ACS Final Flashcards
(227 cards)
1
Q
Law of Conservation of Mass
A
In a chemical reaction, matter is neither created nor destroyed
2
Q
Thompson Cathode Ray Experiment
A
Showed that toms were divisible into charged particles by generating a stream of electrons coming from atoms.
3
Q
Pure Substances
A
Crystalline Sugar and Copper Wire
4
Q
Homogeneous Mixture
A
Gatorade
5
Q
What mass(in kg) does 5.84 moles of titanium(Ti) have?
A
- 280kg
(5. 84mol Ti)(47.867g/1mol Ti)(1kg/1000g) = 0.2795kg
6
Q
Elements in the same group of the periodic table typically have
A
similar physical and chemical properties
7
Q
Diatomic molecules in elemental form
A
H,N,O,F,Cl,Br,I
8
Q
Diatomic molecules in elemental form
A
H,N,O,F,Cl,Br,I
9
Q
Calculate % by mass of lead in Pb(NO3)2
A
62.6%
Molar mass: 207.2 + (214.01)+(316.00) = 331.22
% mass = (207.2/331.22)(100) = 62.6%
10
Q
Calculate % by mass of lead in Pb(NO3)2
A
62.6%
Molar mass: 207.2 + (214.01)+(316.00) = 331.22
% mass = (207.2/331.22)(100) = 62.6%
11
Q
What is the volume of osmium in the tip of a fountain pen containing 152mg osmium and density of 22.59 g/cm^3?
A
- 73 micro liters(uL)
(152mg) (1g/1000mg) = 0.152g
d=m/v
22.59g/cm^3=0.152g/v
v = (0.006729cm^3)(1000uL/1mL)
= 6.7 uL
12
Q
What is the volume of osmium in the tip of a fountain pen containing 152mg osmium and density of 22.59 g/cm^3?
A
- 73 micro liters(uL)
(152mg) (1g/1000mg) = 0.152g
d=m/v
22.59g/cm^3=0.152g/v
v = (0.006729cm^3)(1000uL/1mL)
= 6.7 uL
13
Q
transition metals and chemical formula for osmium (VIII) oxide
A
Os^8+ and O^2-
use empirical formula for ionic compounds:
OsO4 = answer
14
Q
190^Os^6+
A,Z, n, e
A
Z = 76 A = 190 n = 114 e = 70
15
Q
190^Os^6+
A,Z, n, e
A
Z = 76 A = 190 n = 114 e = 70
16
Q
Vanadium has two naturally occurring isotopes: ^50V(49.9472amu) and 51^V(50.9440). The atomic weight of Vanadium is 50.9415.
What are the % abundances of 50^V and 51^V?
A
- 25(50^V) and 99.75%(51^V)
- 9415 = 49.9472x + (1-x)(50.9440)
- 0.0025 = -.9968x
50^V=x = 0.0025(100) = 0.25%
51^V = (1-0.0025)(100) = 99.75%
17
Q
Vanadium has two naturally occurring isotopes: ^50V(49.9472amu) and 51^V(50.9440). The atomic weight of Vanadium is 50.9415.
What are the % abundances of 50^V and 51^V?
A
- 25(50^V) and 99.75%(51^V)
- 9415 = 49.9472x + (1-x)(50.9440)
- 0.0025 = -.9968x
50^V=x = 0.0025(100) = 0.25%
51^V = (1-0.0025)(100) = 99.75%
18
Q
Gallium is solid at room temp with a melting point of 302.91K. Compared to the human body temperature of 98.6C what will happen?
A
melting point = 302.91 - 273.15 = 29.76C
body temp = 5/9(98.6-32) = 37C
Gallium melts at a lower temperature than body temperature, so it could melt if you hold it.
19
Q
A Freon leak in the air conditioning system of an old car releases 25g CF2Cl2. How many molecules of Freon are emitted?
A
- 2 * 10^23 molecules
(25g) (1mol CF2Cl2/120.912g)(6.022*10^23molecules/1mol)
= 1.245 * 10^23 molecules CF2Cl2
20
Q
A Freon leak in the air conditioning system of an old car releases 25g CF2Cl2. How many molecules of Freon are emitted?
A
- 2 * 10^23 molecules
(25g) (1mol CF2Cl2/120.912g)(6.022*10^23molecules/1mol)
= 1.245 * 10^23 molecules CF2Cl2
21
Q
A sample of Equilin is found to contain 80.546% carbon, 7.510% hydrogen, and 11.943% oxygen
What is the empirical formula?
A
C9H10O
C = (80.546g C)(1mol C/12.01g) = 6.707 mol C
H = (7.510g)(1mol H/1.0079g) = 7.451 mol H
O = (11.943g)(1mol O/16.00 g) = 0.746 mol O
(C6.707H7.451O0.746)/0.745
= C9H10O
22
Q
A sample of Equilin is found to contain 80.546% carbon, 7.510% hydrogen, and 11.943% oxygen
What is the empirical formula?
A
C9H10O
C = (80.546g C)(1mol C/12.01g) = 6.707 mol C
H = (7.510g)(1mol H/1.0079g) = 7.451 mol H
O = (11.943g)(1mol O/16.00 g) = 0.746 mol O
(C6.707H7.451O0.746)/0.745
= C9H10O
23
Q
molar mass of Equilin is 268.34g/mol.
What is molecular formula is empircal is C9H10O?
A
C18H20O2
n = (268.34g/mol)/(134.169 g/mol) = 2
(2)(C9H10O) = C18H20O2
24
Q
molar mass of Equilin is 268.34g/mol.
What is molecular formula is empircal is C9H10O?
A
C18H20O2
n = (268.34g/mol)/(134.169 g/mol) = 2
(2)(C9H10O) = C18H20O2
25
Balance the following and find coefficient of sulfur dioxide:
__PbS(s) + __O2(g) -> __PbO(s) + __SO2(g)
2,3,2,2
2SO2(g)
26
Systemic name for Fe3P2
iron(II) phosphide
27
Chemical formula for calcium nitrate
Ca(NO3)2
28
Chemical formula for calcium nitrate
Ca(NO3)2
29
N2O5 name
dinitrogen pentoxide
30
HNO3 name
nitric acid
31
Example of a chemical change
Sugar burns when heated on a skillet
32
Example of a chemical change
Sugar burns when heated on a skillet
33
an Aluminum sphere contains 8.55*10^22 aluminum atoms. What is the spheres radius in centimeters?
