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ACS Chem Final > ACS Final > Flashcards

ACS Final Flashcards

1
Q

Law of Conservation of Mass

A

In a chemical reaction, matter is neither created nor destroyed

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2
Q

Thompson Cathode Ray Experiment

A

Showed that toms were divisible into charged particles by generating a stream of electrons coming from atoms.

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3
Q

Pure Substances

A

Crystalline Sugar and Copper Wire

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4
Q

Homogeneous Mixture

A

Gatorade

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5
Q

What mass(in kg) does 5.84 moles of titanium(Ti) have?

A
  1. 280kg

(5. 84mol Ti)(47.867g/1mol Ti)(1kg/1000g) = 0.2795kg

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6
Q

Elements in the same group of the periodic table typically have

A

similar physical and chemical properties

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7
Q

Diatomic molecules in elemental form

A

H,N,O,F,Cl,Br,I

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8
Q

Diatomic molecules in elemental form

A

H,N,O,F,Cl,Br,I

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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%

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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%

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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
  1. 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

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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
  1. 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

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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

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14
Q

190^Os^6+

A,Z, n, e

A 
Z = 76
A = 190
n = 114
e = 70
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15
Q

190^Os^6+

A,Z, n, e

A 
Z = 76
A = 190
n = 114
e = 70
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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
  1. 25(50^V) and 99.75%(51^V)
  2. 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%

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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
  1. 25(50^V) and 99.75%(51^V)
  2. 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%

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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.

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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
  1. 2 * 10^23 molecules
    (25g) (1mol CF2Cl2/120.912g)(6.022*10^23molecules/1mol)

= 1.245 * 10^23 molecules CF2Cl2

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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
  1. 2 * 10^23 molecules
    (25g) (1mol CF2Cl2/120.912g)(6.022*10^23molecules/1mol)

= 1.245 * 10^23 molecules CF2Cl2

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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

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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

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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

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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

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25
Q

Balance the following and find coefficient of sulfur dioxide:

__PbS(s) + __O2(g) -> __PbO(s) + __SO2(g)

A

2,3,2,2

2SO2(g)

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26
Q

Systemic name for Fe3P2

A

iron(II) phosphide

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27
Q

Chemical formula for calcium nitrate

A

Ca(NO3)2

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28
Q

Chemical formula for calcium nitrate

A

Ca(NO3)2

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29
Q

N2O5 name

A

dinitrogen pentoxide

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30
Q

HNO3 name

A

nitric acid

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31
Q

Example of a chemical change

A

Sugar burns when heated on a skillet

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32
Q

Example of a chemical change

A

Sugar burns when heated on a skillet

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33
Q

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

A

0.697cm = radius

mass of sphere:
(8.5510^22atoms)(1mole/6.02210^23atoms)(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

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34
Q

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

A

0.697cm = radius

mass of sphere:
(8.5510^22atoms)(1mole/6.02210^23atoms)(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

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35
Q

speed of light

c =

A

c = 2.9979*10^8 m/s

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36
Q

h =

A

6.6261-^-34Js

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37
Q

Rh =

A

1.097 * 10^7m^-1

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38
Q

delta E = -hcRh(1/nf-1/ni)

A 
h = 6.626*1-^-34J*s
c = 2.9979*10^8 m/s
Rh = 1.097 * 10^7m^-1
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39
Q

delta E = -hcRh(1/nf-1/ni)

A 
h = 6.626*1-^-34J*s
c = 2.9979*10^8 m/s
Rh = 1.097 * 10^7m^-1
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40
Q

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?

A

1.46g

M = moles solute/liter soln

0.100M = x moles/0.250L

moles = 0.0250

0.0250mol(58.44g/1mol) = 1.46g

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41
Q

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?

