# Chemistry homework ```Chemistry homework
7.76) For the unbalanced combustion reaction shown below, 1 mol of ethanol,
C2H5OH+O CO2 + H2O
(a) Write a balanced equation for the combustion reaction.
(b) What is the sign of ΔH for this reaction?
(c )How much heat (in kilocalories) is released from the combustion of 5.00 g of ethanol?
(d)How many grams of C2H5OH must be burned to raise the temperature of 500.0mL of
water from 20.0C to 100.0C?(the specific heat of water is 1.00 cal/g*C.
(e)If the density of ethanol is .789g/mL, calculate the combustion energy of ethanol in
kilocalories/milliliter.
(a)
C2H5OH(l) + 3O2(g)-----&gt; 2CO2(g) + 3H2O(g)
(b)
Combustion releases energy, therefore the reaction is exothermic.
H has a negative sign
(c)
1) Heat released = 1.44 kcal/mole * (1 mole/18 g H2O) * 29 g water = 2.32 kcal = 2.3
kcal
(d)
Density of water = 1.00 g/mL
Mass of water = density of water * volume of water = 1.00 * 500.0 = 500 g
From the balanced equation, 1 mole of C2H5OH produces 3 moles of H2O,
Therefore, 500 g H2O = (500/3) g C2H5OH = 166.67 g C2H5OH
(e)
Density = .789g/mL
Mass = density/volume = 0.789 g/mL
Relative molecular mass of ethanol = 46.1 g/mol
Number of moles = mass/relative molecular mass = 0.789 mL/46.1 = 0.017 mL
Energy = specific heat capacity of water x mass of water x change in water temperature
Energy = 1.00 * 500 * (100 – 20) = 40 kcal
Assume that all heat produced from burning ethanol has gone into heating the water, i.e.
no heat has been wasted.
0.017 mL ethanol produced 40 kcal of heat
Therefore, 1 mole of ethanol would produce 40 kcal/0.017 mL = 2352.94 kcal/mL
The heat of combustion of ethanol is 2352.94 kcal/mL
7.77)For the product of ammonia from its elements, ΔH= -22kcal/ mol
(a) Is this process endothermic or exothermic?
(b) How many kilocalories are involved in the production of .700 mol of NH3?
(a)
The process is exothermic because H is negative.
(b)
H evolved  0.700 mol NH 3 x
 22 kcal
 15.4 kcal
1 mol NH 3
7.78) Magnetite, an iron ore with formula Fe3O4, can be reduced by treatment of
hydrogen yield iron metal and water vapor.
(a) Write the balanced equation
(b)This process requires 36k cal of every 1.00 mol of Fe3O4 reduced. How much
energy(in kilocalories) is required to produce 55 g of iron?
(c ) How many grams of hydrogen are needed to produce 75 grams of iron?
(d)This reaction has K= 2.3 x 10^-18. Are the reactants or the products favored?
(a)
Balanced equation:
Fe3O4 + 4H2 = 3Fe + 4H2O
(b)
Heat required 
(1 mole Fe 3 O 4 )
1mole
36 kcal
x
x
x 55 g Fe  11.82 kcal
1mole Fe3 O 4
3 moles Fe
55.85 g Fe
(c)
Grams of H 2  75 g Fe x
1mole Fe 4 moles H 2
2.02 g
x
x
 3.62 g H 2
55.85 g Fe 3 mole Fe mole H 2
(d)
The concentration of products is much less than that of reactants at equilibrium.
The reaction is strongly reactant-favored.
7.79) Hemoglobin (Hb) reacts reversibly with O2 to form HbO2, a substance that
transfers oxygen to tissues:
Hb(aq) + O2 (aq) &lt; -- &gt;HbO2 (aq)
Carbon monoxide(CO) is attracted to Hb 140 times more strongly O2 and establishes
another equilibrium.
(a) Explain, using chatelier’s priciple, why inhalation of CO can cause weakening and
eventual death.