Density is 2.70g/cm^3
V = 4/3(pie)(r)^3
0.697cm = radius
mass of sphere:
(8.55*10^22atoms)(1mole/6.022*10^23atoms)(26.98g/1mole) = 3.83g Al
d=m/v
2.70g/cm^3 = 3.83g/v = 1.4187cm^3
1.4187cm^3 = 4/3(pie)(r)^3
r = 0.697cm
34
an Aluminum sphere contains 8.55*10^22 aluminum atoms. What is the spheres radius in centimeters?
Density is 2.70g/cm^3
V = 4/3(pie)(r)^3
0.697cm = radius
mass of sphere:
(8.55*10^22atoms)(1mole/6.022*10^23atoms)(26.98g/1mole) = 3.83g Al
d=m/v
2.70g/cm^3 = 3.83g/v = 1.4187cm^3
1.4187cm^3 = 4/3(pie)(r)^3
r = 0.697cm
35
speed of light
c =
c = 2.9979*10^8 m/s
36
h =
6.626*1-^-34J*s
37
Rh =
1.097 * 10^7m^-1
38
delta E = -hcRh(1/nf-1/ni)
```
h = 6.626*1-^-34J*s
c = 2.9979*10^8 m/s
Rh = 1.097 * 10^7m^-1
```
39
delta E = -hcRh(1/nf-1/ni)
```
h = 6.626*1-^-34J*s
c = 2.9979*10^8 m/s
Rh = 1.097 * 10^7m^-1
```
40
In your chemistry lab, you prepare a 0.100M solution of NaCl by dissolving NaCl(58.44g/mol) and adding water to a volume of 250. mL
What mass of NaCl was dissolved?
1.46g
M = moles solute/liter soln
0.100M = x moles/0.250L
moles = 0.0250
0.0250mol(58.44g/1mol) = 1.46g
41
In your chemistry lab, you prepare a 0.100M solution of NaCl by dissolving NaCl(58.44g/mol) and adding water to a volume of 250. mL
if you diluted 10.00 mL of the NaCl solution to a total volume of 25.0mL what will be the concentration of the diluted solution?
0.0400M
M1V1 = M2V2
(10.00mL)(0.100M) = M2(25.0mL)
M2 = 0.0400M
42
In your chemistry lab, you prepare a 0.100M solution of NaCl by dissolving NaCl(58.44g/mol) and adding water to a volume of 250. mL
you add Pb(NO3)2 to the NaCl solution(25mL and 0.0400M)
What happens?
a precipitation reaction with spectator ions NO3^-(aq) and Na^+(aq)
Pb^2+(aq) + 2Cl^-(aq) -> PbCl2(s)
43
In your chemistry lab, you prepare a 0.100M solution of NaCl by dissolving NaCl(58.44g/mol) and adding water to a volume of 250. mL
you add Pb(NO3)2 to the NaCl solution(25mL and 0.0400M)
What happens?
a precipitation reaction with spectator ions NO3^-(aq) and Na^+(aq)
Pb^2+(aq) + 2Cl^-(aq) -> PbCl2(s)
44
What volume of 0.2500M H2SO4(98.08g*mol^-1) is required to neutralize 25.0mL of 0.200M LiOH(23.96g*mol^-1)?
10. 00 mL
0. 200M LiOH = xmoles/0.025L = .005mol LiOH
(0. 005mol LiOH)(1mol H2SO4/2mol LiOH) = 0.0025mol H2SO4
0. 25M H2SO4=0.0025mol H2SO4/X L = 0.0100L
10. 00mL
45
What volume of 0.2500M H2SO4(98.08g*mol^-1) is required to neutralize 25.0mL of 0.200M LiOH(23.96g*mol^-1)?
10. 00 mL
0. 200M LiOH = xmoles/0.025L = .005mol LiOH
(0. 005mol LiOH)(1mol H2SO4/2mol LiOH) = 0.0025mol H2SO4
0. 25M H2SO4=0.0025mol H2SO4/X L = 0.0100L
10. 00mL
46
In which compound does vanadium have the lowest oxidation state?
VO = 2^+
47
What is happening here:
Fe(s) + Au(NO3)3(aq) -> Fe(NO3)3(aq)+Au(s)
Metallic iron is the reducing agent
48
A mixture of 57.33g PbO(223.2g/mol) and 33.80g PbS(239.3g/mol) react.
Identify the limiting reagent and determine the maximum mass of lead(207.2 g/mol) that can be obtained.
2PbO(s)+PbS(s) -> 3Pb(s)+SO2(g)
PbO limiting; 79.83g Pb obtained.
(57. 33g PbO)(1mol/223.2g)(3mol Pb/2mol PbO)(207.2g/1mol Pb) = 79.83g Pb
(33. 80g PbS)(1mol PbS/239.3g)(3mol Pb/1mol PbS)(207.2g Pb/1mol Pb) = 87.80g Pb
49
HS^-(aq)+NH4^+(aq) -> H2S(g)+NH3(g)
NH4^+ is a proton donor for HS^-
50
HS^-(aq)+NH4^+(aq) -> H2S(g)+NH3(g)
NH4^+ is a proton donor for HS^-
51
What is the percentage yield for a reaction in which 1.40g of Sb2S3(339.7g/mol) is obtained from 1.73g antimony(121.8g/mol) reacting and a slight excess of sulfur(32.07g/mol?
58. 0%
(1. 73g Sb)(1mol Sb/121.8g)(1mol Sb3S3/2mol Sb)(339.7g/1molSb2S3) = 2.412g Sb3S3 theoretical yield
(1. 40g/2.412g)(100) = 58%
52
What is the percentage yield for a reaction in which 1.40g of Sb2S3(339.7g/mol) is obtained from 1.73g antimony(121.8g/mol) reacting and a slight excess of sulfur(32.07g/mol?
58. 0%
(1. 73g Sb)(1mol Sb/121.8g)(1mol Sb3S3/2mol Sb)(339.7g/1molSb2S3) = 2.412g Sb3S3 theoretical yield
(1. 40g/2.412g)(100) = 58%
53
Calculate deltaHrxn:
CS2(l)+6H2O(l)->CO2(g)+6H2O(l)+2SO2(g)
Given:
CS2(l)+3O2(g)->CO2(g)+2SO2(g) = -1077kJ
H2(g)+O2(g)->H2O2(l) = -188kJ
H2(g)+1/2O2(g)->H2O(l) = -286kJ
-1665 kJ/mol
f1: leave alone(-1077kJ)
f2: reverse and multiple by 6(1128)
f3: multiply by 6(-1716)
54
Calculate deltaHrxn:
CS2(l)+6H2O(l)->CO2(g)+6H2O(l)+2SO2(g)
Given:
CS2(l)+3O2(g)->CO2(g)+2SO2(g) = -1077kJ
H2(g)+O2(g)->H2O2(l) = -188kJ
H2(g)+1/2O2(g)->H2O(l) = -286kJ
-1665 kJ/mol
f1: leave alone(-1077kJ)
f2: reverse and multiple by 6(1128)
f3: multiply by 6(-1716)
55
The combustion of ammonia is represented by:
4NH3(g) + 5O2(g) -> 4NO(g)+6H2O(g)
deltaH=-904.8kJ
what is the enthalpy of formation of NH3(g)?