A

0.0400M

M1V1 = M2V2

(10.00mL)(0.100M) = M2(25.0mL)

M2 = 0.0400M

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42
Q

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

a precipitation reaction with spectator ions NO3^-(aq) and Na^+(aq)

Pb^2+(aq) + 2Cl^-(aq) -> PbCl2(s)

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43
Q

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

a precipitation reaction with spectator ions NO3^-(aq) and Na^+(aq)

Pb^2+(aq) + 2Cl^-(aq) -> PbCl2(s)

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44
Q

What volume of 0.2500M H2SO4(98.08gmol^-1) is required to neutralize 25.0mL of 0.200M LiOH(23.96gmol^-1)?

A
  1. 00 mL
  2. 200M LiOH = xmoles/0.025L = .005mol LiOH
    (0. 005mol LiOH)(1mol H2SO4/2mol LiOH) = 0.0025mol H2SO4
  3. 25M H2SO4=0.0025mol H2SO4/X L = 0.0100L
  4. 00mL
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45
Q

What volume of 0.2500M H2SO4(98.08gmol^-1) is required to neutralize 25.0mL of 0.200M LiOH(23.96gmol^-1)?

A
  1. 00 mL
  2. 200M LiOH = xmoles/0.025L = .005mol LiOH
    (0. 005mol LiOH)(1mol H2SO4/2mol LiOH) = 0.0025mol H2SO4
  3. 25M H2SO4=0.0025mol H2SO4/X L = 0.0100L
  4. 00mL
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46
Q

In which compound does vanadium have the lowest oxidation state?

A

VO = 2^+

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47
Q

What is happening here:

Fe(s) + Au(NO3)3(aq) -> Fe(NO3)3(aq)+Au(s)

A

Metallic iron is the reducing agent

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48
Q

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)

A

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

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49
Q

HS^-(aq)+NH4^+(aq) -> H2S(g)+NH3(g)

A

NH4^+ is a proton donor for HS^-

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50
Q

HS^-(aq)+NH4^+(aq) -> H2S(g)+NH3(g)

A

NH4^+ is a proton donor for HS^-

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51
Q

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?

A
  1. 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%
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52
Q

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?

A
  1. 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%
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53
Q

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

A

-1665 kJ/mol

f1: leave alone(-1077kJ)
f2: reverse and multiple by 6(1128)
f3: multiply by 6(-1716)

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54
Q

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

A

-1665 kJ/mol

f1: leave alone(-1077kJ)
f2: reverse and multiple by 6(1128)
f3: multiply by 6(-1716)

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55
Q

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
A
  • 46.1kJ/mol

- 904.8 = (4(90.4)+6(-241.8))-(4(x)+5(0))

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56
Q

When dry ice, CO2(s) sublimes to the gas phase to form CO2(g)

A

the process is endothermic and heat is absorbed from the surroundings.

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57
Q

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?

A

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

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58
Q

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?

A

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

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59
Q

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?

A

-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

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60
Q

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?

A

-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

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61
Q

Which set of quantum numbers if correct and consistent for an electron with n=3?

A

I=2

mj = +2

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62
Q

Which set of quantum numbers if correct and consistent for an electron with n=3?

A

l=2

mj = +2

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63
Q

Which of the following illustrates an orbital with quantum numbers n=5, l=1?

A

one vertical barbell

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64
Q

Which of the following illustrates an orbital with quantum numbers n=5, l=1?

A

one vertical barbell

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65
Q

List in order of increasing wavelength:

radio,x-ray,ultraviolet,visible light

A

x ray

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66
Q

List in order of increasing wavelength:

radio,x-ray,ultraviolet,visible light

A

x ray

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67
Q

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?

A

6.0*10^13Hz

frequency(u symbol) = c/wavelength(inverted y)

=(3.0010^8m/s)/(5.01-^-6m)

= 6.0*10^13Hz

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68
Q

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?

A

4.0*10^-20J

Frequency=6.0*10^13Hz

frequency(u symbol) = c/wavelength(inverted y)

E = hu(frequency)

=(6.62610^-34js)(6.0*10^13x^-1)

= 3.9756*10^-20J

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69
Q

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?