(b) Still another equilibrium is established when both O2 and CO are present:
Hb(CO)(aq) + O2 (aq)&lt; -- &gt; HbO2(aq) + CO (aq)
Explain , using Le Chaterliers principle, why pure oxygen is oftern administered to
victims of CO poisoning.
(a)
O2 has a partial pressure of 0.2 atm, which is sufficient enough to allow the molecules of
O2 to be taken up by haemoglobin. As a result of this, the red pigment of blood becomes
loosely bound in a complex known as oxyhaemoglobin. The concentration of O2 is
reduced to almost 50% at the ends of the capillaries that deliver the blood to the tissues as
a result of its consumption by the cells. The equilibrium is then shifted to the left, which
releases oxygen in order for it to be diffused into the cells.
(b)
Carbon monoxide is a product of incomplete combustion. It basically blocks the transport
and uptake of oxygen by setting up a complete equilibrium. CO poisoning is often treated
by administering pure O2 because the administration of pure O2 promotes the shifts the
equilibrium to the left.
7.80) Many hospitals administer glucose intravenously to patients. If 3.8 kcal is provided
by each gram of glucose, how many grams must be administered to maintain a persons
normal basal metabolic needs of about 1700kcal/day?
1 gram = 3.8 kcal
 1700/3.8 = 447.37 kcal
Therefore, 447.37 kcal/day must be administered to maintain a person’s normal basal
metabolic needs of 1700 kcal/day
7.81 For the evaporation of water, H2O(l)  H2O(g), at 100C, ΔH = +9.72kcal/mol.
(a) How many kilocalories are needed to vaporize 10.0g of H2O (l)?
(b)How many kilocalories are released when 10.0g of H2O(g) is condensed?
(a)
H vaporization  0.56 mol H 2 O (l ) x
9.72 kcal
 5.44 kcal
1 mol H 2 O (l )
(b)
H released  0.56 mol H 2 O ( g ) x
9.72 kcal
 5.44 kcal
1 mol H 2 O ( g )
7.82) Ammonia reacts slowly in air to produce nitrogen monoxide and water vapor:
NH3(g) + O2(g) &lt; -- &gt;NO(g) + Heat
(a) balance the equation
(b) write the equilibrium equation
(c ) Explain the effect on the equilibrium of:
(1) Raising the pressure
(3) Decreasing the concentration of NH3
(4) Lowering the temperature
(a)
Balanced equation:
4NH3 + 5O2 -----&gt; 4NO + 6H2O
(b)
4NH3(g) + 5O2(g) &lt;-----&gt; 4NO(g) + 6H2O(g)
(c)
(1) Raising the pressure:
According to Le Chatelier, the position of equilibrium will move so that the pressure is
reduced again. The reaction will reduce the pressure by reacting in such a way as to
produce fewer molecules. The equilibrium will shift to the left because an increase in
pressure on a gas reaction shifts the position of equilibrium towards the side with fewer
molecules.
According to Le Chatelier, the position of equilibrium will move in such a way as to
counteract the change causing it. Therefore, the position of equilibrium will shift to the
left so that the concentration of NO decreases again.
(3) Decreasing the concentration of NH3
More NO and H2O will react to replace the NH3 that has been removed. The position of
equilibrium will shift to the left.
(4) Lowering the temperature
The equilibrium will move in such a way that the temperature increases again. The
position of equilibrium will move to the right so as to absorb the heat.
7.83) Methanol, CH3OH, is used as race car fuel.
(a) Write the balanced equation for the combustion of methanol.
(b) ΔH = -174 kcal/mol methanol for the process. How many kilocalories are released by
burning 50.0g of methanol?
(a)
Balanced equation for the combustion of methanol:
3
CH3OH (l) +
O2(g) -----&gt; CO2(g) + 2H2O(g)
2
(b)
Balanced equation indicates that -174 kcal/mol are needed to combust 1 mol CH3OH (l)
Using this relationship, the conversion factor is
 174 kcal
1 mol CH 3 OH (l )
Calculate the number of moles of methanol as follows;
1 mol
50.0 g x
 1.5625 mol
32 g
 174 kcal
H released  1.5625 mol CH 3OH (l ) x
 271.875 kcal
1 mol CH 3OH (l )
7.84 Sketch an energy diagram for a system in which the forward reaction has Eact = +25
kcal/mol and the reverse reaction has Eact = +35 kcal/mol.
