & Diffusion & Effusion 24 iii.
Thermodynamics 27 iv.
Hess’s Law 30 Exam V 31 i.
Heating Curve
ii.
Equilibrium 33 iii.
Heat of Formation 35 iv.
Bond Energies 36 Challenge Questions 37
Chemistry I Final Exam Review —2— Exam I Significant Figures, Dimensional Analysis, Isotopes, & Light Energy
Significant Figures Rules: Addition & subtraction:
Your significant answer will contain the same number of decimal places as the value with the least number of decimal places from your problem. Multiplication & division:
Your significant answer will contain the same number of sig. figs. as the value with the least number of sig. figs. from your problem. These rules still apply to computations where PEMDAS is necessary, however, you want to keep all integers within your computation, then go through the problem again, and apply the sig. fig. rules to determine the correct number of sig. figs. in your final answer.
1.
Compute the problems below. If x
is a variable in the equation, solve for it. a.
((23.56 + 42.001)(2.34))/1.965 = b.
(5423
x
/100.)(12) + 4 = 56 c.
(11.2 – 4.06)(5.632)(0.200)(20.)(20) = d.
x
/40 = 233
Chemistry I Final Exam Review —3— Dimensional Analysis Conversion factors have an infinite number of significant figures; only numbers from measurements will carry the sig. figs. 1.
Convert 6.0 years into nanoseconds (ns). {1 month = 28 days}. Remember to carry your sig. figs. and round at the end of your computation. 2.
A car has two axles. Each axle has two wheels, one on either end. Hypothetically, how many cars can be built if there are only 67 wheels available? 3.
Assume a car gets 32.0 miles per gallon when burning regular gasoline at a cost of $3.65 per gallon. A novel fuel known as 2-methylhexane (2Mh), has been synthesized that allows this same car to get 38.2 miles per gallon. The cost of 2Mh is $3.82 per gallon. How much will it cost to run this car for 300. miles? 4.
If the density (
!
) of 24 karat gold is 19.32 kilograms per liter at 23°C. What is the volume in inches of a 21.2 milligram block of gold?
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —4— Isotopes Symbol: "
!
"
{A = mass number = protons + neutrons, Z = atomic number = protons} 1.
For the species below, write their atomic symbol (
"
!
"
) given the number of protons, neutrons, and electrons (P, N, e
-
) a.
(P, N, e
-
) = (16, 18, 18) b.
(P, N, e
-
) = (1,0,1) c.
(P, N, e
-
) = (1,1,1) d.
(P, N, e
-
) = (1,2,1) e.
(P, N, e
-
) = (22, 26, 22) f.
(P, N, e
-
) = (19, 21, 18) 2.
From the above species, which three are isotopes? 3.
Write the atomic symbol for a fluoride ion.
Chemistry I Final Exam Review —5— Light Energy N
A
= 6.022E23 c = 2.998E8 ms
-1
h = 6.626E-34 Js E = h
ν
c = λν
E = h(c
λ
-1
) dE = -2.178E-18 J ((n
2
f
)
-1
– (n
2
i
)
-1
) λ
= h(mv)
-1
E = 0.5mv
2 1.
What is the frequency (
ν
) in inverse seconds (s
-1
) of each option below: a.
Blue light b.
Light radiation with λ
= 8242 Å
c.
88.9 MHz d.
550 nm radiation 2.
What is the change in energy for an electron transitioning from n = 6 to n = 3? a.
What color of light would this produce? 3.
What color of light is observed by a proton traveling at a velocity of 0.566 ms
-1
? 4.
What is the velocity of an electron with an energy of 9.576E-19 J?
Chemistry I Final Exam Review —6— Exam II Quantum Numbers, Polarity & Atomic Radii, Electron Configuration, Compound Nomenclature, Lewis Structure, & VSEPR Theory
Quantum Numbers Pattern n 1 2 3 l 0
s
0
s
1
p 0
s
1
p
2
d
m
l
0 0, -1.0.1 0, -1.0.1, -2.-1.0.1.2 m
s ↑↓
↑↓
,
↑↓
.
↑↓
.
↑↓
↑↓
,
↑↓
.
↑↓
.
↑↓
,
↑↓
.
↑↓
.
↑↓
.
↑↓
.
↑↓
n
= principal number, describes size/energy l = angular momentum number; codes for the subshell/shape (s, p, d, or f) = (0, 1, 2, or 3) m
l
= magnetic number; orbital/orientation, two e
-
per orbital: Pauli’s Exclusion Principal m
s
= magnetic spin number, denoted as (
↑
) for one e
-
with spin +0.5, another e
-
(
↓
) in the same orbital (m
l
) has spin -0.5
1.
If n = 2, what is the maximum number of e
-
available? a.
How many total orbitals/orientations are there? b.
When l = 1, how many e
-
are spinning at -0.5? 2.
If n = 4,
what is the largest l
value? a.
What subshell does this largest value denote? b.
How many +2 m
l
values are there? c.
How many |1| m
l
values are there? 3.
If l
= 1, what is the lowest n
value? a.
What is the highest n
value for Xenon? b.
How many e
-
does carbon have in its p
subshell? 4.
If n = 3 and l = 2
: a.
How many orientations are there? b.
How many +0.5 electrons are there?
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —7— Polarity, Ionization Energy, Electronegativity, & Atomic Radii Ionization energy is the energy required to remove one electron; fifth ionization energy = energy to remove the fifth highest electron. Z
eff
= Z – C
e-
Z
eff
= (protons – core e
-
) Z
eff
∝
radii
-1
Core Electrons: Electrons that the atom has with the previous noble gas and any completed transition series. P
[Ne]3s
2
3p
3
10 core e
-
Mn
[Ar]4s
2
3d
5
18 core e
-
Te
[Kr]5s
2
4d
10
5p
4
46 core e
-
Valence Electrons: Electrons that occupy the highest n energy level, for transition elements, they include any (n – 1)d electrons as well. P
[Ne]3s
2
3p
3
5 ve
-
Mn
[Ar]4s
2
3d
5
7 ve
-
Te
[Kr]5s
2
4d
10
5p
4
6 ve
-
Electronegativity table:
Chemistry I Final Exam Review —8— 1.
Which ionization energy for Barium would be the greatest? 2.
What is the Δ
EN for an O—H bond? a.
Draw the structure for dihydrogen monoxide and state the net polar vector. 3.
What is the Δ
EN for an N—H bond? a.