A

4.0*10^-20J

Frequency=6.0*10^13Hz

frequency(u symbol) = c/wavelength(inverted y)

E = hu(frequency)

=(6.62610^-34js)(6.0*10^13x^-1)

= 3.9756*10^-20J

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70
Q

Photoelectric Effect

A

high intensity, high energy light can eject electrons from a metal plate. This means energy can be transferred from light to electrons.

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71
Q

Heisenbergs Uncertainty Principle

A

The more precisely you know the position of an electron, the less precisely you can know its velocity

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72
Q

Atomic Emission Spectra

A

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.

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73
Q

Atomic Emission Spectra

A

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.

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74
Q

Electron Diffraction

A

When firing electrons one at a time toward a grating, an interference pattern is generated. This indicates that electrons behave like waves.

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75
Q

Electron Diffraction

A

When firing electrons one at a time toward a grating, an interference pattern is generated. This indicates that electrons behave like waves.

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76
Q

Arrange the elements in order of increasing energy required to remove the first electron of their atoms:

Mg, Ne, Na, Ar

A

Na

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77
Q

Arrange the elements in order of increasing energy required to remove the first electron of their atoms:

Mg, Ne, Na, Ar

A

Na

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78
Q

which of the following is an isoelectronic series listed with decreasing radius?

A

Se^2- > Br^- > Kr

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79
Q

Why does the atomic radius decrease down the list for the elements Na(r=0.186nm), Mg(r=0.160nm), Al(r=0.146nm)?

A

the effective nuclear charge increases down the list.

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80
Q

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?

A

X^2+

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81
Q

Choose the bond below that is least polar.

C-Cl, C-Br,C-F, C-I

A

C-I

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82
Q

Choose the ground state electron configuration for Ti^2+

A

[Ar]4s^03d^2

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83
Q

Choose the ground state electron configuration for Ti^2+

A

[Ar]4s^03d^2

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84
Q

How many of the following species are paramagnetic?

Se^3+, Br^-, Mg^2+, Se

A

1

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85
Q

place the following in order of decreasing magnitude of lattice energy:

KCl, MgO, RbI

A

MgO>KCl>RbI

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86
Q

place the following in order of decreasing magnitude of lattice energy:

KCl, MgO, RbI

A

MgO>KCl>RbI

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87
Q

In the generation of most anions, the energy change(kJ/mol) that _______ an electron is _______. This energy is called______.

A

Adds, negative, electron affinity

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88
Q

In the generation of most anions, the energy change(kJ/mol) that _______ an electron is _______. This energy is called______.

A

Adds, negative, electron affinity

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89
Q

Which is true of the resonance structures of NO2^-?

A

the bond lengths of N-O bonds in NO2^- are equal.

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90
Q

Which molecule of C and N will have the shortest C-N bond?

A

more paired electrons = shorter bond length

triple>double>single

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91
Q

Which molecule of C and N will have the shortest C-N bond?

A

more paired electrons = shorter bond length

triple>double>single

92
Q

Calculate deltaHrxn with bond energies

A

Add up all the bonds of the reactants and then subtract all the bonds formed

delta H = sum(bonds broken) - sum(bonds formed)

93
Q

Which compounds violate the octet rule?

A

NO2 and XeF4(4 bonds + 2 extra electron pairs)

94
Q

Lewis Structure for Allene(C3H4)

How many “o” and “pie” bonds?

95
Q

Choose the best Lewis structure for ICl5

A

I with 5 paired single bonded Cl(each with 6 electrons) and one extra lone pair.