```
NO(g) = +90.4kJ/mol
H2O(g) = -241.8kJ/mol
```
- 46.1kJ/mol
| - 904.8 = (4(90.4)+6(-241.8))-(4(x)+5(0))
56
When dry ice, CO2(s) sublimes to the gas phase to form CO2(g)
the process is endothermic and heat is absorbed from the surroundings.
57
When a 45.0g sample of an alloy at 100 degrees C is dropped into 100.0g of water(Cs of 4.18J/gC) the final temperature is 37C. What is the specific heat of the alloy with a starting temperature of 25C?
1.77J/gC
qsys=-qsurr
q=(mass)(Cs)(delta T)
(45.0g)(Cs)(37-100)=-(100.0g)(4.18J/gC)(37.0-25.0)
Cs(-2835gC)=-5016J
Cs=1.7693J/gC
58
When a 45.0g sample of an alloy at 100 degrees C is dropped into 100.0g of water(Cs of 4.18J/gC) the final temperature is 37C. What is the specific heat of the alloy with a starting temperature of 25C?
1.77J/gC
qsys=-qsurr
q=(mass)(Cs)(delta T)
(45.0g)(Cs)(37-100)=-(100.0g)(4.18J/gC)(37.0-25.0)
Cs(-2835gC)=-5016J
Cs=1.7693J/gC
59
A 35.6g sample of ethanol(C2H5OH 46.07g/mol) is burned in a bomb calorimeter, according to the following reaction:
C2H5OH(l) +3O2(g)->2CO2(g)+3H2O(g)
if the temperature rose from 35.0C to 76.0C and the heat capacity is 23.3kJ/C what is the value of delta E?
-1.24*10^3 kJ/mol
qrxn=-Ccal=-(23.3kJ/C)(76.0-35.0C) = -955.3kJ
report delta E in kJ/mol!
(35.6g)(1mol/46.07g) = 0.7727mol C2H5OH
delta E = (-955.3kJ/0.7727mol) = -1236.3kJ/mol
60
A 35.6g sample of ethanol(C2H5OH 46.07g/mol) is burned in a bomb calorimeter, according to the following reaction:
C2H5OH(l) +3O2(g)->2CO2(g)+3H2O(g)
if the temperature rose from 35.0C to 76.0C and the heat capacity is 23.3kJ/C what is the value of delta E?
-1.24*10^3 kJ/mol
qrxn=-Ccal=-(23.3kJ/C)(76.0-35.0C) = -955.3kJ
report delta E in kJ/mol!
(35.6g)(1mol/46.07g) = 0.7727mol C2H5OH
delta E = (-955.3kJ/0.7727mol) = -1236.3kJ/mol
61
Which set of quantum numbers if correct and consistent for an electron with n=3?
I=2
| mj = +2
62
Which set of quantum numbers if correct and consistent for an electron with n=3?
l=2
| mj = +2
63
Which of the following illustrates an orbital with quantum numbers n=5, l=1?
one vertical barbell
64
Which of the following illustrates an orbital with quantum numbers n=5, l=1?
one vertical barbell
65
List in order of increasing wavelength:
radio,x-ray,ultraviolet,visible light
x ray
66
List in order of increasing wavelength:
radio,x-ray,ultraviolet,visible light
x ray
67
in guided missile technology, "heat seeking" missiles are directed to target exhaust from jets which emit infrared radiation with wavelength 5.0*10^-6m.
What is the frequency of the infrared light described above?
6.0*10^13Hz
frequency(u symbol) = c/wavelength(inverted y)
=(3.00*10^8m/s)/(5.0*1-^-6m)
= 6.0*10^13Hz
68
in guided missile technology, "heat seeking" missiles are directed to target exhaust from jets which emit infrared radiation with wavelength 5.0*10^-6m.
What is the energy of the infrared light described above?
4.0*10^-20J
Frequency=6.0*10^13Hz
frequency(u symbol) = c/wavelength(inverted y)
E = hu(frequency)
=(6.626*10^-34j*s)(6.0*10^13x^-1)
= 3.9756*10^-20J
69
in guided missile technology, "heat seeking" missiles are directed to target exhaust from jets which emit infrared radiation with wavelength 5.0*10^-6m.
What is the energy of the infrared light described above?
4.0*10^-20J
Frequency=6.0*10^13Hz
frequency(u symbol) = c/wavelength(inverted y)
E = hu(frequency)
=(6.626*10^-34j*s)(6.0*10^13x^-1)
= 3.9756*10^-20J
70
Photoelectric Effect
high intensity, high energy light can eject electrons from a metal plate. This means energy can be transferred from light to electrons.
71
Heisenbergs Uncertainty Principle
The more precisely you know the position of an electron, the less precisely you can know its velocity
72
Atomic Emission Spectra
When argon gas is heated, the gas emits light with discrete energy levels, rather than a continuous spectrum. This indicates that electrons are not found at random energy levels in an atom, but at energy levels that are quantized.
73
Atomic Emission Spectra
When argon gas is heated, the gas emits light with discrete energy levels, rather than a continuous spectrum. This indicates that electrons are not found at random energy levels in an atom, but at energy levels that are quantized.
74
Electron Diffraction
When firing electrons one at a time toward a grating, an interference pattern is generated. This indicates that electrons behave like waves.
75
Electron Diffraction
When firing electrons one at a time toward a grating, an interference pattern is generated. This indicates that electrons behave like waves.
76
Arrange the elements in order of increasing energy required to remove the first electron of their atoms:
Mg, Ne, Na, Ar
Na
77
Arrange the elements in order of increasing energy required to remove the first electron of their atoms:
Mg, Ne, Na, Ar
Na
78
which of the following is an isoelectronic series listed with decreasing radius?