Draw the structure of ammonia and state the net polar vector. 4.
What is the Δ
EN for a F—C bond? 5.
Which atomic radii is larger, C or Na? 6.
Which atomic radii is larger, Cl or Ca? 7.
Which atomic radii is larger, Cl
-
or Ca
2+
? 8.
The anionic form of Chlorine and the cationic form of Magnesium are isoelectric. T or F 9.
K is larger than K
+
. T or F 10.
F is larger than F
-
. T or F 11.
What is the Z
eff
for I
-
? 12.
I
-
is larger than Cs
+
. T or F 13.
Which element is Ca
2+
isoelectric with?
Chemistry I Final Exam Review —9— Electron Configuration 1.
Write the full electron config. for Gallium a.
Write the condensed electron config. b.
How many ve
-
are available? c.
How many core e
-
are there? 2.
Write the full electron config. for Copper and Chromium a.
Write their condensed configs. b.
How many ve
-
are available for each? c.
How many core e
-
are there for each? 3.
Write the full electron config, for Indium a.
Write the condensed config. b.
What is the highest principal quantum number? c.
How many e
-
are in l = 2
?
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —10— Compound Nomenclature Magnesium Sulfite MgSO
4
Potassium hydrogen sulfate Calcium hydrogen phosphate LiH
2
PO
4 Calcium carbonate Aluminum hydrogen carbonate Nickle oxalate Cobalt(II) chlorate CsOCl NaClO
2
Sodium thiosulfate AgCN Ti
3
(PO
4
)
2 Titanium(IV) sulfate SnCl
4 Gallium phosphate Zinc acetate AgPF
6
This one is tricky
!
Lead(IV) sulfide Arsenic mononitride Rb
2
O HgCl
2 Dihydrogen monoxide Zinc bromide Platinum(II) fluoride Pt(CH
3
COO)
2
NaCl Manganese(II) phosphate Co(CN)
2 Ammonium oxide Potassium perchlorate FePO
4 MgCr
2
O
7 Sodium chromate Carbon dioxide H
2
O
2 KF SnO
2 Lead(II) nitrate Ca(SCN)
2 Gold(III) chloride Potassium permanganate NaNO
2
Chemistry I Final Exam Review —11— Lewis Structure, Resonance, & VSEPR Geometry = electron geometry—what the shape is if you include all lone pairs and bonds. Shape = molecular geometry—what the shape is if you exclude lone pairs and only look at bonds Electron groups: The bonds and lone pairs associated with a specific atom. Single, double, and triple bonds
are one electron group
, lone pairs
are one as well.
Chemistry I Final Exam Review —12— 1.
Draw the Lewis structure, then give the geometry, shape, and polarity (polar (P) or nonpolar (NP)) for each of the following species. If the compound is charged, write the resonance structures for it as well: Glycine is a tricky one—it is a general Lewis structure for a non-metallic (organic) compound—it will come up soon in lecture (not necessary for exam II)! AsBr
3
PI
3
Angles: N
2
gas Angles: SiCl
4
Angles: HNO Angles: AsN FCN Angles: H
2
NCH
2
COOH (glycine) IBr
2
+
ClO
2
-
Angles: SO
4
2-
Angles: NO
3
-
Angles:
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —13— Exam III Hybridization, Stoichiometry, Electrolytes, Intermolecular Forces, & Dilutions
Hybridization, Sigma (
σ
) & Pi (
π
) Bonds, & Lone Pairs Sigma bond (
σ
-bond)
: One hybridized bond between two atoms. Pi bond (
π
-bond)
: A bond between two atoms with at least one unhybridized p-orbital; a double bond has one pi-bond, a triple bond has two pi-bonds. 1.
Adenine is one of four main nucleotides present in DNA, add all π
-bonds and lone pairs to complete its structure.
“R” is how organic chemists refer to the “Rest of the molecule”; where “R”, in this case, is the ribose sugar and phosphate backbone, giving DNA the extended name of deoxyribose nucleic acid. Line Structure Lewis Structure a.
How many π
-bonds and lone pairs are in adenine? b.
What is the hybridization of each nitrogen atom (Note: two amines are different from all the rest)? Write your answer directly on each respective atom within either structure. c.
How many electron groups are in adenine, include the N—R bond. N
N
N
N
NH
2
R
C
C
N
C
N
C
N
C
N
N
R
=
H
H
H
H
Chemistry I Final Exam Review —14— C
C
C
C
C
C
H
H
H
H
H
H
=
H
2
N
O
OH
OH
=
H
H
H
2.
Benzene is an aromatic compound and has three pi-bonds. These bonds can interact with a base pair such as adenine to intercalate with DNA, causing errors in replication and potentially mutations that confer cancer. Draw all π
-bonds for both structures below: Line Structure Lewis Structure 3.
Serine is a nonessential common amino acid found in the human body and can be obtained from the diet. An incomplete line structure
for Serine is given below. Draw the Lewis structure with all π
-bonds and lone pairs for Serine. Line Structure Lewis Structure a.
How many π
-bonds, σ
-bonds, and lone pairs are in serine? b.
What is the hybridization of each atom?
Chemistry I Final Exam Review —15— 4.
Cyanoacrylate is the main compound in super glue. The incomplete
line and Lewis structures for Cyanoacrylate are given below. Add the π
-bonds and lone pairs to both structures. Line Structure Lewis Structure a.
How many π
- and σ
-bonds, and lone pairs are in Cyanoacrylate? b.
What is the hybridization for the O
a
atom. Draw its unhybridized, and hybridized electron diagrams below: O
a c.
What is the hybridization for the O
b
atom. Draw its unhybridized, and hybridized electron diagrams below: O
b C
e
O
b
N
d
C
c
H
2
O
a
C
C
N
d
C
c
H
H
C
e
O
b
O
a
C
C
H
H
H
H
H
=
s
0 s
p
-1 p
0 p
+1 p
relative energy (e)
hybridization/bonding relative energy (e)
s
0 s
p
-1 p
0 p
+1 p
relative energy (e)
hybridization/bonding relative energy (e)
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —16— d.
What is the hybridization for the C
c
atom. Draw its unhybridized, and hybridized electron diagrams below: C
c
e.
What is the hybridization for the N
d
atom. Draw its unhybridized, and hybridized electron diagrams below: N
d f.