96
Q

Choose the orbital diagram that represents the ground state of N

A

1s(2), 2s(2), 2p(3, 1 in each level all pointing up)

97
Q

Choose the orbital diagram that represents the ground state of N

A

1s(2), 2s(2), 2p(3, 1 in each level all pointing up)

98
Q

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)

A

n=2
l=1
ml=0
ms=+1/2

99
Q

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)

A

n=2
l=1
ml=0
ms=+1/2

100
Q

Determine the electron geometry(eg), molecular geometry(mg), and hybridization of XeF4

A

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
Q

Which of the following is true of the resonance structures for OCl2

A

the resonance with a formal charge of 0 on each atom is more accurate.

102
Q

Which of the following is true of the resonance structures for OCl2

A

the resonance with a formal charge of 0 on each atom is more accurate.

103
Q

What IMFs exist between CH3F?

A

dipole-dipole and london dispersion forces.

104
Q

List the following compounds in order of increasing boiling point:

CH3CH2Cl, NH3,CH2,Cl,NH3CH2OH,CH3CH3

105
Q

Under what conditions is the solubility of nitrogen in water the greatest?

A

low temperature and high pressure

106
Q

CO2 in the form of dry ice would be classified as

A

a molecular solid

107
Q

CO2 in the form of dry ice would be classified as

A

a molecular solid

108
Q

How much energy is required to vaporize 48.7g of dichloromethane(CH2Cl2 84.9g/mol) at its boiling point with deltaHvap 31.6kJ/mol?

A

18.1kJ

q=deltaH(n)

n = 48.7g(1mol/84.9g/mol) = 0.574 mol

q = (31.6kJ)(0.574) = 18.1 kJ

109
Q

How much energy is required to vaporize 48.7g of dichloromethane(CH2Cl2 84.9g/mol) at its boiling point with deltaHvap 31.6kJ/mol?

A

18.1kJ

q=deltaH(n)

n = 48.7g(1mol/84.9g/mol) = 0.574 mol

q = (31.6kJ)(0.574) = 18.1 kJ

110
Q
  1. g of ethanol(C2H5OH 46.07 g/mol) was added to 8.010^3g water. How much would this lower the freezing point? The kf for water is 1.86Cm^-1 and.
A

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
Q
  1. g of ethanol(C2H5OH 46.07 g/mol) was added to 8.010^3g water. How much would this lower the freezing point? The kf for water is 1.86Cm^-1 and.
A

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
Q

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

A

22.5mmHg

Xwater = (3.6mol)/(0.200mol+3.6mol) = 0.947

Pwater = (0.947)(23.8mmHg) = 22.5mmHg

113
Q

What is NOT true of surface tension?

A

surface tension is the resistance of a liquid to flow

this is viscosity

114
Q

What is true of surface tension?

A
  • the energy required to increase the surface area of a liquid
  • decreases as IMFs decrease
  • more bonds = higher surface tension
115
Q

Phase diagrams

A

A->B->D

the compound would undergo fusion then vaporization

116
Q

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?

A

335K

ln(p2/p1)=(-deltaHvap)/R(1/T2-1/T1)

ln(760/445) = -(30.7210^3J/mol/8.314J/molk)(1/353-1/T1)

T1 = 335K

117
Q

Face Centered Cubic

A

1/8atom in each corner and 1/2atom in each face

118
Q

Face Centered Cubic

A

1/8atom in each corner and 1/2atom in each face

119
Q

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?

A

167pm

SCC - 1 atom per unit cell

1atom(1mol/6.02210^23atoms)(208.98g/1mol) = 3.4710^-22g

d=m/v
9.30g/cm^3 = (3.4710^-22g/v) = 3.7310^-23cm^3

v = l^3
3.7310^-23cm^3=l^3
l = 3.3410^-8cm

l=2r

r = 1.67*10^-8cm = 167pm

120
Q

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?

A

167pm

SCC - 1 atom per unit cell

1atom(1mol/6.02210^23atoms)(208.98g/1mol) = 3.4710^-22g

d=m/v
9.30g/cm^3 = (3.4710^-22g/v) = 3.7310^-23cm^3

v = l^3
3.7310^-23cm^3=l^3
l = 3.3410^-8cm

l=2r

r = 1.67*10^-8cm = 167pm

121
Q

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?

