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Homework 5 Chapter 05
Homework 5 Chapter 05
Due: 11:59pm on Wednesday, November 2, 2016
You will receive no credit for items you complete after the assignment is due. Grading Policy
± Work and Energy
An object can possess kinetic energy, the energy of motion, and/or potential energy, the energy of position. Energy can be
transferred in many ways, one of which is in the form of work. Work, w, is the energy transferred when an object is caused to
move a distance, d, against a force, F . Mathematically, we would write this as
w = Fd
Force is the product of mass, m, and acceleration, a:
F = ma
When the acceleration is due to gravity, the symbol g is used instead of a, and its value is g = 9.81 m ⋅ s−2 .
Part A
How much work is done when a 205 g tomato is lifted 13.1 m ?
Express your answer with the appropriate units.
Hint 1. Determine the relevant work equation
Given that w = F d and F
acceleration of gravity, g?
= mg
, determine the formula for work, w, in terms of mass, m, distance, d, and the
Express work in terms of the variables m, d, and g.
ANSWER:
= w
mgd
ANSWER:
= 26.3 J
w
Correct
The energy transferred to the tomato by doing work on it increases its potential energy because the tomato is now at a
greater height or position relative to the ground.
Kinetic energy
Kinetic energy, Ek , is given by the formula Ek
meters per second.
=
1
2
2
mv
where m is mass is the mass in kilograms and v is velocity in
Part B
The tomato is dropped. What is the velocity, v , of the tomato when it hits the ground? Assume 90.3 % of the work done in
Part A is transferred to kinetic energy, E , by the time the tomato hits the ground.
Express your answer with the appropriate units.
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Hint 1. Calculate the kinetic energy
A total of 90.3 % of the work done in Part A (26.3 J ) is transferred to kinetic energy. Calculate the kinetic energy.
Express your answer with the appropriate units.
ANSWER:
E
= 23.8 J
Hint 2. Rearrange the formula for kinetic energy
Using the symbol E to represent kinetic energy, rearrange the formula
E=
to isolate v .
1
2
2
mv
Express the velocity in terms of the variables E and m.
ANSWER:
v
= −
−
−
√
2E
m
ANSWER:
v
= 15.2 m
s
Correct
Once the tomato hits the ground, it stops moving and its kinetic energy goes to zero. Some of the kinetic energy is
transferred into the work involved in smashing the tomato, while the rest is dissipated to the surroundings as heat.
Sample Exercise 5.1 Practice Exercise 1 with feedback
Part A ­ Describing and Calculating Energy Changes
Which of the following objects has the greatest kinetic energy?
ANSWER:
a 500­kg motorcycle moving at 100 km/h
a 1,000­kg car moving at 50 km/h
a 1,500­kg car moving at 30 km/h
a 5,000­kg truck moving at 10 km/h
a 10,000­kg truck moving at 5 km/h
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Correct
Using the kinetic energy formula, a 500­kg motorcycle moving at 100 km/h has the greatest kinetic energy. Without
doing calculations, this could have been deduced by looking at each value of mass and velocity. While all the other
options have much higher masses, this option has the highest velocity. In the kinetic energy equation, the velocity is
squared therefore, it has a larger impact in these particular calculations than mass does.
Problem 5.91
At 20 ∘ C (approximately room temperature) the average velocity of N2 molecules in air is 1050 mph.
Part A
What is the average speed in m/s?
Express your answer using four significant figures.
ANSWER:
v
= 469.4 m/s Correct
Part B
What is the kinetic energy (in J) of an N2 molecule moving at this speed?
Express your answer using four significant figures.
ANSWER:
K
= 5.124×10−21 J All attempts used; correct answer displayed
Part C
What is the total kinetic energy of 1 mol of N2 molecules moving at this speed?
Express your answer using four significant figures.
ANSWER:
K
= 3.086 kJ/mol Correct
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Sample Exercise 5.2 Practice Exercise 1 with feedback
Part A ­ Relating Heat and Work to Changes of Internal Energy
Consider the following four cases
i. A chemical process in which heat is absorbed.
ii. A change in which q = 30 J, w = 44 J.
iii. A process in which a system does work on its surroundings with no change in q.
iv. A process in which work is done on a system and an equal amount of heat is withdrawn.
In how many of these cases does the internal energy of the system decrease?