What is the hybridization for the C
e
atom. Draw its unhybridized, and hybridized electron diagrams below: C
e
s
0 s
p
-1 p
0 p
+1 p
relative energy (e)
hybridization/bonding relative energy (e)
s
0 s
p
-1 p
0 p
+1 p
relative energy (e)
hybridization/bonding relative energy (e)
s
0 s
p
-1 p
0 p
+1 p
relative energy (e)
hybridization/bonding relative energy (e)
Chemistry I Final Exam Review —17— Stoichiometry, Limiting Reactant, Empirical & Molecular Formula, Percent Mass, Molarity, & Balancing Combustion Reaction 1.
An isothermal titration calorimeter (ITC) is an instrument used to measure the heat that is released when two species are mixed. A standard bomb calorimeter is used to measure the heat released for combustion reactions under constant volume (dV = zero). 700 milligrams (mg) of the organic compound 9-hydroxyanthracene are combusted with two grams of pure oxygen gas and produces gaseous water and carbon dioxide. Complete and balance the combustion reaction below
: a.
What is the limiting reactant? b.
How many grams (g) of water and carbon dioxide can be theoretically produced? c.
If only 297 mg of water was produced, what is the percent yield? d.
If you had 100 g of 9-hydroxyanthracene, what is the percent mass for the following elements: Carbon: Hydrogen: Oxygen: C
14
H
10
O
OH
+
Δ
spark
(s)
+
Chemistry I Final Exam Review —18— 2.
Reactant X contains some amount of carbon, hydrogen, oxygen, and nitrogen. Analysis with 3.50 g of X suggests that it adequately reacted with oxygen gas and produced 1.68 g of carbon dioxide and 0.98 g of water. Later, it was determined that the percent nitrogen was 18%. What is the empirical formula for reactant X? 3.
How many micromoles (
μ
M = μ
mol L
-1
) for 72 micrograms (
μ
g) of ethanol (CH
3
CH
2
OH) in 4 milliliters (mL) of water. 4.
What is the molarity of the Cl
-
ion when you add 50. mL of 0.50 M NaCl to 100. mL of 1.50 M KCl?
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —19— Electrolytes, Balancing Ionic Equations, Salts & Solubility Strong electrolyte: Strong acids and bases
, and salts (especially group IA cations
) Weak electrolytes: Weak acids and bases Non-electrolytes: Organic compounds 1.
Tetrakis(acetonitrile)copper(I) hexafluorophosphate or Cu(CH
3
CN)
4
PF
6
dissociates in the reaction below: a.
How many moles of PF
6
-
ions would be in a 300. mL solution of 0.25 M Cu(CH
3
CN)
4
PF
6
? 2.
Copper(I) sulfate dissociate as shown in the dissociation reaction below: a.
How many moles of Cu
+
ions would be in a 500. mL solution of 0.50 M Cu
2
SO
4
? 3.
Complete and balance the following precipitation reactions. Write the ionic equation, circle the spectator ions, and include the net ionic equation. a.
b.
Cu(MeCN)
4
PF
6
(s)
Cu(MeCN)
4
+
(aq) + PF
6
-
(aq)
Cu
2
SO
4
(s)
2 Cu
+
(aq) + SO
4
-
(aq)
Na
2
S
(aq)
+ NiSO
4(aq)
Al(NO
3
)
3(aq)
+ Na
3
PO
4(aq)
Chemistry I Final Exam Review —20— 4.
Determine if the following species are strong (S), weak (W), or non (N)-electrolytes: a.
NaCl b.
KOH c.
Acetic acid = vinegar (CH
3
COOH) d.
Acetone (CH
3
COCH
3
) e.
Benzoic acid (C
6
H
5
COOH) = f.
HF g.
HCl h.
H
2
SO
4
i.
Glucose (C
6
H
12
O
6
) = sugar =
j.
HNO
2
k.
KI l.
NaHCO
3
m.
HBr n.
K
3
PO
4
o.
Benzene (C
6
H
6
) O
OH
O
OH
HO
HO
HO
OH
Chemistry I Final Exam Review —21— Intermolecular Forces & Boiling Point Dispersion: Occurs with any polyatomic nonpolar compound
; monoatomic species do not have dispersion. *Note: induced dipole can occur where electron density is not equally distributed throughout the compound, resulting in
#
+
and
#
-
poles.
ie., diatomic oxygen gas (O
2
): Δ
EN = zero
, implying nonpolar Induced dipole in oxygen gas, creating poles of different electron density. Dipole-dipole: Occurs only with polar compounds
; if it is polar, it has dipole-dipole. ie., water (H
2
O): Hydrogen-bonding: Occurs between two molecules
that each have an H— (O, N, F)
. ie., water and ammonia (NH
3
): O
O
O
O
H
O
H
δ
+
δ
-
H
O
H
1
2
2
1
3
3
H
O
H
H
O
H
H
O
H
H
O
H
H
O
H
H
N
H
H
H
N
H
H
H
N
H
H
H
N
H
H
Net polar vector, implying polar molecule.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —22— 1.
Label the following compounds from lowest (#7) to highest (#1) boiling point: a.
HF b.
CH
3
CH
2
CH
2
CH
2
CH
3
c.
H
2
O
2
d.
SeF
6
e.
CH
3
OH f.
CH
3
I g.
CH
3
CH
2
OH Dilutions 1.
What molarity for a stock solution of benzene dissolved in 7 mL of tetrahydrofuran (THF) is required to make a 2000. μ
L of 0.50 molar aliquot? a.
How many mL of THF are required for a 10:1 dilution of that 0.50 M aliquot?
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —23— Exam IV Solubility & Acid-Base Reactions, Gases & Gas Kinetics, Diffusion & Effusion, Thermodynamics, & Hess’s Law Solubility Rules 1.
Most nitrate, acetate, chlorate, & bicarbonate (hydrogen carbonate) salts are soluble. 2.
Most salts of alkali metals and ammonium cations are soluble. 3.
Most chloride, bromide and iodide salts are soluble. Exceptions: salts containing Ag
+
, Pb
2+
and Hg
2
2+
ions are insoluble 4.
Most sulfate salts are soluble. Exceptions: sulfates containing Ca
2+
, Sr
2+
, Ba
2+
, Pb
2+
,Hg
2
2+
and Ag
+
ions are insoluble 5.
Most hydroxide salts are insoluble. Exceptions: hydroxides containing alkali metals, ammonium ions, Ba
2+
, Sr
2+
, and Ca
2+
ions are soluble 6.