A

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
Q

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?

A

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
Q

N2(g) + 3H2(g) -> 2NH3

A

N2 and 3H2 decrease while 2NH3 increase

H2 will decrease 3 times as quickly as N2

124
Q

Catalyzed reactions

A

the activation energy of the catalyzed reaction is less than an uncatalyzed reaction

125
Q

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

A

to act as a catalyst

126
Q

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

A

to act as a catalyst

127
Q

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?

A

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
Q

zero, first, and second order reactions and their graphs

A

look this up

129
Q

zero, first, and second order reactions and their graphs

A

look this up

130
Q

2NO(g)+O2(g)->2NO2(g)

s1: NO(g)+ O2(g)->NO2(g)+O(g) slow
s2: O(g)+NO(g)->NO2(g) fast

A

rate = k[NO][O2]

s1 is rate limiting

rate = k[NO][O2]

131
Q

2NO(g)+O2(g)->2NO2(g)

s1: NO(g)+ O2(g)->NO2(g)+O(g) slow
s2: O(g)+NO(g)->NO2(g) fast

A

rate = k[NO][O2]

s1 is rate limiting

rate = k[NO][O2]

132
Q

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)

A

rate = 36 M^-1s^-1[S2O8^2-][I^-]

review this

133
Q

What is Kc(equilibrium constant) for:

Mg(s)+2HCl(aq) MgCl2(aq)+H2(g)

A

Kc = [H2][MgCl2]/[HCl]^2

134
Q

3A(aq) B(aq) + 2C(aq)

What is the equilibrium constant?

A

K = 0.14

K = [B][C]^2/[A]^3
K = [1][3]^2/[4]^3
K = 0.14
135
Q

3A(aq) B(aq) + 2C(aq)

What is the equilibrium constant?

A

K = 0.14

K = [B][C]^2/[A]^3
K = [1][3]^2/[4]^3
K = 0.14
136
Q

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 = ?

A

Kc = 0.91

Kc = (0.24)(3.8)
Kc = 0.91
137
Q

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 = ?

A

Kc = 0.91

Kc = (0.24)(3.8)
Kc = 0.91
138
Q

I2(g) 2I(g) Kp = 0.209

A reaction contains 0.89atm I2 and 1.77atm I.

Which statement is true of the system?

A

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
Q

A(g) + 2B(g) 2C(g) delta H = -495kJ

What favors the production of more C?

A

decrease the temperature.

removing heat from an exothermic reaction shifts the reaction to the right.

140
Q

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?

A

3.6*10^-2

Kp = Kc(RT)^delta(n)
n = mol prod = mol react

Kp = (61)[(0.0821Latm/molK)(500K)]^-2

Kp = 0.036

141
Q

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?

A

3.6*10^-2

Kp = Kc(RT)^delta(n)
n = mol prod = mol react

Kp = (61)[(0.0821Latm/molK)(500K)]^-2

Kp = 0.036

142
Q

What is dynamic equilibrium?

A

when the rate of the forward reaction equals the rate of the reverse reaction.

Exchange still happens but no net change

143
Q

What is NOT true of dynamic equilibrium?

A

equilibrium constants for forward and reverse reactions are equal.

Q> K and reaction favors products

the amount of reactants and products is equal

144
Q

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.

A

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
Q

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.

A

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
Q

A Bronstead-Lowry base is

A

a proton acceptor

147
Q

A Lewis Acid is

A

an electron pair acceptor

148
Q

An Arrehenius base produces

A

OH^- ions in water

149
Q

For this reaction:

CH3NH2(aq) + HCN(aq) CH3NH3^+(aq) + CN^-(aq)

A

HCN is the acid
CN^- is the conjugate base

CH3NH2 is the base
CH3NH3^+ is the conjugate acid

150
Q

What is the conjugate base of H2PO4^-

151
Q

Which acid is the weakest?