ANSWER:
0
1
2
3
4
Correct
The internal energy changes for each case can be summarized as follows:
i. A chemical process in which heat is absorbed results in an increase in internal energy.
ii. A change in which q = 30 J, w = 44 J. Since the change in internal energy is the sum of the heat
added to or liberated from the system, q, and the work done on or by the system, w, this change results
in an increase in internal energy.
iii. A process in which a system does work on its surroundings with no change in q results in a decrease in
the internal energy because the systems is essentially losing energy to the surroundings.
iv. A process in which work is done on a system and an equal amount of heat is withdrawn has no net
change in the internal energy.
Thus out of the four processes, only (iii) results in a decrease in internal energy.
PV Work
The joule (J) is a unit of energy. Recall that energy may be converted between many different forms such as mechanical
energy, thermal energy (heat), chemical energy, electrical energy, and light.
The mechanical energy produced by a system is called work. When work is accomplished through the changing volume of a
gas, it is called PV work and is given by the formula w = −P ΔV , where w is the work, P is the external pressure, and ΔV is the change in volume. If the volume change and pressure are in liters and atmospheres respectively, then the work will
have units of liter­atmospheres, which can be converted to joules using the conversion factor 1 L ⋅ atm = 101.3 J.
Work can also be expressed as force multiplied by distance: w = F
newtons, and d is the distance in meters. Note that 1 N ⋅ m = 1 J .
, where w is the work in joules, F is the force in
×d
Part A
A piston has an external pressure of 8.00 atm. How much work has been done in joules if the cylinder goes from a volume
of 0.180 liters to 0.590 liters?
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Express your answer with the appropriate units.
Hint 1. Calculate the change in volume
What is the value of the change in volume ΔV ?
Express your answer with the appropriate units.
ANSWER:
ΔV
= 0.410 L
Hint 2. Calculate the work done
Calculate the amount of work done in units of L ⋅ atm. Notice that the units are already entered for you.
Express your answer numerically in liter­atmospheres.
ANSWER:
­3.28 L ⋅ atm Hint 3. Conversion factor
Use the relation 101.3 J
.
= 1 L ⋅ atm
One way to remember this is to divide the two common values of the gas constant R:
8.314 J/(mol⋅K)
0.0821 L⋅atm/(mol⋅K)
=
101.3 J
1 L⋅atm
ANSWER:
= ­332 J
w
Correct
Sample Exercise 5.3 Practice Exercise 1 with feedback
Part A
If a balloon is expanded from 0.055 L to 1.403 L against an external pressure of 1.02 atm, how many L⋅atm of work is
done?
ANSWER:
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­0.056 L⋅atm
­1.37 L⋅atm
1.43 L⋅atm
1.49 L⋅atm
1.39 L⋅atm
Correct
A balloon that is expanded from 0.055 L to 1.403 L against an external pressure of 1.02 atm does ­1.37 L⋅atm
worth of work. Since the volume is expanding, work has to be a negative value.
Problem 5.38 with feedback
A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in the following figure. When 0.480 heat is added to the gas, it expands and does 214 J of work on the surroundings.
You may want to reference (
kJ
of
pages 175 ­ 179) Section 5.3 while completing this problem.
Part A
What is the value of ΔH for this process?
Express the energy in kilojoules to three significant digits.
ANSWER:
ΔH
= 0.480 kJ Correct
For a system under constant pressure, the change in enthalpy is equal to the heat, and the work energy (w) can be
ignored because it effectively cancels. The relationship relating the change in enthalpy (ΔH ) to the change in internal
energy (ΔE ), heat (qp , under constant pressure), and work (= −P ΔV ) is
ΔHsys = (ΔE) + P ΔV = (qp + wsys ) − wsys = qp
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Part B
What is the value of ΔE for this process?
Express the energy in kilojoules to three decimal places.
ANSWER:
ΔE
= 0.266 kJ Correct
The change in internal energy is the sum of heat and work. Heat is gained by the system (gas), thus q = 0.480 kJ.
Additionally, the system (gas) performs work, so w = −0.214 kJ. Summing the two terms provides the value for the
change in internal energy.