Most sulfide, carbonate, chromate and phosphate salts are insoluble. Exceptions: salts of alkali metals and ammonium cations are soluble Solubility, Precipitation, & Acid-Base Chemistry 1.
Circle the compounds that are soluble in water: ZnCl
2
AgBr AlPO
4
Ba(OH)
2
BaCO
3
LiCl RbNO
3
NH
4
CH
3
COO 2.
Write the formula equation, ionic, and net ionic equations for the following reactions: a.
(NH
4
)
2
SO
4(aq)
+ BaCl
2(aq)
→
b.
HI
(aq)
+ NaOH
(aq)
→
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —24— Gas Laws, Arithmetic Relationships, Gas Kinetics, & Diffusion/Effusion
Ideal Gas Law: %& = ()*
%
#
&
#
= ()*
%
$
&
$
= ()*
⇒ %
#
&
#
= %
$
&
$
) = 0.08206 12345
%#
467
%#
) = 8.314 ;467
%#
5
%#
Boyle’s Law: %
#
&
#
= %
$
&
$
∴ % ∝
#
&
Charle’s Law: '
!
&
!
=
'
"
&
"
Avogadro’s Law: '
!
(
!
=
'
"
(
"
Dalton’s Law of Partial Pressure: Σ% = %
#
+ %
$
+ %
)
+
…
+ %
+
= (
#
)*
&
+ (
$
)*
&
+ (
)
)*
&
+
…
+ (
+
)*
&
Under constant T & V, % ∝ ( ⇒ 467@ AB2C3D6( (F) =
(
#
(
$%$&'
=
,
#
,
$%$&'
∵ % ∝ (
Total atmospheric pressure (%
-.-/0
) = %
1/2
+ %
3/-45(1)
⇒ %
1/2
= %
-.-/0
− %
3/-45(1)
Kinetic Molecular Theory: 1.
volume of particles is negligible 2.
constant motion 3.
no particle-particle IMFs 4.
5
/84
= *
∴ 5
/84
=
)
$
)*; Kℎ@B@ (M4@B236B 6A 3 ⇒ ℝ
)
(3-space) O = P
592
= Q
):&
9
∵ 5
/84
=
)
$
)* =
#
$
4O
$
⇒ O
$
=
):&
9
Diffusion: Transport of one species through a permeable membrane; driven by concentration gradient (high to low) Effusion: Transport of one species through a pin hole; driven by vacuum (high to none) B23@ 6A Cℎ2(R@ D( S6T
#
B23@ 6A Cℎ2(R@ D( S6T
$
=
UDT32(C@
3
#
UDT32(C@
3
$
=
V
3
#
V
3
$
= W
4
$
4
#
=
3
$
3
#
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —25— 1.
What is the velocity of xenon gas at 72°C? 2.
What is the velocity of acetylene (C
2
H
2
) at 48°C? 3.
Hydrogen gas traveled through a pin hole over 26 seconds. How long will it take for argon gas to travel through the same pin hole? 4.
You wish to grill some yummy hotdogs on a nice and toasty 32°C summer day! What is the pressure of 2 kg of propane (CH
3
CH
2
CH
3
) in a 20 liter propane tank? 5.
Consider three different gases: He, Br
2
, and H
2
. a.
Order the gases by increasing mass. b.
Order the gases by increasing velocity. c.
Order the gases by increasing average kinetic energy. 6.
The pressure of argon gas is 2 atm at 40°C and in 5 liters. What will the pressure be if the volume increased to 8.6 L.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —26— 7.
Consider the reaction below: Inside of an enclosed cylindrical chamber filled with pure oxygen gas, octanitrocubane (oNC, C
8
N
8
O
16
) receives enough activation energy from an ignition spark that it combusts into carbon monoxide, nitrogen gas, and oxygen gas. a.
How many moles of gas are produced via this combustion reaction if 300. g of oNC are used? b.
What is the final pressure of the 250 mL reaction system after this amount of gas moles are produced at 68°C? c.
Was the reaction chamber able to contain the combustion of oNC if it had a structural integrity to withstand 1000 atm of internal pressure? NO
2
O
2
N
O
2
N
NO
2
NO
2
NO
2
O
2
N
O
2
N
O
2(g)
∆
spark
8 CO
(g)
+
+
4 N
2(g)
C
8
N
8
O
16
MW = 464.13
5 O
2(g)
+
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —27— Thermodynamics Internal Energy: XY = XZ = [ + \ = [ + %U&
Heat & Work Signs: Heat released by system on surroundings = -q Heat gained by system from surroundings = +q Work done by system on surroundings = -W = -PdV for expansion Work done on system by surroundings = +W = +PdV for compression Enthalpic Change: X]
5;(
=
<
()*
9.0
Calorimetry: Coffee-cup Bomb C6(T32(3 % ⇒ U% = ^@B6
C6(T32(3 % ⇒ [
=
= U]
[
5;(
= 4CX*
[
2>2
= −[
2?55
[
94-/0
= −4
2.0?-@.(
C
2.0?-@.(
X*
C6(T32(3 & ⇒ U& = ^@B6
[
5;(
= _
A/0.5@94-45
X* = _X*
[
2>2
= −[
2?55
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —28— 8.
Consider an engine piston transitioning from state one to state two from the combustion of benzene and straight-chain alkanes (hydrocarbon chains) within the combustion chamber: State 1 →
State 2 Combustion of such fuel caused the piston to travel down its combustion chamber, effectively rotating the crankshaft and driving the rotation of a car’s wheels. For this instance, please consider only the piston and the combustion chamber as your system, and the rotation of the crankshaft as your surroundings. Over this transition, the temperature of chamber increased from 23°C to 111°C, and the pressure remained at 1.8 atm. a.
What ideal gas parameters (P, V, n, R, T) are constant and nonconstant for this system? b.
For the internal energy formula, ∆Z = [ + \
, what are the signs of heat and work? c.
Is this reaction exothermic or endothermic? +
+
14 O
2(g)
∆
spark
10 CO
2(g)
8 H
2
O
(g)
+
+
E
(q+w)
(s)
(l)
C
6
H
6
MW = 78.11
C
4
H
10
MW = 58.12
∆
H = -600 kJmol
-1
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —29— d.
What is the value of heat of reaction in joules (J) if 5 L of butane is combusted. The density of butane (!
B?-/(4
) = 0.580 R41
%#
. e.
What is the specific heat capacity of butane (
C
B?-/(4
)? 9.
You place dry ice into a balloon and tie it closed at room temperature. a.