Se^2- > Br^- > Kr
79
Why does the atomic radius decrease down the list for the elements Na(r=0.186nm), Mg(r=0.160nm), Al(r=0.146nm)?
the effective nuclear charge increases down the list.
80
The first three ionization energies of an element X are 590, 1145, 10,912kJ/mol.
What is the likely formula for a stable ion of X?
X^2+
81
# Choose the bond below that is least polar.
C-Cl, C-Br,C-F, C-I
C-I
82
Choose the ground state electron configuration for Ti^2+
[Ar]4s^03d^2
83
Choose the ground state electron configuration for Ti^2+
[Ar]4s^03d^2
84
How many of the following species are paramagnetic?
Se^3+, Br^-, Mg^2+, Se
1
85
place the following in order of decreasing magnitude of lattice energy:
KCl, MgO, RbI
MgO>KCl>RbI
86
place the following in order of decreasing magnitude of lattice energy:
KCl, MgO, RbI
MgO>KCl>RbI
87
In the generation of most anions, the energy change(kJ/mol) that _______ an electron is _______. This energy is called______.
Adds, negative, electron affinity
88
In the generation of most anions, the energy change(kJ/mol) that _______ an electron is _______. This energy is called______.
Adds, negative, electron affinity
89
Which is true of the resonance structures of NO2^-?
the bond lengths of N-O bonds in NO2^- are equal.
90
Which molecule of C and N will have the shortest C-N bond?
more paired electrons = shorter bond length
triple>double>single
91
Which molecule of C and N will have the shortest C-N bond?
more paired electrons = shorter bond length
triple>double>single
92
Calculate deltaHrxn with bond energies
Add up all the bonds of the reactants and then subtract all the bonds formed
delta H = sum(bonds broken) - sum(bonds formed)
93
Which compounds violate the octet rule?
NO2 and XeF4(4 bonds + 2 extra electron pairs)
94
Lewis Structure for Allene(C3H4)
How many "o" and "pie" bonds?
6o
| 2pie
95
Choose the best Lewis structure for ICl5
I with 5 paired single bonded Cl(each with 6 electrons) and one extra lone pair.
96
Choose the orbital diagram that represents the ground state of N
1s(2), 2s(2), 2p(3, 1 in each level all pointing up)
97
Choose the orbital diagram that represents the ground state of N
1s(2), 2s(2), 2p(3, 1 in each level all pointing up)
98
Give the set of four quantum numbers that could represent the electron last added to the configuration of N.
based on:
1s(2), 2s(2), 2p(3, 1 in each level all pointing up)
n=2
l=1
ml=0
ms=+1/2
99
Give the set of four quantum numbers that could represent the electron last added to the configuration of N.
based on:
1s(2), 2s(2), 2p(3, 1 in each level all pointing up)
n=2
l=1
ml=0
ms=+1/2
100
Determine the electron geometry(eg), molecular geometry(mg), and hybridization of XeF4
eg = octahedral
mg = square planar
sp^3d^2 hybridization
Xe bonded to 4 F(6 electrons each) with 2 extra pairs on Xe. 6 bonds total
101
Which of the following is true of the resonance structures for OCl2
the resonance with a formal charge of 0 on each atom is more accurate.
102
Which of the following is true of the resonance structures for OCl2
the resonance with a formal charge of 0 on each atom is more accurate.
103
What IMFs exist between CH3F?
dipole-dipole and london dispersion forces.
104
List the following compounds in order of increasing boiling point:
CH3CH2Cl, NH3,CH2,Cl,NH3CH2OH,CH3CH3
CH3CH3
105
Under what conditions is the solubility of nitrogen in water the greatest?
low temperature and high pressure
106
CO2 in the form of dry ice would be classified as
a molecular solid
107
CO2 in the form of dry ice would be classified as
a molecular solid
108
How much energy is required to vaporize 48.7g of dichloromethane(CH2Cl2 84.9g/mol) at its boiling point with deltaHvap 31.6kJ/mol?
18.1kJ
q=deltaH(n)
n = 48.7g(1mol/84.9g/mol) = 0.574 mol
q = (31.6kJ)(0.574) = 18.1 kJ
109
How much energy is required to vaporize 48.7g of dichloromethane(CH2Cl2 84.9g/mol) at its boiling point with deltaHvap 31.6kJ/mol?
18.1kJ
q=deltaH(n)
n = 48.7g(1mol/84.9g/mol) = 0.574 mol
q = (31.6kJ)(0.574) = 18.1 kJ
110
800. g of ethanol(C2H5OH 46.07 g/mol) was added to 8.0*10^3g water. How much would this lower the freezing point? The kf for water is 1.86C*m^-1 and.
4.0 degrees C
moles ethanol = 800.g(1mol/46.07g) = 17.36mol
molarity(m) = (17.36mol/8.0 kg water) = 2.17m
delta T = (1.86C)(2.17m) = 4.0C
111
800. g of ethanol(C2H5OH 46.07 g/mol) was added to 8.0*10^3g water. How much would this lower the freezing point? The kf for water is 1.86C*m^-1 and.
4.0 degrees C
moles ethanol = 800.g(1mol/46.07g) = 17.36mol
molarity(m) = (17.36mol/8.0 kg water) = 2.17m
delta T = (1.86C)(2.17m) = 4.0C
112
If 0.200 mol of propylene glycol is dissolved in 3.60 mol water what is the vapor pressure of the resulting solution? The vapor pressure of pure water is 23.8mmHg at 25C
22.5mmHg
Xwater = (3.6mol)/(0.200mol+3.6mol) = 0.947
Pwater = (0.947)(23.8mmHg) = 22.5mmHg
113
What is NOT true of surface tension?
surface tension is the resistance of a liquid to flow
this is viscosity
114
What is true of surface tension?
- the energy required to increase the surface area of a liquid
- decreases as IMFs decrease
- more bonds = higher surface tension
115
Phase diagrams
A->B->D
the compound would undergo fusion then vaporization
116
Benzene has a Hvap of 30.72kJ/mol and a normal boiling point of 80.1C.
At what temperature does benzene boil when the external pressure is 445 torr?
335K
ln(p2/p1)=(-deltaHvap)/R(1/T2-1/T1)
ln(760/445) = -(30.72*10^3J/mol/8.314J/mol*k)(1/353-1/T1)
T1 = 335K
117
Face Centered Cubic
1/8atom in each corner and 1/2atom in each face
118
Face Centered Cubic
1/8atom in each corner and 1/2atom in each face
119
Polonium(208.98g*mol^-1) crystallizes with a simple cubic structure and d=9.30g/cm^3.