A

the one with the smallest Ka

HC3H3O3, Ka = 1.4*10^-4

152
Q

HC2O4^- has a larger Kb and is a stronger base than

153
Q

Determine the pH of a 0.023 M HNO3 solution

A

1.64

HNO3 is a strong acid

[H3O^+] = -log(0.023)

= 1.64

154
Q

Determine the pH of a 0.023 M HNO3 solution

A

1.64

HNO3 is a strong acid

[H3O^+] = -log(0.023)

= 1.64

155
Q

Calculate the pH of a solution that contains 7.8*1-^-6M NaOH at 25C

A

8.89

NaOH = strong base = [OH^-]

pOH = -log(7.8*1-^-6) = 5.108

pH = 14-5.108

= 8.89

156
Q

What is the equation for the ion product constant of water at 25C?

A

[H3O^+][OH^-]=10^-14

157
Q

Which species will produce a basic solution when dissolved in water?

NH3Cl
NaNO3
K2CO3
Co(ClO4)2

A

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
Q

Which of the following is the strongest acid?

HBrO4
HBrO3
HBrO2
HBr

A

HBrO4

all other factors equal then more O = stronger acid

159
Q

Which of the following is the strongest acid?

HBrO4
HBrO3
HBrO2
HBr

A

HBrO4

all other factors equal then more O = stronger acid

160
Q

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?

A

pH = 3.08, 0.70%

(ICE)C5H5NH^+ + H2O H3O^+ + C5H5NH
0.12-x x x

  1. 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
Q

The conjugate base of a weak acid C5H5N. What is the Kb for this with a Kb of 5.88*10^-6?

A

1.7*10^-9

(Ka)(Kb)=Kw

Ka = Kw/Kb

110^-14/5.8810^-6 = 1.7*10^-9

162
Q

The conjugate base of a weak acid C5H5N. What is the Kb for this with a Kb of 5.88*10^-6?

A

1.7*10^-9

(Ka)(Kb)=Kw

Ka = Kw/Kb

110^-14/5.8810^-6 = 1.7*10^-9

163
Q

what is the pKs of a weak acid-base titration?

A

pKa = 4.8

1/2 the eq point

164
Q

what is the equivalence point of a titration of HNO3 and a weak base?

165
Q

what produces a buffer soln when equal volumes of the solutions are combined?

A

0.1M HOCl and 0.1M NaOCl

166
Q

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

A

an increase in [C7H5O2]

more reactants = shift right to balance

167
Q

What is the pH of a solution prepared by adding 2.00g of solid CH3CH2NH2(45.09 gmol^-1) to 100.0mL of 0.50M aqueous CH3CH2NH3Cl(81.55gmol^-1)?

The Kb of ethyl amine(CH3CH2NH2) is 5.6*10^-4.

A

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
Q

What is the pH of a solution prepared by adding 2.00g of solid CH3CH2NH2(45.09 gmol^-1) to 100.0mL of 0.50M aqueous CH3CH2NH3Cl(81.55gmol^-1)?

The Kb of ethyl amine(CH3CH2NH2) is 5.6*10^-4.

A

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
Q

What would be a suitable acid/base pair for preparing a buffer with pH 5.3?

A

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
Q

What would be a suitable acid/base pair for preparing a buffer with pH 5.3?

A

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
Q

What would be a suitable acid/base pair for preparing a buffer with pH 5.3?

A

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
Q

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?

A

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
Q

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?

A

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
Q

Which indicator is the most appropriate for the detection in the titration of a weak base by a strong acid?

A

Bromphenol Blue(pH range 3.8-4.8)

Strong acid will make the pH slightly acidic

175
Q

Which indicator is the most appropriate for the detection in the titration of a weak base by a strong acid?