What happens to the volume of the balloon? 10.
You fill a balloon with helium, tie it closed, and let it go into the atmosphere; never to be seen again. b.
What will happen to the balloon as it continues to rise; does is pop or continue travelling upwards into the outer rim of our atmosphere and eventually space? c.
Does the volume of the balloon increase or decrease? d.
What causes the volume to change? e.
What is the sign of work? 11.
A closed can of Coke at 1.15 atm is submerged to the bottom of the ocean. The external pressure from the ocean being exerted on the can is 1200 atm. a.
What happens to the can at the bottom of the ocean? b.
Does the volume of the can change? c.
What is the sign of dV? d.
What is the sign of work?
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —30— Hess’s Law
Net enthalpic change (dH
net
) = ∑
dH 1.
What is the total enthalpic change (
dH
net
) for the net reaction below: 2.
What is the enthalpic change for the first reaction when given dH
net
? C
4
H
8(g) + 6 O
2(g)
∆
spark
4 CO
2(g) + 4 H
2
O
(l) : ࠵?
H
net
= x
2 H
2(g) + O
2(g)
2 H
2
O
(g)
: ࠵?
H = -600 kJ
C
4
H
8(g) + H
2(g)
2 C
4
H
10(g) + 13 O
2(g)
8 CO
2(g) + 10 H
2
O
(l)
: ࠵?
H = -4040 kJ
C
4
H
10(g) : ࠵?
H = -140 kJ
NO
2(g) + N
2
O
(g)
3 NO
(g) : ࠵?
H
net
= 155.5 kJ
2 N
2(g) + O
2(g)
2 N
2
O
(g)
: ࠵?
H = x
kJ
N
2(g) + O
2(g)
N
2(g) + 2 O
2(g)
2 NO
2(g)
: ࠵?
H = 66.4 kJ
2 NO
(g) : ࠵?
H = 180.5 kJ
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —31— Exam V Heating Curve, Equilibrium, Bond Energies Heating Curve States of Matter Conversions: 1.
Answer the questions regarding the heating curve of Gold below: Note: Δ]
C?2
= 13.2
DE
9.0
, Δ]
8/=
= 310.9
DE
9.0
, C
2.0@F
=
G.#$IE
1℃
, C
0@<?@F
=
G.G)KKE
1℃
, C
1/2
=
G.#)LE
1℃
a.
At what temperature does gold undergo fusion? b.
At what temperature does gold undergo vaporization? c.
What temperature range does gold exist as a solid? Gas
Liquid
Solid
Condensation
Vaporization
Freezing
Fusion
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
Temperature (°C)
Heat Added
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —32— d.
What temperature range does gold exist as a liquid? e.
At what temperature does gold exist as a gas? f.
What is the energy for 2 moles of gold as a solid in kJ? g.
What is the energy for 2 moles of gold associated as a liquid in kJ? h.
What is the energy for 2 moles of gold associated as a gas in kJ? i.
What is the energy for 2 moles of gold during the transition from solid to liquid in kJ? j.
What is the energy for 2 moles of gold during the transition from liquid to gas in kJ? k.
What is the total energy in kJ required to convert 2 moles of solid gold at 0°C to gaseous gold at 8000°C? l.
What is the sign for the total energy and why? 2.
Draw the heating curve of water with temperature vs. heat added. a.
What is the energy required to heat 5 L of liquid water? Assume dT is all temperatures where water exists as a liquid.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —33— Equilibrium, Le Chatelier, Equilibrium Constant & Reaction Quotient, & ICE Tables Only applies to reactions involving gases K > Q favors products K < Q favors reactants K = Q equilibrium, no shift K values are multiplicative K
p
= K(RT)
dn
{dn = change in gas moles = (n
f gas
– n
i gas
)}
5% Rule: K < 1E-4 Favors reactants Very little product is formed relative to the starting concentration [Reactant] is relatively large, loss of product is close to zero, x becomes negligible Which direction does the reaction shift according to the changes in the reaction: 1.
2SO
3(g)
⇌
2SO
2(g)
+ O
2(g)
a.
Add SO
3
b.
Add O
2
c.
Remove SO
2
d.
Add Argon gas e.
Decrease the volume of the system 2.
2Ag
(s)
+ 4H
+
(aq)
+ O
2(g)
⇌
2Ag
+
(aq)
+ 2H
2
O
(l)
∆
H = +40 kJ a.
Add silver ions b.
Remove liquid water c.
Increase the temperature of the system d.
Write the expression of K
eq
for this reaction at equilibrium e.
What is the numerical value for K
eq
if the concentrations of each species is 2 M at equilibrium?
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —34— f.
If the concentrations for H
+
(aq)
, O
2(g)
, and Ag
+
(aq)
are 1, 0.250, and 0.5, respectively, does the reaction favor products or reactants? 3.
Consider the equilibrium expressions below: HF
(aq)
⇌
H
+
(aq)
+ F
-
(aq)
K
c1
= 0.0055 H
2
C
2
O
4(aq)
⇌
2H
+
(aq)
+ C
2
O
4
2-
(aq) K
c2
= 0.00028 a.
Determine the net value of K
c
for the net reaction below: 2HF
(aq)
+ C
2
O
4
2-
(aq)
⇌
2F
-
(aq)
+ H
2
C
2
O
4(aq)
Π
K
c
= ? 4.
Consider the reactions below at 400°C: a.
CO
(g)
+ H
2
O
(g)
⇌
CO
2(g)
+ H
2(g)
K
c
= 0.04 If the initial concentrations for the reactants are 1.500 M, what are the equilibrium concentrations? b.
H
2(g)
+ F
2(g)
⇌
2HF
(g)
K
c
= 0.43 If the initial molar concentrations for nitrogen gas, oxygen gas, and nitrous oxide are 2.00, 4.00, and 6.00 M, respectively, what are the equilibrium concentrations?
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —35— 5.
Consider the combustion of methanol (MeOH = CH
3
OH) below: 2MeOH
(l)
+ 3O
2(g)
⇌
2CO
2(g)
+ 4H
2
O
(g)
Rxn P
O2 P
CO2 P
H2O I 0.02 0.04 0.14 II 0.01 0.03 x What is the partial pressure of water vapor for rxn II? Compute the equilibrium constant. 6.
Determine the equilibrium concentrations for the reaction below if you add 654.6 g of NOCl to 4000 mL reaction flask: 2NOCl
(g)
⇌
2NO
(g)
+ Cl
2(g)
K
c
= 1E-6 Heat of Formation Reaction enthalpy: dH
rxn
= ndH
f, products
– ndH
f, reactants
1.