1m = 10^12pm
What is the radius of the polonium atom?
167pm
SCC - 1 atom per unit cell
1atom(1mol/6.022*10^23atoms)(208.98g/1mol) = 3.47*10^-22g
d=m/v
9.30g/cm^3 = (3.47*10^-22g/v) = 3.73*10^-23cm^3
v = l^3
3.73*10^-23cm^3=l^3
l = 3.34*10^-8cm
l=2r
r = 1.67*10^-8cm = 167pm
120
Polonium(208.98g*mol^-1) crystallizes with a simple cubic structure and d=9.30g/cm^3.
1m = 10^12pm
What is the radius of the polonium atom?
167pm
SCC - 1 atom per unit cell
1atom(1mol/6.022*10^23atoms)(208.98g/1mol) = 3.47*10^-22g
d=m/v
9.30g/cm^3 = (3.47*10^-22g/v) = 3.73*10^-23cm^3
v = l^3
3.73*10^-23cm^3=l^3
l = 3.34*10^-8cm
l=2r
r = 1.67*10^-8cm = 167pm
121
4NH3(g)+3O2(g) -> 2N2(g) + 6H2O(l)
if the rate of formation of N2 is 0.10M*sec^-1, what is the corresponding rate of disappearance of O2?
0.15M*sec^-1
(deltaN2)/(delta t) = 0.10*M/s
1/2(deltaN2/delta t) = 1/3(delta O2/delta t)
(delta O2)/(delta t) = 3/2(delta N2/delta t)
(delta O2/delta t) = 0.15M/s
122
4NH3(g)+3O2(g) -> 2N2(g) + 6H2O(l)
if the rate of formation of N2 is 0.10M*sec^-1, what is the corresponding rate of disappearance of O2?
0.15M*sec^-1
(deltaN2)/(delta t) = 0.10*M/s
1/2(deltaN2/delta t) = 1/3(delta O2/delta t)
(delta O2)/(delta t) = 3/2(delta N2/delta t)
(delta O2/delta t) = 0.15M/s
123
N2(g) + 3H2(g) -> 2NH3
N2 and 3H2 decrease while 2NH3 increase
H2 will decrease 3 times as quickly as N2
124
Catalyzed reactions
the activation energy of the catalyzed reaction is less than an uncatalyzed reaction
125
s1: H2O2(aq)+I^-(aq)->H2O(l)+IO^-(aq)
s2: H2O2(aq)+IO^-(aq)->H2O(l)+O2(g)+I^-(aq)
the purpose of I^- is
to act as a catalyst
126
s1: H2O2(aq)+I^-(aq)->H2O(l)+IO^-(aq)
s2: H2O2(aq)+IO^-(aq)->H2O(l)+O2(g)+I^-(aq)
the purpose of I^- is
to act as a catalyst
127
The first order decomposition of N2O5 at 328K has a rate constant of 1.70*10^-3s^-1. if the initial concentration of N2O5 is 2.88M what is the concentration of N2O5 after 12.5minutes?
0.805M
ln[A]t = -kt+ln[A]0
ln[A]t=-(1.70*10^-3s^-1)(750seconds) + ln[2.88M]
[A]t = 0.805M
128
zero, first, and second order reactions and their graphs
look this up
129
zero, first, and second order reactions and their graphs
look this up
130
2NO(g)+O2(g)->2NO2(g)
s1: NO(g)+ O2(g)->NO2(g)+O(g) slow
s2: O(g)+NO(g)->NO2(g) fast
rate = k[NO][O2]
s1 is rate limiting
rate = k[NO][O2]
131
2NO(g)+O2(g)->2NO2(g)
s1: NO(g)+ O2(g)->NO2(g)+O(g) slow
s2: O(g)+NO(g)->NO2(g) fast
rate = k[NO][O2]
s1 is rate limiting
rate = k[NO][O2]
132
Determine the rate law and the value of k for the following reaction using the data provided:
S2O8^2-(aq) + 3I^-(aq)->2SO4^2-(g)+I3^-(aq)
rate = 36 M^-1s^-1[S2O8^2-][I^-]
***review this***
133
What is Kc(equilibrium constant) for:
Mg(s)+2HCl(aq) MgCl2(aq)+H2(g)
Kc = [H2][MgCl2]/[HCl]^2
134
3A(aq) B(aq) + 2C(aq)
What is the equilibrium constant?
K = 0.14
```
K = [B][C]^2/[A]^3
K = [1][3]^2/[4]^3
K = 0.14
```
135
3A(aq) B(aq) + 2C(aq)
What is the equilibrium constant?
K = 0.14
```
K = [B][C]^2/[A]^3
K = [1][3]^2/[4]^3
K = 0.14
```
136
Find equilibrium constant for the following:
A(g)+B(g) AB(g) Kc = 0.24
AB(g)+A(g) A2B(g) Kc = 3.8
2A(g)+B(g) A2B(g) Kc = ?
Kc = 0.91
```
Kc = (0.24)(3.8)
Kc = 0.91
```
137
Find equilibrium constant for the following:
A(g)+B(g) AB(g) Kc = 0.24
AB(g)+A(g) A2B(g) Kc = 3.8
2A(g)+B(g) A2B(g) Kc = ?
Kc = 0.91
```
Kc = (0.24)(3.8)
Kc = 0.91
```
138
I2(g) 2I(g) Kp = 0.209
A reaction contains 0.89atm I2 and 1.77atm I.
Which statement is true of the system?
The reaction will shift in the direction of the reactants.
```
Q = pI^2/pI2
Q = (1.77)^2/0.89 = 3.52
```
Q(3.52)>K(0.209)
reaction proceeds in reverse so more reactants.
139
A(g) + 2B(g) 2C(g) delta H = -495kJ
What favors the production of more C?
decrease the temperature.
removing heat from an exothermic reaction shifts the reaction to the right.
140
N2(g) + 3H2(g) 2NH3(g)
The reaction has a Kc value of 61. What is the value of Kp for this reaction at 500K?
3.6*10^-2
```
Kp = Kc(RT)^delta(n)
n = mol prod = mol react
```
Kp = (61)[(0.0821L*atm/mol*K)(500K)]^-2
Kp = 0.036
141
N2(g) + 3H2(g) 2NH3(g)
The reaction has a Kc value of 61. What is the value of Kp for this reaction at 500K?
3.6*10^-2
```
Kp = Kc(RT)^delta(n)
n = mol prod = mol react
```
Kp = (61)[(0.0821L*atm/mol*K)(500K)]^-2
Kp = 0.036
142
What is dynamic equilibrium?
when the rate of the forward reaction equals the rate of the reverse reaction.