A

Bromphenol Blue(pH range 3.8-4.8)

Strong acid will make the pH slightly acidic

176
Q

Determine the equilibrium constant for the following reaction at 298K

Cl(g)+O3(g)-> ClO(g) + O2(g)

Delta G = -34.5kJ

A

1.12*10^6

deltaG = -RT(lnK)

lnK = -34.5kJ/((.008314kj/mol)(298))

ln K = 13.92

K = 1.12*10^6

177
Q

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

A

524.1kJ

flip the second equation and add together

178
Q

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

A

524.1kJ

flip the second equation and add together

179
Q

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

A

-1052.6J/K

delta S = nSprods -nSreact

[4(364.6)]-[1(280.0)+10(223.1)]

= -1052.6 J/K

180
Q

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

A

-1052.6J/K

delta S = nSprods -nSreact

[4(364.6)]-[1(280.0)+10(223.1)]

= -1052.6 J/K

181
Q

Place the following in order of increasing molar entropy at 298K:

CO2, C2H8, SO

182
Q

Place the following in order of increasing molar entropy at 298K:

CO2, C2H8, SO

183
Q

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

A

632K

delta G = delta H - T(deltaS)

0 = -11kJ - T(-0.0174kJ/k)
11kJ = T(0.0174kJ/K)

T = 632K

184
Q

Consider a reaction that has a positive delta H and a positive delta S.

When will the reaction be spontaneous?

A

the reaction will be spontaneous at high temperatures

delta G must be negative for spontaneous

185
Q

Can endothermic processes be spontaneous?

186
Q

For a spontaneous process detla S is

A

greater than 0

187
Q

Exothermic processes generally increase the

A

entropy of the surroundings

188
Q

When will S = 0?

A

a perfect crystal at 0K

189
Q

NaCl dissolving in water will have a ______ delta S

190
Q

NaCl dissolving in water will have a ______ delta S

191
Q

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

A

Ba^2+

BaSO4 has a smaller Ksp and will precipitate first

192
Q

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

A

Ba^2+

BaSO4 has a smaller Ksp and will precipitate first

193
Q

The solubility product constant for Fe(OH)2 is 4.87*10^-17.

What is the molar solubility?

A

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
Q

The solubility product constant for Fe(OH)2 is 4.87*10^-17.

What is the molar solubility?

A

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
Q

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?

A

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^+] = 510^-5mol HCl/(0.025L + 0.0575L + 0.005L) = 5.7110^-4M

pH = -log(5.71*10^-4) = 3.24

196
Q

Reduction occurs when

A

electrons are gained

oxidation number decreases

197
Q

balance the equation and what is the coefficient Fe^2+

Al(s)+Fe^2+(aq)->Al^3+(aq)+Fe(s)

A

2Al(s)+3Fe^2+(aq)->2Al^3+(aq)+2Fe(s)

3 on Fe^2+

198
Q

balance the equation and what is the coefficient Fe^2+

Al(s)+Fe^2+(aq)->Al^3+(aq)+Fe(s)

A

2Al(s)+3Fe^2+(aq)->2Al^3+(aq)+2Fe(s)

3 on Fe^2+

199
Q

how many moles of electrons are transferred in the following rxn:

CoCl3(aq)+H2(g)->Co(s)+HCl(aq)

A

6 moles e^-

2CoCl3(aq) + 3H2(g) -> 2Co(s) + 6HCl(aq)

200
Q

2CoCl3(aq) + 3H2(g) -> 2Co(s) + 6HCl(aq)

What is the oxidizing agent?

A

Co^3+

H2 was oxidized and is reducing agent

Co^3+ is reduced and is oxidizing agent

201
Q

2CoCl3(aq) + 3H2(g) -> 2Co(s) + 6HCl(aq)

What is the oxidizing agent?