One mole of glycylglycine can be hydrolyzed into two moles of glycine. The enthalpy of reaction (dH
rxn
) for glycylglycine hydrolysis: glycylglycine
(aq)
+ H
2
O
(l)
→
2 glycine
(aq) is dH
rxn
= -43.36 kJ mol
-1
. The heats of formation (dH
f
) for liquid water and aqueous glycylglycine are -68.39 kJ mol
-1
and -178.29 kJ mol
-1
, respectively. Compute the dH
f
of glycine.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —36— Bond Energies Sum: ∑
BE = dH = ∑
nBE
reactants
- ∑
nBE
products
1.
2H
2(g)
+ O
2(g)
⇌
2 H
2
O
(l)
∆
H = -400 kJ a.
What is the total bond energy? b.
If the bond energy for hydrogen gas is one-third the bond energy of liquid water, and gZ
M$
= 150
kJ, what is gZ
N$
and gZ
N$M
in kJ? 2.
What is the enthalpy change for the decomposition of hydrochloric acid? 2HCl →
H
2
+ Cl
2
H—Cl 427 kJ mol
-1 H—H 436 kJ mol
-1
C—C 347 kJ mol
-1
Cl—Cl 242 kJ mol
-1
C—Cl 339 kJ mol
-1
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —37— Challenge Questions 1.
David J. Van Wazer, 2. John M. Van Wazer 1.
The SARS-CoV-2 and HIV-1 viral spike proteins are highly antigenic, prompting the production of antibodies from B-cells as part of the adaptive immune response. Each spike protein exists as a homotrimer, and each antigen binding fragment (Fab) of an antibody recognizes one of the three spike subunits composing the homotrimer. Consider the binding of Fab-antibody to the HIV spike protein: You are working at the Vaccine Research Center at the National Institute of Allergy and Infectious Diseases (NIAID) overlooking a clinical trial for a potential HIV vaccine. Four weeks after receiving the vaccine, patient VRC01 was challenged with live HIV and has since seemed to completely recover from the infection. You took a blood serum sample of patient VRC01 and purified 1 mg of Fab. (FYI...a patient would never be challenged with live virus that there is no cure for... this is theoretical) MW of Fab = 50,000 g/mol (50 kDa for you biology/biochemistry majors) MW of HIV homotrimer spike protein = 230,000 g/mol (225 kDa) a.
If you have 100 micrograms (ug) of HIV spike protein, how many moles of Fab are required to make spike-Fab complexes? b.
How many molecules of Fab did you use to make the spike-Fab complexes in part a? c.
How many micrograms of spike-Fab complex did you make? +
3
HIV-spike homotrimer
Antibody
Fab
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —38— 2.
As an undergraduate research assistant in the Burke Lab at the University of Illinois Urbana-Champaign, you are studying the step-wise binding of a novel small-molecule compound capable of positive allosteric cooperativity. Positive allosteric cooperativity refers to when the binding of one species promotes the binding of subsequent species; where the second binding is more favorable once the first binding event has occurred. You have successfully synthesized and modeled the small-molecule’s theoretical two-step binding cooperativity in Spartan modeling software, and now wish to test the small-
molecule for its positive cooperativity in a laboratory environment to confirm its unique binding properties. Via isothermal titration calorimetry (ITC), a technique to measure the enthalpic change upon binding events at constant pressure and volume, you titrate (add over time) ligand L into the reaction chamber containing the small-molecule that you have named tetra-
carboxyl bis-benzene (t-Cbb). You obtain an ITC binding curve below: The negative slope of each binding event is your binding constant K
c
. Each binding event has a respective binding constant K
1
and K
2
. The two-step binding reaction of L to t-Cbb is below: 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
0
1
2
3
4
5
6
Q(kcal/mol)
[L]/[t-Cbb]
O
O
O
O
O
O
O
O
+ 2 L
K
1
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
K
2
L
L
L
+ L
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Chemistry I Final Exam Review —39— a. Consider the step-wise binding of L to t-Cbb, what is the dH in kJ for each binding event, dH
1
and dH
2
? Hint: Consider the relationship of enthalpy and heat given the ITC conditions. 1 kcal mol
-1
= 4184 J mol
-1 b. On ITC, what are the binding constants for each binding event in kcal mol
-1
? Which K
c
is greater? Note: you do not need to write the K
c binding expression. c. Does t-Cbb possess the positive allosteric cooperativity you had hypothesized?
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Related Documents
Related Questions
BMy Dashboard myBinghX
BUnit 9- Molecular Geom
X
LON-CAPA 10-043: E-G
O Not secure loncapa5.chem.binghamton.edu/res/fsu/silberbe
M Lumen OHM Asse...
Apps
G Gmail
Jazz Notes Exam 1...
Pt Dynamic Periodic...
Draw Lewis structures for each of the following and use them to answer the next 3 questions.
1) O3
2) NF3
3) NO2
4) SO2
5) CF4
6) N20 (N central)
Match each molecule or ion above to the correct electron-group arrangement from the list below.
NF3
A T shape
03
B tetrahedral
N20 (N central)
CF4
C square planar
D linear
NO2
D SO2
trigonal pyramidal
F bent
G trigonal planar
Tries 0/5
Submit Answer
Now match the sames molecules or ions above to the correct molecular shape.
N2O (N central)
A T shape
NO2
NF3
B linear
C square planar
CF4
D trigonal planar
trigonal pyramidal
SO2
O3
tetrahedral
F
G bent
Tries 0/5
Submit Answer
Finally, match the same molecules and ions above to the correct ideal bond angles from the list below.
NO2
03
1800
A
B 600
CF 4
C 109.50
NF3
D 900
SO2
30°
E
N20 (N…
arrow_forward
A east.cengagenow.com
CHM 103 003 General Chemistry 1
OWLV2 | Online teaching and learning resource from Cengage Learning
-35
[References)
pt
Use the References to access important values if needed for this que
pt
To answer the questions, interpret the following Lewis diagram for NO,".
pt
"pt
pt
1. For the central nitrogen atom:
pt
The number of non-bonding electrons =
pt
The number of bonding electrons
The total number of electrons
pt
pt
2. The central nitrogen atom
pt
A. obeys the octet rule.
B. has less than an octet.
C. has more than an octet.
Submit Answer
Try Another Version
5 item attempts remaining
Cengage Learning Cengage Technical Support
arrow_forward
i need help with my vsepr assignment.
arrow_forward
geNSUIortheastern St
OWLv2 LOnline teaching.