Exchange still happens but no net change
143
What is NOT true of dynamic equilibrium?
equilibrium constants for forward and reverse reactions are equal.
Q> K and reaction favors products
the amount of reactants and products is equal
144
CuS(s)+O2(g) Cu(s) + SO2(g)
A mixture initially contains 2.9M O2. Determine the equilibrium concentration of O2 if Kc for the reaction is 1.5.
1.2M
```
Kc = [SO2]/[O2]
Kc = x/(2.9 - X)=1.5
Kc = 1.74
```
(for above)ICE table O2 and SO2
O2 = 2.9-x
SO2 = 0 +x
[O2] = 2.9-x = 2.9 - 1.74 = 1.16
145
CuS(s)+O2(g) Cu(s) + SO2(g)
A mixture initially contains 2.9M O2. Determine the equilibrium concentration of O2 if Kc for the reaction is 1.5.
1.2M
```
Kc = [SO2]/[O2]
Kc = x/(2.9 - X)=1.5
Kc = 1.74
```
(for above)ICE table O2 and SO2
O2 = 2.9-x
SO2 = 0 +x
[O2] = 2.9-x = 2.9 - 1.74 = 1.16
146
A Bronstead-Lowry base is
a proton acceptor
147
A Lewis Acid is
an electron pair acceptor
148
An Arrehenius base produces
OH^- ions in water
149
For this reaction:
CH3NH2(aq) + HCN(aq) CH3NH3^+(aq) + CN^-(aq)
HCN is the acid
CN^- is the conjugate base
CH3NH2 is the base
CH3NH3^+ is the conjugate acid
150
What is the conjugate base of H2PO4^-
H2PO4^2-
151
Which acid is the weakest?
the one with the smallest Ka
HC3H3O3, Ka = 1.4*10^-4
152
HC2O4^- has a larger Kb and is a stronger base than
C2O4^2-
153
Determine the pH of a 0.023 M HNO3 solution
1.64
HNO3 is a strong acid
[H3O^+] = -log(0.023)
= 1.64
154
Determine the pH of a 0.023 M HNO3 solution
1.64
HNO3 is a strong acid
[H3O^+] = -log(0.023)
= 1.64
155
Calculate the pH of a solution that contains 7.8*1-^-6M NaOH at 25C
8.89
NaOH = strong base = [OH^-]
pOH = -log(7.8*1-^-6) = 5.108
pH = 14-5.108
= 8.89
156
What is the equation for the ion product constant of water at 25C?
[H3O^+][OH^-]=10^-14
157
Which species will produce a basic solution when dissolved in water?
NH3Cl
NaNO3
K2CO3
Co(ClO4)2
K2CO3 - Co3^2- is a base
- NH4Cl - NH4^+ is a weak acid
- NaNO3 - both ions are pH neutral
- Co(ClO4)2 - Co^2+ is a Lewis acid
158
Which of the following is the strongest acid?
HBrO4
HBrO3
HBrO2
HBr
HBrO4
all other factors equal then more O = stronger acid
159
Which of the following is the strongest acid?
HBrO4
HBrO3
HBrO2
HBr
HBrO4
all other factors equal then more O = stronger acid
160
A weak acid solution containing 0.12M C5H5NH^+. The Ka of the weak acid is 5.88*10^-6.
What is the pH and percent ionization?
pH = 3.08, 0.70%
(ICE)C5H5NH^+ + H2O H3O^+ + C5H5NH
0.12-x x x
5. 88*10^-6 = (x)^2/(0.12-x)
- assume x is small
```
x = 8.4*10^-4 = [H3O^+]
pH = -log(8.4*10^-4) = 3.08
```
% ionization = (8.4*10^-4/0.12)(100) = 0.70%
161
The conjugate base of a weak acid C5H5N. What is the Kb for this with a Kb of 5.88*10^-6?
1.7*10^-9
(Ka)(Kb)=Kw
Ka = Kw/Kb
1*10^-14/5.88*10^-6 = 1.7*10^-9
162
The conjugate base of a weak acid C5H5N. What is the Kb for this with a Kb of 5.88*10^-6?
1.7*10^-9
(Ka)(Kb)=Kw
Ka = Kw/Kb
1*10^-14/5.88*10^-6 = 1.7*10^-9
163
what is the pKs of a weak acid-base titration?
pKa = 4.8
1/2 the eq point
164
what is the equivalence point of a titration of HNO3 and a weak base?
below 7.0
165
what produces a buffer soln when equal volumes of the solutions are combined?
0.1M HOCl and 0.1M NaOCl
166
A pH 4.4 buffer is prepared by dissolving equimolar amounts of benzoic acid(HC7H5O2) and sodium benzoate(C7H5O2^-)
HC7H5O2+H2O C7H5O2^- + H3O^+
The addition of KOH to this buffer results in
an increase in [C7H5O2]
more reactants = shift right to balance
167
What is the pH of a solution prepared by adding 2.00g of solid CH3CH2NH2(45.09 g*mol^-1) to 100.0mL of 0.50M aqueous CH3CH2NH3Cl(81.55g*mol^-1)?
The Kb of ethyl amine(CH3CH2NH2) is 5.6*10^-4.
pH = 10.70
moles CH3CH2NH2 = 2.00(1mol/45.09g) = 0.0444mol
Molarity CH3CH2NH3 = 0.0444mol/0.100L = 0.444M
pH = pKa + log[base]/[acid]
pH = 10.74 +log(0.444/0.500)
pH = 10.70
168
What is the pH of a solution prepared by adding 2.00g of solid CH3CH2NH2(45.09 g*mol^-1) to 100.0mL of 0.50M aqueous CH3CH2NH3Cl(81.55g*mol^-1)?
The Kb of ethyl amine(CH3CH2NH2) is 5.6*10^-4.
pH = 10.70
moles CH3CH2NH2 = 2.00(1mol/45.09g) = 0.0444mol
Molarity CH3CH2NH3 = 0.0444mol/0.100L = 0.444M
pH = pKa + log[base]/[acid]
pH = 10.74 +log(0.444/0.500)
pH = 10.70
169
What would be a suitable acid/base pair for preparing a buffer with pH 5.3?
HC2H3O2/C2H3O2^- with Ka 1.8*10^-5
pH = -log(1.8*10^-5) = 4.74
5.3+/- 1 = range of 4.3-6.3
170
What would be a suitable acid/base pair for preparing a buffer with pH 5.3?