A

Co^3+

H2 was oxidized and is reducing agent

Co^3+ is reduced and is oxidizing agent

202
Q

What is the Estandard of:

Sn^4+(aq) + Ni(s)->Ni^2+(aq)+Sn^2+(aq)

Sn Estnard = 0.15V
Ni Estandard = -0.23

A

Spontaneous

0.15 -(-0.23) = 0.38V

203
Q

What is the Estandard of:

Sn^4+(aq) + Ni(s)->Ni^2+(aq)+Sn^2+(aq)

Sn Estnard = 0.15V
Ni Estandard = -0.23

A

Spontaneous

0.15 -(-0.23) = 0.38V

204
Q

What is Ecell of:

Cd(s)+2Ag^+(aq)-> Cd^2+(aq)+2Ag(s)

Cd^2+ = E = -0.40V
Ag^+ = E = 0.80V
A
  1. 20V

0. 80-(-0.40) = 1.20V

205
Q

What is deltaG of:

Cd(s)+2Ag^+(aq)-> Cd^2+(aq)+2Ag(s)

Cd^2+ = E = -0.40V
Ag^+ = E = 0.80V
A

-232kJ

deltaG=-nFE
deltaG = -(2)(96485)(1.20V)

= -231564J
= -232kJ

206
Q

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

A

-232kJ

deltaG=-nFE
deltaG = -(2)(96485)(1.20V)

= -231564J
= -232kJ

207
Q

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
A

3.9*10^40

deltaG = -RT(lnK)
-231,564J = -(8.314J)(298)(lnK)

K = 3.9*10^40

208
Q

What is not true for a spontaneous reaction?

A

Edegreecell

209
Q

What is true of a spontaneous reaction?

A

deltaG1

deltaSuniv>0

210
Q

Where does oxidation occur in an electrochemical cell?

211
Q

Where does oxidation occur in an electrochemical cell?

212
Q

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
A

0.86V

E = Edegree - (0.0592V/n)(log Q)

E = 0.94 -(0.0592/2)(log(1.8/0.055)^2)

213
Q

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

A

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
Q

write a nuclear equation for the alpha decay of 241/95Am

A

241/95Am->4/2He+237/93Np

215
Q

Determine the identity of the daughter nuclide from the beta decay of 14/6C

A

14/7N

14/6C -> 0/-1e+14/7N

216
Q

the following reaction represents what nuclear process?

137/55Cs+0/-1e->137/54Xe

A

Electron capture

217
Q

Stable isotopes, with low atomic numbers, have a N/Z ratio of 1. What does this imply?

A

the number of neutrons equals the number of protons

218
Q

Stable isotopes, with low atomic numbers, have a N/Z ratio of 1. What does this imply?

A

the number of neutrons equals the number of protons

219
Q

Nuclices above the valley of stability can become more stable through which of the following processes?

A

Beta emission

too many neutrons can be reduced by turning a neutron into a proton via B decay

220
Q

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?

A

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
Q

Nuclear fusion reactions produce energy by

A

combining light nuclei into a heavier nucleus

222
Q

Nuclear fission reactions produce energy by

A

splitting a nucleus into lighter ones

223
Q

Unlike rate constants for chemical reactions, nuclear decay reactions

A

are independent of temperature

224
Q

Radiocarbon dating is accomplished by

A

measuring the amount of C-14 in organic samples

225
Q

Elements with atomic numbers greater than 83 are

A

radioactive

226
Q

Elements with atomic numbers greater than 83 are

A

radioactive

227
Q

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

A
  1. 65*10^13J/mol U
  2. 05258-235.869181 = 0.183399amu = deltam
    (0. 183399amu)(1.6605410^-27kg/1amu)=3.04510^-28kg
E = mc^2
E = (3.045*10^-28kg)(2.9979*10^8m/s)^2 = 2.737*10^-11

(2.73710^-11)(6.02210^23)=1.65*10^13 J/mol U

ACS Chem Final (1 decks)

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