Chemistry Questions eA.
vLAShort Paner: Alcohol/Foo.
Centent
O X
Screenshot 2020-11-15 at 11.03....
O O Q Q
ĮKererencesĮ
a. Predict the molecular structure and bond angles for SeCle. Approximate bond angles are sufficient.
Molecular Structure =
Bond Angles
b. Predict the molecular structure and bond angles for IC15. Approximate bond angles are sufficient.
Molecular Structure =
Bond Angles
linear
bent
Try A
3 item attempts remaining
ubmit Answer
trigonal planar
T-shaped
trigonal pyramidal
tetrahedral
square pyramidal
square planar
octahedral
11:04
arrow_forward
Can you help?
arrow_forward
1. Draw the complete Lewis Structure for each. Then determine the electron pair geometry (EPG), make a
3-dimensional representation (drawing), and determine the molecular geometry (MG) for all of them.
Determine whether each molecule is polar or non-polar.
Formula
Lewis structure
EPG
3-D drawing
MG
polar?
a. PF3
b. CH;F2
Cl3
d. SC4
e. XeF4
arrow_forward
8. For each molecular fomula, detem ine the electron pair geom etry (EPG), make a 3-dimensional
representation, and determ ine the molecular geometry. Determine whether the molecule is polar or non-
polar. All of these have central atoms surounded by term inal atom s/one pairs only.
Formula
A. CHF3
Lewis structure
EPG
3-D drawing
Mol Geom.
Polar molecule?
B. SF2
arrow_forward
Need complete solution please.
arrow_forward
OM
-. CH₁
a. Lewis structure
c. Electronic geometry_
d. Molecular shape_
e. Bond angle?
2. NH3
b. Draw the different bonds and label
their polarity.
arrow_forward
MOLECULAR MODELS *VSEPR INTRODUCTION LABORATORY SIMULATION Calculate the number of
valence electrons in nitrogen trichloride, NCI (3). valence electrons Identify the Lewis structure that best
represents nitrogen trichloride, NCI (3).Se A. B. C.
arrow_forward
Molecule
or lon
SnCl4
#valence e
y
**NO3
#valence e
6
**PF6
#valence e
6
SbCl5
#valence e
**N3
#valence e
Lewis
Structure
No. of
Holes
Needed
in the
Central
Atom
No. of
Regions of
Electron
Density on
Central
Atom
Electron
Geometry
Molecular
Geometry
Sketch
Molecule/lon
Showing Molecular
Geometry
(show atoms only!)
** lons are treated in the same way as neutral molecules. (The charge is, of course, used in
determining the number of electrons in the valence shell of the ion.)
arrow_forward
2req
Please note that "geometry" refers to the molecular or ionic geometry.
:0–Xe–6:
:0:
A. The Lewis diagram for XeO3 is:
The electron-pair geometry around the Xe atom in XeO3 is
There are
lone pair(s) around the central atom, so the geometry of XeO3 is
B. The Lewis diagram for AsO₂ is: Ö-As=0
Submit Answer
Recall that for predicting geometry, double and triple bonds count as only one electron pair.
The electron-pair geometry around the As atom in AsO₂ is
There are lone pair(s) around the central atom, so the geometry of AsO₂ is
question.
Retry Entire Group 9 more group attempts remaining
arrow_forward
ne-Student
Merced College - Login to On
* Molecular Geometry Lab
W GOB_4803_L06_Molecular_G X
Sheets & Slides chrome-extension://bpmcpldpdmajfigpchkicefoigmkfalc/views/app.html
Geometry.docx
nal
BIU A A -
EXERCISE 1: MOLECULAR MODELS
Data Sheet
Table 1. Lewis Dot Structures
Element
Symbol Group Number Valence Electrons
Lewis Dot Structure
Hydrogen
Carbon
Chlorine
Aluminum
Oxygen
&
%23
$4
8
4.
arrow_forward
General Chemistry II, Spring 2021
Data Sheet IV
Lewis Structure and VB Diagram
Species
CH,0
formaldehyde
VSEPR
SN:
EGA:
# Valence
electrons:
Shape:
CO2
SN:
carbon
dioxide
EGA:
# Valence
electrons:
Shape:
HCN
hydrogen
cyanide
SN:
EGA:
# Valence
electrons:
Shape:
arrow_forward
provide th following information for NCl3 and HCN including the Lewis structure and 3D sketch
arrow_forward
Molecular geometry—don’t know why I’m incorrect all of a sudden. Might be one shape needed that is not given in the option choices.
arrow_forward
1. What is the relationship between the following compounds?
H :0: H
H:0:
H
H-C=C-C-C-C-H
H-C-C-C=Ċ-H
H HH
H.
H-C-H
B. constitutional isomers
C. the same structure
A. isotopes
D. composed of different elements
E. no relationship
2. What is the correct Lewis structure for HN3, including the formal charges?
CIH
arrow_forward
shift
um Chemical X um Dashboar X um Chemical x Chem Bo X Modeling X OWLV2 x Cengage x G periodic x
← → C esciencelabs.com/sites/default/files/V3/Digital Labs/Chem_Bonds_Valence/story_html5.html
Modeling lonic and Covalent Bonds 1
Nitrous Oxide
A
In each box, enter the appropriate number of valence electrons for each atom and the number of bonds
formed. Submit your choice when you are confident you have the correct answer.
N-N Bonds N-O Bonds
the option
W
S
X
command
80
F3
E
Charges 0
D
4
C
888
FA
R
F
N
0₁
%
5
*****
V
Electrons
(N)
24
FS
0
T
G
+1
^
6
0
Ν Ο
O
MacBook Air
Y
0
B
&
79
H
44
F7
-1
Electrons
(O)
Ü
N
0
+
8
J
DII
FA
I
M
(
9
K
MOSISO
Dashboar X Shopping x
DD
0
0
L
A
P
command
Submit
.
d
1
M
New Window Help
{
[
3.5: lonic x +
option
➤30
$42
1
S
delete
arrow_forward
US
108 Laboratory Manual for General, Organic, and Biological Chemistry
3. Names of molecular compounds
Formula
Name
Formula
Name
CIF
SF
CS2
N203
PCIS
SeF,
F. Electron-Dot Formulas and Shape
Formula 1. Lewis
2. Number of
5. Shape 6. Polar or
Nonpolar?