HC2H3O2/C2H3O2^- with Ka 1.8*10^-5
pH = -log(1.8*10^-5) = 4.74
5.3+/- 1 = range of 4.3-6.3
171
What would be a suitable acid/base pair for preparing a buffer with pH 5.3?
HC2H3O2/C2H3O2^- with Ka 1.8*10^-5
pH = -log(1.8*10^-5) = 4.74
5.3+/- 1 = range of 4.3-6.3
172
A titration is performed on 0.350g of a weak monoprotic acid. it is titrated to equivalence with 22.7mL of 0.250M NaOH.
What is the molar mass of the acid?
61.7 g*mol^-1
at equivalence, moles base = moles acid
(0.250mol/L)(0.0227) = 0.005675mol
molar mass = 0.350g/0.005675mol = 61.67g/mol
173
A titration is performed on 0.350g of a weak monoprotic acid. it is titrated to equivalence with 22.7mL of 0.250M NaOH.
What is the molar mass of the acid?
61.7 g*mol^-1
at equivalence, moles base = moles acid
(0.250mol/L)(0.0227) = 0.005675mol
molar mass = 0.350g/0.005675mol = 61.67g/mol
174
Which indicator is the most appropriate for the detection in the titration of a weak base by a strong acid?
Bromphenol Blue(pH range 3.8-4.8)
Strong acid will make the pH slightly acidic
175
Which indicator is the most appropriate for the detection in the titration of a weak base by a strong acid?
Bromphenol Blue(pH range 3.8-4.8)
Strong acid will make the pH slightly acidic
176
Determine the equilibrium constant for the following reaction at 298K
Cl(g)+O3(g)-> ClO(g) + O2(g)
Delta G = -34.5kJ
1.12*10^6
deltaG = -RT(lnK)
lnK = -34.5kJ/((.008314kj/mol)(298))
ln K = 13.92
K = 1.12*10^6
177
Calculate delta Grxn for:
ClO(g)+O3(g) -> Cl(g) +2O2(g) delta G = ?
2O3(g)-> 3O2(g) delta G = 489.6kJ
Cl(g)+O3(g)->ClO(g)+O2(g) delta G = -34.5kJ
524.1kJ
flip the second equation and add together
178
Calculate delta Grxn for:
ClO(g)+O3(g) -> Cl(g) +2O2(g) delta G = ?
2O3(g)-> 3O2(g) delta G = 489.6kJ
Cl(g)+O3(g)->ClO(g)+O2(g) delta G = -34.5kJ
524.1kJ
flip the second equation and add together
179
Calculate deltaS(degree)rxn for the following reaction:
P4(g) + 10Cl2(g) -> 4PCl5(g)
Sdegree(J/mol*K) = 280.0, 223.1, and 364.6
-1052.6J/K
delta S = nSprods -nSreact
[4(364.6)]-[1(280.0)+10(223.1)]
= -1052.6 J/K
180
Calculate deltaS(degree)rxn for the following reaction:
P4(g) + 10Cl2(g) -> 4PCl5(g)
Sdegree(J/mol*K) = 280.0, 223.1, and 364.6
-1052.6J/K
delta S = nSprods -nSreact
[4(364.6)]-[1(280.0)+10(223.1)]
= -1052.6 J/K
181
Place the following in order of increasing molar entropy at 298K:
CO2, C2H8, SO
SO
182
Place the following in order of increasing molar entropy at 298K:
CO2, C2H8, SO
SO
183
Above what temperature does the following reaction become nonspontaneous?
FeO(s)+CO(g)->CO2(g)+Fe(s)
delta H = -11.0 kJ
delta S = 17.4 J/K
632K
delta G = delta H - T(deltaS)
```
0 = -11kJ - T(-0.0174kJ/k)
11kJ = T(0.0174kJ/K)
```
T = 632K
184
Consider a reaction that has a positive delta H and a positive delta S.
When will the reaction be spontaneous?
the reaction will be spontaneous at high temperatures
delta G must be negative for spontaneous
185
Can endothermic processes be spontaneous?
Yes!
186
For a spontaneous process detla S is
greater than 0
187
Exothermic processes generally increase the
entropy of the surroundings
188
When will S = 0?
a perfect crystal at 0K
189
NaCl dissolving in water will have a ______ delta S
positive
190
NaCl dissolving in water will have a ______ delta S
positive
191
As aqueous solution is 0.20M in bother Ba^2+ and Pb^2+. As K2SO4 is added which will precipitate first?
BaSO4 Ksp = 1.07*10^-10
PbSO4 Ksp = 1.82 * 10^-8
Ba^2+
BaSO4 has a smaller Ksp and will precipitate first
192
As aqueous solution is 0.20M in bother Ba^2+ and Pb^2+. As K2SO4 is added which will precipitate first?
BaSO4 Ksp = 1.07*10^-10
PbSO4 Ksp = 1.82 * 10^-8
Ba^2+
BaSO4 has a smaller Ksp and will precipitate first
193
The solubility product constant for Fe(OH)2 is 4.87*10^-17.
What is the molar solubility?
2.30*10^-6
Fe(OH)2 Fe^2++2OH^-
solid - x, x, 2x
Ksp = [Fe^2+][OH^-]^2
4.87*10^-17 = (x)(2x)^2 = 4x^3
x = 2.3*10^-6
194
The solubility product constant for Fe(OH)2 is 4.87*10^-17.
What is the molar solubility?
2.30*10^-6
Fe(OH)2 Fe^2++2OH^-
solid - x, x, 2x
Ksp = [Fe^2+][OH^-]^2
4.87*10^-17 = (x)(2x)^2 = 4x^3
x = 2.3*10^-6
195
A titration of 25.0mL 0.0230M NH3 with titrant 0.0100M HCl is over-titrated by 5.00mL past the equivalence point.
