ACS GEN CHEM 2 EXAM REVIEW WITH
VERIFIED QUESTIONS WITH 100%
ANSWERS
I. 800 g ethanol was added to 8*10^3 g water. How much would this lower the
freezing point? (Kf = 1.86)
- ANSWER-(800 g ethanol) (1 mol / 46 g ) = 17.39 mol / 8 kg H2O
II. = 2.17 m
III. ∆Tf = (1.86)(2.17)
IV. X - 0 = 4.04
V. x = 4.04 degrees celsius
VI. a gas or vapor may be liquified only at temperatures
- ANSWER-at or below the critical temperature
VII. -above the critical temperature it is not possible to liquify gas
VIII. a mixture of 100 g of K2Cr2O7 and 200 g of water is stirred until no more of the
salt dissolves. The resulting solution is poured off, leaving the undissolved
behind. the solution is now cooled to 20C. What mass of K2Cr2O7 crystallizes
from the solution during the cooling?
- ANSWER-1. the solubility of K2Cr2O7 at 60C is 40 per 100 so 80 grams of
K2Cr2O7 will remain in solution. 20 grams will precipitate out.
IX. 2 at 20C only ~9 g per 100 dissolves so ~18 grams total remain in solution
X. you have 80 g and 18 g dissolves so you are left with 62 grams that didn't
dissolve
XI. a particular chemical reaction has a negative ∆H and a negative ∆S. What is true
- ANSWER-the reaction becomes spontaneous as temperature decreases
XII. ∆G = ∆H - T∆S = -∆H - T-∆S = -∆H + T∆S
XIII. if T is increased it will be more positive, if T decreased it becomes more negative
XIV. spontaneous when G < 0
XV. a particular solid is soft, poor conductor of heat and electricity, and has low
melting point. such solid is classified as
- ANSWER-molecular
XVI. -KNOW TABLE that summarizes bonds and properties!!
, XVII. -molecular solid has low conductivity and low melting point
XVIII. soft, low to moderate melting point, poor thermal and electrical conductors
XIX. a simple method of showing experimentally that a solid substance may be ionic is
to show that it
- ANSWER-conducts electricity in when dissolved in water
XX. -unique to ionic compounds
XXI. A2 + B2 > 2 AB
XXII. if the concentrations of both a2 and b2 are doubled the the reaction rate will
change by a factor of
- ANSWER-4
XXIII. according to the phase diagram showing the gas, liquid and solid phases of a
pure substance what phase or phases can be present at point x (phase diagram
where x lies on boiling point curve)
- ANSWER-liquid and gas
XXIV. -know where each state lies on diagram
XXV. three phases are always in same relative positions with varying slopes. if point
lies on the phase boundary line both phases can occur together
XXVI. an atom of the element of atomic number 84 and mass number 199 emits an
alpha particle. the residual atom after this change has an atomic number of
_______ and a mass number of _______.
- ANSWER-when an atom emits an alpha it looses 2 protons and 2 neutrons
XXVII. So atomic number Z which represents the number of protons is decreased by 2 :
here 84-2 =82
XXVIII. and The mass number A which represents number of protons + neutrons is
dereassed by 4: here 199-4= 195
XXIX. side note: beta particles result in gain of one proton and loss of one neutron, so
mass stays the same and atomic number increases by one
XXX. calculate the rate law for the reaction
XXXI. 2N2O5>< 4NO2 + O2
XXXII. 1 .... .15 ..... .30 ..... 46
XXXIII. 2 .... .20 .... .60 .... 61
XXXIV. 3...... .20 .... .30 .... 61
- ANSWER-1. use 2 and 3 to solve for exponent of [O2] = 0
XXXV. 2 use 1 and 3 to solve for exponent of [N2O5] = 1
VERIFIED QUESTIONS WITH 100%
ANSWERS
I. 800 g ethanol was added to 8*10^3 g water. How much would this lower the
freezing point? (Kf = 1.86)
- ANSWER-(800 g ethanol) (1 mol / 46 g ) = 17.39 mol / 8 kg H2O
II. = 2.17 m
III. ∆Tf = (1.86)(2.17)
IV. X - 0 = 4.04
V. x = 4.04 degrees celsius
VI. a gas or vapor may be liquified only at temperatures
- ANSWER-at or below the critical temperature
VII. -above the critical temperature it is not possible to liquify gas
VIII. a mixture of 100 g of K2Cr2O7 and 200 g of water is stirred until no more of the
salt dissolves. The resulting solution is poured off, leaving the undissolved
behind. the solution is now cooled to 20C. What mass of K2Cr2O7 crystallizes
from the solution during the cooling?
- ANSWER-1. the solubility of K2Cr2O7 at 60C is 40 per 100 so 80 grams of
K2Cr2O7 will remain in solution. 20 grams will precipitate out.
IX. 2 at 20C only ~9 g per 100 dissolves so ~18 grams total remain in solution
X. you have 80 g and 18 g dissolves so you are left with 62 grams that didn't
dissolve
XI. a particular chemical reaction has a negative ∆H and a negative ∆S. What is true
- ANSWER-the reaction becomes spontaneous as temperature decreases
XII. ∆G = ∆H - T∆S = -∆H - T-∆S = -∆H + T∆S
XIII. if T is increased it will be more positive, if T decreased it becomes more negative
XIV. spontaneous when G < 0
XV. a particular solid is soft, poor conductor of heat and electricity, and has low
melting point. such solid is classified as
- ANSWER-molecular
XVI. -KNOW TABLE that summarizes bonds and properties!!
, XVII. -molecular solid has low conductivity and low melting point
XVIII. soft, low to moderate melting point, poor thermal and electrical conductors
XIX. a simple method of showing experimentally that a solid substance may be ionic is
to show that it
- ANSWER-conducts electricity in when dissolved in water
XX. -unique to ionic compounds
XXI. A2 + B2 > 2 AB
XXII. if the concentrations of both a2 and b2 are doubled the the reaction rate will
change by a factor of
- ANSWER-4
XXIII. according to the phase diagram showing the gas, liquid and solid phases of a
pure substance what phase or phases can be present at point x (phase diagram
where x lies on boiling point curve)
- ANSWER-liquid and gas
XXIV. -know where each state lies on diagram
XXV. three phases are always in same relative positions with varying slopes. if point
lies on the phase boundary line both phases can occur together
XXVI. an atom of the element of atomic number 84 and mass number 199 emits an
alpha particle. the residual atom after this change has an atomic number of
_______ and a mass number of _______.
- ANSWER-when an atom emits an alpha it looses 2 protons and 2 neutrons
XXVII. So atomic number Z which represents the number of protons is decreased by 2 :
here 84-2 =82
XXVIII. and The mass number A which represents number of protons + neutrons is
dereassed by 4: here 199-4= 195
XXIX. side note: beta particles result in gain of one proton and loss of one neutron, so
mass stays the same and atomic number increases by one
XXX. calculate the rate law for the reaction
XXXI. 2N2O5>< 4NO2 + O2
XXXII. 1 .... .15 ..... .30 ..... 46
XXXIII. 2 .... .20 .... .60 .... 61
XXXIV. 3...... .20 .... .30 .... 61
- ANSWER-1. use 2 and 3 to solve for exponent of [O2] = 0
XXXV. 2 use 1 and 3 to solve for exponent of [N2O5] = 1