(a) Is the forward process endergonic or exergonic?
(b)What is the value of ΔG for the reaction?
(a)
The forward process is endergonic
(b)
ΔGforward = +25 kcal
ΔGreverse = +35 kcal
7.85) The thermite reaction, in which aluminum metal reacts with iron (III) oxide to
produce a spectacular display of sparks, is so exothermic that the product (iron) is the
molten state:
2Al(s) + Fe2O3(s) 2Al2O3(s) + 2 Fe (l)
ΔH= -202.9 kcal/mol
How much heat ( in kilocalories) is released when 5.00 g of Al is used in the reaction?
Balanced equation indicates that 2 moles of Al (s) is reacted and -202.9 kcal/mol are
needed.
Using this relationship, the conversion factor is
 202.9 kcal
2 mol Al ( s )
Calculate the number of moles of methanol as follows;
1 mol
5.00 g x
 0.185322 mol
26.98 g
 202.9 kcal
H rxn  0.185322 mol Al ( s ) x
 18.80 kcal
2 mol Al ( s )
8.108) If 3.0L of hydrogen and 1.5L of oxygen at STP react to yield water, how many
moles of water are formed? What gas volume does the water have at a temperature of
100C and 1 atm pressure?
The chemical reaction between Hydrogen gas (H2) and Oxygen gas (O2) is as follows,
H2(g) + &frac12;O2(g) -----&gt; H2O(g)
1 moles of Hydrogen gas plus 0.5 mole of Oxygen gas, reaction to form 1 mole of water.
For every 1 mole of Oxygen gas reacted, 2 moles of Hydrogen gas are required.
1.5L of Oxygen gas, so 1.5 * 2 = 3.0L of Hydrogen gas are required.
3.0L Hydrogen gas, so 3*1 = 3L of Oxygen gas required.
But there is only 1.5L of Oxygen gas, so Oxygen gas is then the limiting reagent and
Hydrogen gas is in excess.
Therefore, there are 3.0 L of water.
Relative molecular mass of H2O = 18
i.e. 1 mole of H2O = 18 g
1 mole of water occupies 22.4 L at s.t.p
Density = mass/volume = 18/22.4 = 0.804 g/L
Mass = 0.804*3 = 2.412 g
Number of moles = 2.412/18 = 0.134
PV = nRT
P = 1 atm = 101.325 kPa
n = 0.134
R = 8.3 J/mol/K
T = 100oC = 373K
 V = nRT/P = (0.134*8.3*373)/101.325 = 4.09 L
8.109)Approximately 240 mL/min of CO2 is exhaled by an average adult at rest.
Assuming a temperature of 37C and 1 atm pressure, how many moles of CO2 is this?
PV = nRT
P = 1 atm = 101.325 kPa
n=?
V = 240 mL/min = 240 mL * 60 = 14.4 L
R = 8.3 J/mol/K
T = 37oC = 310K
 n = (PV)/(RT) = (101.325*14.4)/(8.3*310) = 0.567
8.110 How many grams of CO2 are exhaled by an average resting adult in 23 hrs? (see
problem 8.109)
Mass = number of moles * relative molecular mass
Relative molecular mass of CO2 = 12+32 = 44g
 mass = 0.576*44*23= 573.804 g/23 hr
8.111 Imagine that you have two identical containers, one containing hydrogen at STP
and the other containing oxygen at STP. How can you tell which is which without
opening them?
You can tell which is which without opening them by calculating the densities of the two
containers and then comparing them:
1 mole of any gas occupies 22.4L at STP
Density = mass/volume
1 mole of H2 = 2 g
Density of H2 @ STP = 2/22.4 = 0.089 g/L
1 mole of O2 = 32 g
Density of O2 @ STP = 32/22.4 = 1.43 g/L
Therefore, the container with the greater density will be identified to contain oxygen.