What is the pH of the resulting solution?
pH = 3.24
Find volume at equivalence:
(0.025L)(0.0230mol/L) = 5.75*10^-4mol NH3 = mol NH4
volume HCL = (5.75*10^-4 mol/(0.0100mol/L)) = 0.0575L
Moles excess HCL = (0.0100mol/L)(0.005L) = 5*10^-5mol HCl
[H^+] = 5*10^-5mol HCl/(0.025L + 0.0575L + 0.005L) = 5.71*10^-4M
pH = -log(5.71*10^-4) = 3.24
196
Reduction occurs when
electrons are gained
oxidation number decreases
197
balance the equation and what is the coefficient Fe^2+
Al(s)+Fe^2+(aq)->Al^3+(aq)+Fe(s)
2Al(s)+3Fe^2+(aq)->2Al^3+(aq)+2Fe(s)
3 on Fe^2+
198
balance the equation and what is the coefficient Fe^2+
Al(s)+Fe^2+(aq)->Al^3+(aq)+Fe(s)
2Al(s)+3Fe^2+(aq)->2Al^3+(aq)+2Fe(s)
3 on Fe^2+
199
how many moles of electrons are transferred in the following rxn:
CoCl3(aq)+H2(g)->Co(s)+HCl(aq)
6 moles e^-
2CoCl3(aq) + 3H2(g) -> 2Co(s) + 6HCl(aq)
200
2CoCl3(aq) + 3H2(g) -> 2Co(s) + 6HCl(aq)
What is the oxidizing agent?
Co^3+
H2 was oxidized and is reducing agent
Co^3+ is reduced and is oxidizing agent
201
2CoCl3(aq) + 3H2(g) -> 2Co(s) + 6HCl(aq)
What is the oxidizing agent?
Co^3+
H2 was oxidized and is reducing agent
Co^3+ is reduced and is oxidizing agent
202
What is the Estandard of:
Sn^4+(aq) + Ni(s)->Ni^2+(aq)+Sn^2+(aq)
Sn Estnard = 0.15V
Ni Estandard = -0.23
Spontaneous
0.15 -(-0.23) = 0.38V
203
What is the Estandard of:
Sn^4+(aq) + Ni(s)->Ni^2+(aq)+Sn^2+(aq)
Sn Estnard = 0.15V
Ni Estandard = -0.23
Spontaneous
0.15 -(-0.23) = 0.38V
204
What is Ecell of:
Cd(s)+2Ag^+(aq)-> Cd^2+(aq)+2Ag(s)
```
Cd^2+ = E = -0.40V
Ag^+ = E = 0.80V
```
1. 20V
| 0. 80-(-0.40) = 1.20V
205
What is deltaG of:
Cd(s)+2Ag^+(aq)-> Cd^2+(aq)+2Ag(s)
```
Cd^2+ = E = -0.40V
Ag^+ = E = 0.80V
```
-232kJ
deltaG=-nFE
deltaG = -(2)(96485)(1.20V)
= -231564J
= -232kJ
206
What is deltaG of:
Cd(s)+2Ag^+(aq)-> Cd^2+(aq)+2Ag(s)
```
Cd^2+ = E = -0.40V
Ag^+ = E = 0.80V
```
Ecell = 1.20V
-232kJ
deltaG=-nFE
deltaG = -(2)(96485)(1.20V)
= -231564J
= -232kJ
207
What is equilibrium constant @298K of:
Cd(s)+2Ag^+(aq)-> Cd^2+(aq)+2Ag(s)
```
Cd^2+ = E = -0.40V
Ag^+ = E = 0.80V
```
```
Ecell = 1.20V
deltaG = -232kJ
```
3.9*10^40
```
deltaG = -RT(lnK)
-231,564J = -(8.314J)(298)(lnK)
```
K = 3.9*10^40
208
What is not true for a spontaneous reaction?
Edegreecell
209
What is true of a spontaneous reaction?
deltaG1
| deltaSuniv>0
210
Where does oxidation occur in an electrochemical cell?
The Anode
211
Where does oxidation occur in an electrochemical cell?
The Anode
212
calculate the cell potential of the nonstandard reaction:
Sn(d)/Sn^2+(aq,1.8M)//Ag^+(aq,0.055M/Ag(s)
```
Sn^2+ = E = -0.140V
Ag^+ = E = 0.80V
```
0.86V
E = Edegree - (0.0592V/n)(log Q)
E = 0.94 -(0.0592/2)(log(1.8/0.055)^2)
213
When aq soln copper(II) sulfate is electrolyzed, copper metal is deposited
Cu^2+(aq)+2e^-
If a constant current of 0.0681 A is applied for 5.00 hours, what mass of copper is deposited? Molar mass = 63.6g/mol
0.404g
moles = (i)(t)/(n)(f)
((0.0681A)(5.00)(60)(60))/((2)(96485))
= 0.00635moles
0.00635 mol(63.6g) = 0.404g
214
write a nuclear equation for the alpha decay of 241/95Am
241/95Am->4/2He+237/93Np
215
Determine the identity of the daughter nuclide from the beta decay of 14/6C
14/7N
14/6C -> 0/-1e+14/7N
216
the following reaction represents what nuclear process?
137/55Cs+0/-1e->137/54Xe
Electron capture
217
Stable isotopes, with low atomic numbers, have a N/Z ratio of 1. What does this imply?
the number of neutrons equals the number of protons
218
Stable isotopes, with low atomic numbers, have a N/Z ratio of 1. What does this imply?
the number of neutrons equals the number of protons
219
Nuclices above the valley of stability can become more stable through which of the following processes?
Beta emission
too many neutrons can be reduced by turning a neutron into a proton via B decay
220
Flourine-18 undergoes positron emission with a half life of 1.1*10^2minutes. If a patient is given a 248mg dose for a PET scane, how long will it take for the amount of Flourine-18 to drop to 83mg?
1.74*10^2minutes
t1/2=0.693/k
```
k = 0.693/1.1*10^2min
k= 0.0063
```
ln(83/248) = -0.0063(t)
t = 174mins
221
Nuclear fusion reactions produce energy by
combining light nuclei into a heavier nucleus
222
Nuclear fission reactions produce energy by
splitting a nucleus into lighter ones
223
Unlike rate constants for chemical reactions, nuclear decay reactions
are independent of temperature
224
Radiocarbon dating is accomplished by
measuring the amount of C-14 in organic samples
225
Elements with atomic numbers greater than 83 are
radioactive
226
Elements with atomic numbers greater than 83 are
radioactive
227
in a nuclear fission reaction, the following process occurs:
1/0n+235/92U->139/56Ba+94/36Kr+(3)1/0n
how much energy(J) is released per mole of Uranium?
mass reactants = 236.05258
Mass products = 235.869181
1. 65*10^13J/mol U
236. 05258-235.869181 = 0.183399amu = deltam
(0. 183399amu)(1.66054*10^-27kg/1amu)=3.045*10^-28kg
```
E = mc^2
E = (3.045*10^-28kg)(2.9979*10^8m/s)^2 = 2.737*10^-11
```
(2.737*10^-11)(6.022*10^23)=1.65*10^13 J/mol U