Structure
Electron
Groups
H20
SF,
NI3
SO3
CO2
Copyright © 2014 Pearson Education, Inc.
arrow_forward
I only want help with the marked ones please!
arrow_forward
Decide whether the Lewis structure proposed for each molecule is reasonable or not.
molecule proposed Lewis structure
OF 2
BeH₂
PF4
||
:O:
H - Be
H
F - P F
W
...
:F:
:F:
Is this a reasonable structure?
If not, why not?
Yes, it's a reasonable structure.
O No, the total number of valence electrons is wrong.
The correct number is:
No, some atoms have the wrong number of electrons around them.
The symbols of the problem atoms are:
O Yes, it's a reasonable structure.
No, the total number of valence electrons is wrong.
The correct number is:
O No, some atoms have the wrong number of electrons around them.
The symbols of the problem atoms are:
O Yes, it's a reasonable structure.
No, the total number of valence electrons is wrong.
The correct number is:
No, some atoms have the wrong number of electrons around them.
The symbols of the problem atoms are: 0
* If two or more atoms have the wrong number of valence electrons around them, just enter the chemical symbol
for the atom as many times as…
arrow_forward
SEE MORE QUESTIONS
Recommended textbooks for you
Organic Chemistry: A Guided Inquiry
Chemistry
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Cengage Learning
Related Questions
- BMy Dashboard myBinghX BUnit 9- Molecular Geom X LON-CAPA 10-043: E-G O Not secure loncapa5.chem.binghamton.edu/res/fsu/silberbe M Lumen OHM Asse... Apps G Gmail Jazz Notes Exam 1... Pt Dynamic Periodic... Draw Lewis structures for each of the following and use them to answer the next 3 questions. 1) O3 2) NF3 3) NO2 4) SO2 5) CF4 6) N20 (N central) Match each molecule or ion above to the correct electron-group arrangement from the list below. NF3 A T shape 03 B tetrahedral N20 (N central) CF4 C square planar D linear NO2 D SO2 trigonal pyramidal F bent G trigonal planar Tries 0/5 Submit Answer Now match the sames molecules or ions above to the correct molecular shape. N2O (N central) A T shape NO2 NF3 B linear C square planar CF4 D trigonal planar trigonal pyramidal SO2 O3 tetrahedral F G bent Tries 0/5 Submit Answer Finally, match the same molecules and ions above to the correct ideal bond angles from the list below. NO2 03 1800 A B 600 CF 4 C 109.50 NF3 D 900 SO2 30° E N20 (N…arrow_forwardA east.cengagenow.com CHM 103 003 General Chemistry 1 OWLV2 | Online teaching and learning resource from Cengage Learning -35 [References) pt Use the References to access important values if needed for this que pt To answer the questions, interpret the following Lewis diagram for NO,". pt "pt pt 1. For the central nitrogen atom: pt The number of non-bonding electrons = pt The number of bonding electrons The total number of electrons pt pt 2. The central nitrogen atom pt A. obeys the octet rule. B. has less than an octet. C. has more than an octet. Submit Answer Try Another Version 5 item attempts remaining Cengage Learning Cengage Technical Supportarrow_forwardi need help with my vsepr assignment.arrow_forward
- geNSUIortheastern St OWLv2 LOnline teaching. Chemistry Questions eA. vLAShort Paner: Alcohol/Foo. Centent O X Screenshot 2020-11-15 at 11.03.... O O Q Q ĮKererencesĮ a. Predict the molecular structure and bond angles for SeCle. Approximate bond angles are sufficient. Molecular Structure = Bond Angles b. Predict the molecular structure and bond angles for IC15. Approximate bond angles are sufficient. Molecular Structure = Bond Angles linear bent Try A 3 item attempts remaining ubmit Answer trigonal planar T-shaped trigonal pyramidal tetrahedral square pyramidal square planar octahedral 11:04arrow_forwardCan you help?arrow_forward1. Draw the complete Lewis Structure for each. Then determine the electron pair geometry (EPG), make a 3-dimensional representation (drawing), and determine the molecular geometry (MG) for all of them. Determine whether each molecule is polar or non-polar. Formula Lewis structure EPG 3-D drawing MG polar? a. PF3 b. CH;F2 Cl3 d. SC4 e. XeF4arrow_forward
- 8. For each molecular fomula, detem ine the electron pair geom etry (EPG), make a 3-dimensional representation, and determ ine the molecular geometry. Determine whether the molecule is polar or non- polar. All of these have central atoms surounded by term inal atom s/one pairs only. Formula A. CHF3 Lewis structure EPG 3-D drawing Mol Geom. Polar molecule? B. SF2arrow_forwardNeed complete solution please.arrow_forwardOM -. CH₁ a. Lewis structure c. Electronic geometry_ d. Molecular shape_ e. Bond angle? 2. NH3 b. Draw the different bonds and label their polarity.arrow_forward
- MOLECULAR MODELS *VSEPR INTRODUCTION LABORATORY SIMULATION Calculate the number of valence electrons in nitrogen trichloride, NCI (3). valence electrons Identify the Lewis structure that best represents nitrogen trichloride, NCI (3).Se A. B. C.arrow_forwardMolecule or lon SnCl4 #valence e y **NO3 #valence e 6 **PF6 #valence e 6 SbCl5 #valence e **N3 #valence e Lewis Structure No. of Holes Needed in the Central Atom No. of Regions of Electron Density on Central Atom Electron Geometry Molecular Geometry Sketch Molecule/lon Showing Molecular Geometry (show atoms only!) ** lons are treated in the same way as neutral molecules. (The charge is, of course, used in determining the number of electrons in the valence shell of the ion.)arrow_forward2req Please note that "geometry" refers to the molecular or ionic geometry. :0–Xe–6: :0: A. The Lewis diagram for XeO3 is: The electron-pair geometry around the Xe atom in XeO3 is There are lone pair(s) around the central atom, so the geometry of XeO3 is B. The Lewis diagram for AsO₂ is: Ö-As=0 Submit Answer Recall that for predicting geometry, double and triple bonds count as only one electron pair. The electron-pair geometry around the As atom in AsO₂ is There are lone pair(s) around the central atom, so the geometry of AsO₂ is question. Retry Entire Group 9 more group attempts remainingarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
Organic Chemistry: A Guided Inquiry
Chemistry
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Cengage Learning