8.112 When fully inflated, a hot-air balloon has a volume of 1.6 x 10^5 L at an average
temperature of 175 K and 0.975atm. Assuming that air has an average molar mass of 29
g/mol, what is the density of the air in the hot-air balloon? How does this compare with
the density of air at STP.
PV = nRT
P = 0.975 atm = 101.325*0.975 = 98.79 kPa
n=?
V = 1.6 x 105 L
R = 8.3 J/mol/K
T = 175 K
 n = (PV)/(RT) = (98.79*1.6x105)/(8.3*175) = 10882.2
8.113) A 10.0g sample of an unknown gas occupies 14.7 L at a temperature of 25C and a
pressure of 745 mmHg. What is the molar mass of the gas?
PV = nRT
P = 745 mmHg, 1 mmHg = 0.133 kPa 745 mmHg = 745 * 0.133 = 99.085 kPa
n=?
V = 14.7 L
R = 8.3 J/mol/K
T = 25 oC = 298 K
 n = (PV)/(RT) = (99.085*14.7)/(8.3*298) = 0.589
Number of moles = volume/molar mass
 molar mass = volume/number of mole = 14.7*0.589 = 8.6583
8.114) One mole of any gas has a volume of 22.4 L at STP. What are the molecular
weights of the following gases, and what are their densities in grams per liter of STP?
(a) CH4
(b) CO2
(c ) O2
(a)
Molecular weight of CH4 = [12+(1*4)] = 16
1 mole CH4 = 16 g
i.e. 16 g CH4 occupy 22.4 L at s.t.p.
Density = Mass/volume
 density = 16/22.4 = 0.714 g/L
(b)
Molecular weight of CO2 = [12+ (16*2)] = 44
1 mole CO2 = 44 g
i.e. 44 g g CO2 occupy 22.4 L at s.t.p.
Density = Mass/volume
 density = 44/22.4 = 1.96 g/L
(c)
Molecular weight of CO = (12+16) = 28
1 mole CO = 28 g
i.e. 28 g CO occupy 22.4 L at s.t.p.
Density = Mass/volume
 density = 28/22.4 = 1.25 g/L
8.115) Gas pressure outside the space shuttle is approximately 1x10^-14 mm Hg at a
temperature of approximately 1 K. If the gas is almost entirely hydrogen atoms what
volume of space is occupied by 1 mol of atoms? What is the density of H gas in atoms
per liter?
PV = nRT
P = 1x10-14 mmHg, 1 mmHg = 0.133 kPa
P = 1x10-14 mmHg = 1x10-14 * 0.133 = 1.33x10-14 kPa
n=1
V=?
R = 8.3 J/mol/K
T=1K
 V = (nRT)/P = (1*8.3*1)/(1.33x10-14) = 6.24x1015 L
Density = mass/volume
Mass of H = number of moles * relative molecular mass
=1*1=1g
 density = 1/6.24x1015 = 8.065x10-16 g/L
8.116) Ethylene glycol, C2H6O2, has one OH bonded to each carbon.
(a) Draw the lewis dot structure of ethylene glycol.
(b) Draw the Lewis dot structure of chloromethane, C2H5cl
(c )Chloroethane has a slightly higher molar mass than ethylene glycol, but much lower
boiling point (3C versus 198C) Explain.
(a)
H
H
C
C
H
H
H
H
C
C
H
H
(b)
H
(c)
Chloroethane has much lower boiling point than ethylene glycol because ethers are more
soluble in water than alkanes of similar molecular mass. Ethylene glycol is able to form
hydrogen-bond unlike alkanes which cannot hydrogen-bond to one another.
8.117) A rule of thumb for scuba diving is that the external pressure increases by 1 atm
for every 10 m of depth. A diver using a compressed air tank is planning to descend to a
depth of 25m.
(a) What is the external pressure at this depth?
(b) assuming that the tank contains 20% oxygen and 80% nitrogen, what is the partial
pressure of each gas in the divers lungs at this depth?
(a)
1 atm = 10m
 25m = 2.5 atm
2.5 atm = 2.5*0.133 kPa = 0.3325 kPa
Therefore, the external pressure at this depth is 0.3325 kPa
(b)
20% O2
80% N2
Partial Pressure = 0.20 atm and 0.80 atm
@ 2.atm
Oxygen = 0.20*2.5 = 0.5 pp
Nitrogen = 0.80*2.5 = 2 pp
Total partial pressure = 0.5 + 2 = 2.5 pp
Concentration of the gases inhaled:
50% O2
200% N2
9.101) One test for vitamin C(ascorbic acid, C6H8O6) is based on the reaction of the
vitamin with iodine:
(a) if 25.0 mL of a fruit juice requires 13.0ml of 0.0100 M I2 solution for reaction, what
is the molarity of the ascorbic acid in the fruit juice?
(b) the food and drug administration recommends that 60mg of ascorbic acid be
consumed per day. How many milliliters of the fruit juice in part (a) must a person drink
to obtain the recommended dosage?
(a)
C6H8O6(aq) + I2(aq)  C6H6O6(aq) + 2 HI(aq)
Moles of C6H8O6 = moles of I2 (from the stoichiometry of the balanced chemical
reaction)
Moles I2 = molarity
Volume x Molarity 0.0100 x 13.0
Number of moles of I 2 

 1.3x10  4 mol I 2
1000
1000

mol
C
H
6
8 O6 
1.3x10  4 mol I 2 x
 1.3x10  4 mol C 6 H 8 O6
mol I 2
1.3x10  4 mol C 6 H 8 O6 x 25
Molarity C 6 H 8 O6 
 5.2 x 10 3 M
1000
(b)
5.2 x10
3
 176.13 g C 6 H 8 O6 
 x 1000 mg / g  x L / 1000 mL
mol / L x
 mol C 6 H 8 O6 
915.9 mg C 6 H 8 O6

1000 mL

 1000 mL 

 x 60.0 mg C 6 H 8 O6  65.5 mL
 915.9 mg 
9.102 Ringers solution, used in the treatment of burns and wounds, is prepared by
dissolving 8.6g of NaCl, 0.30g of KCl, and .33g of CaCl2 in water and diluting to a
volume of 1.00L. What is the molarity of each component?
NaCl:
Relative molecular mass of NaCl = 23+35.5 = 58.5
Mass
8.6
Number of moles 

 0.147
Re lative molecular mass 58.5
Volume x Molarity
1000
Moles *1000
 Molarity 
Volume
0.147 *1000
 Molarity of NaCl 
 147 M
1.00
Number of moles 
KCl:
Relative molecular mass of KCl = 39+35.5 = 74.5
Mass
0.3
Number of moles 

 0.004027
Re lative molecular mass 74.5
Volume x Molarity
Number of moles 
1000
Moles * 1000
 Molarity 
Volume
0.004027 *1000
 Molarity of KCl 
 4.027 M
1.00
CaCl2:
Relative molecular mass of CaCl2 = [40+(35.5*2)] = 111
Mass
0.33
Number of moles 

 0.00297
Re lative molecular mass 111
Volume x Molarity
Number of moles 
1000
Moles *1000
 Molarity 
Volume
0.00297 *1000
 Molarity of KCl 
 2.97 M
1.00
9.103 what is osmolarity of ringers solution (see problem9.102)? Is it hypotonic, isotonic,
or hypertonic with blood plasma( 0.30 osmol)?
NaCl:
Molarity = 147
Osmolarity = IM = 2*147 = 294 osmol and hypertonic
KCl:
Molarity = 4.027
Osmolarity = IM = 2*4.027 = 8.054 osmol and hypertonic
CaCl2:
Molarity = 2.97
Osmolarity = IM = 2*2.97 = 5.94 osmol and hypertonic
9.104 The typical dosage of stain drugs for the treatment of high cholesterol is 10mg.
Assuming a total blood volume of 5.0L, calculate the concentration of drug in the blood
in units of (w/v)%.
10 mg = 0.01 g
5.0 L = 5000 mL
 (w/v)% = 0.01/5000 = 0.0002 (w/v)%
9.105 Assuming the density of blood in healthy individuals is approximately 1.05g/ml,
report the concentration of drug in problem 9.104 in units of ppm.
0.01 g per 100 cm3 = 0.01 x 5 g per litre
= 0.05 g/L
= 0.05 x 1000 mg/L
=50 mg/L
= 50 ppm
9.106 In many states, a person with blood alcohol concentration of 0.080% (v/v) is
considered legally drunk. What volume of total alcohol does concentration represent,
assuming a blood volume of 5.0L?
Volume of total alcohol = 5.0 L blood * (1000 mL/L) * (0.080 mL alcohol/100 mL blood
= 4.0 mL
9.107) Ammonia is very soluble in water( 51.8g/L at 20C and 760mmHg).
(a)Show how NH3 can hydrogen bond to water.
(b)What is the solubility of ammonia in water in moles per liter?
(a)
NH3 has five electrons in its outermost shell. Three electrons are already occupied by 3
hydrogens. The remaining lone pair is 1 so this lone pair can attract only one H-atom.
NH3 + H2O -----&gt; NH4+ + OH-
(b)
51.8 g 1 mol
x
 3.047 M
1L
17 g
9.108) Cobalt (II) chloride, a blue solid, can absorb water from the air to form cobalt (II)
chloride hex hydrate, a pink solid. The equilibrium is so sensitive to moisture in the air
that CoCl2 is used as a humidity indicator.
(a) Write a balanced equation for the equilibrium. Be sure to include water as a reactant
to produce the hex hydrate.
(b) How many grams of water are released by the decomposition of 2.50g of cobalt (II)
chloride hexahydrate?
(a)
CO(H2O)4Cl2 + 2H2O &lt;-----&gt; Co(H2O)6Cl2
(b)
Relative molecular mass of Co(H2O)6Cl2 = 237.93
Number of moles of Co(H2O)6Cl2 = 2.50/237.93 = 0.011
 0.011 mol Co(H2O)6Cl2 x (6 mole H2O/1 mole H2O) = 0.066 mole H2O
9.109 How many milliliters of 0.150 M BaCl2 are needed to react completely with
35.0ml of .200 M Na2SO4? How many grams of BaSO4 will be formed?
M 1V1
M V
0.200 * 35.0
 V1  2 2 
 46.67 mL
M 2V2
M1
0.150
BaCl2 + Na2SO4 -----&gt; 2NaCl + BaSO4
46.67 mL BaCl2 = 0.04667 L
35.0 mL Na2SO4 = 0.035 L
0.150 M BaCl2 = moles BaCl2/0.04667 L = 0.007 mole BaCl2
0.200 M Na2SO4 = moles Na2SO4/0.035 L = 0.007 mole Na2SO4
The ratio of BaCl2 to Na2SO4 is one to one;
Therefore, either (BaCl2 or Na2SO4) mole count is capable of driving the reaction.
Using 0.007 mole Na2SO4:
0.007 mole Na2SO4 (1 mole BaSO4/1 mole Na2SO4)(233.37 grams/1 mole BaSO4)
= 1.63 grams of BaSO4 will be formed
9.110 Many compounds are only partially dissociated into ion in aqueous solution.
Trichloroaetic acid ( CCl3CO2H), for instance, is partially dissociated in water according
to the equation. CCL3CO2H(aq) &lt; ---- &gt; H+ (aq) +CCL3CO2-(aq) For a solution
prepared by dissolving 1.00 mol of trichloroacetic acid dissociates to from H+ and
CCL3CO2- ions. (a) what is the total concentration of dissolved ions and molecules in
1kg of water? (b)what is the freezing point of this solution?(the freezing point of 1kg of
water is lowered 1.86C for each mole of solute particles.)