Sample Exercise 5.5 Practice Exercise 1 with feedback
Part A ­ Relating ΔH to Quantities of Reactants and Products
The complete combustion of ethanol, C2 H5 OH (FW = 46.0 g/mol), proceeds as follows:
C2 H 5 OH(l) + 3 O2 (g) → 2 CO2 (g) + 3 H 2 O(l) ΔH = − 555 kJ
What is the enthalpy change for combustion of 15.0 g of ethanol?
ANSWER:
­12.1 kJ
­181 kJ
­422 kJ
­555 kJ
­1700 kJ
Correct
To determine the enthalpy change for combustion of 15.0 g of ethanol follow these steps:
determine moles of ethanol in 15 g : (
mass ethanol
FW ethanol
= 15.0 g
46.0 g
= 0.3261 mol)
find the enthalpy change for 0.3261 mol of ethanol: ΔH = ­555 kJ for the combustion of 1 mol of
ethanol, therefore for 0.3261 mol of ethanol −555 kJ/mol ethanol × 0.03261 mol ethanol = −181 kJ.
Coffee Cup Calorimetry
Calorimetry is a method used to measure enthalpy, or heat, changes that occur during chemical processes. Two common
calorimeters are constant­pressure calorimeters and constant­volume (or "bomb") calorimeters. Bomb calorimeters are used to
measure combustion and other gas­producing reactions, where the reaction is observed in a strong, sealed vessel. A simple
constant­pressure calorimeter can be made from a foam coffee cup and a thermometer; energy changes in a reaction are
observed via a temperature change of the solution in the cup. The idea behind calorimeters is that if they are sufficiently
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insulated from the outside environment, any energy gained or lost in the chemical reaction will be directly observable as a
temperature and/or pressure change in the calorimeter.
Part A
A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 129 g of
solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 ∘ C. What is the enthalpy of this
reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the
solution is the same as that of pure water.
Enter your answer in kilojoules per mole of compound to three significant figures.
Hint 1. How to approach this problem.
Enthalpy is the amount of heat absorbed or produced by a reaction:
heat = mass × specif ic heat × ΔT
Endothermic reactions have a positive enthalpy value. Exothermic reactions have a negative enthalpy value.
Hint 2. Specific heat of water
The specific heat of water is 4.184 J/(g ⋅ ∘ C).
Hint 3. Calculate the temperature change
What is the temperature change ΔT for water in this problem?
Express your answer numerically in degrees Celsius to three significant figures.
ANSWER:
ΔT
= 3.70 ∘ C Hint 4. Calculate the heat absorbed by the water
How many joules of heat did the water gain?
Enter your answer numerically in joules to three significant figures.
ANSWER:
2000 J ANSWER:
ΔH
= ­0.999 kJ/mol Correct
­0.999 kJ/mol has a negative sign to indicate the exothermic nature of the reaction.
Sample Exercise 5.7 Practice Exercise 1 with feedback
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Part A ­ Measuring ΔH Using a Coffee­Cup Calorimeter
When 0.243 g of Mg metal is combined with enough HCl to make 100 mL of solution in a constant­pressure calorimeter,
the following reaction occurs:
Mg(s) + 2 HCl(aq) → MgCl2 (aq) + H 2 (g)
If the temperature of the solution increases from 23.0 ∘ C to 34.1 ∘ C as a result of this reaction, calculate ΔH in kJ/mol
of Mg. Assume that the solution has a specific heat of 4.18 J/g∘ C.
ANSWER:
−19.1 kJ/mol
−111 kJ/mol
−191 kJ/mol
−464 kJ/mol
−961 kJ/mol
Correct
Sample Exercise 5.8 Practice Exercise 1 with feedback
Part A ­ Measuring qrxn Using a Bomb Calorimeter
The combustion of exactly 1.000 g of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat. If the combustion of
0.550 g of benzoic acid causes the temperature of the calorimeter to increase from 22.01∘ C to 24.27∘ C, calculate the
heat capacity of the calorimeter.
ANSWER:
0.660 kJ/∘ C
6.42 kJ/∘ C
14.5 kJ/∘ C
21.2 kJ/∘ C
32.7 kJ/∘ C
Correct
Problem 5.55
When a 6.50­g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee­cup calorimeter (the following
Figure), the temperature rises from 21.6 ∘ C to 37.8 ∘ C.
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Part A
Calculate ΔH (in kJ/mol NaOH) for the solution process +
−
NaOH(s) → Na (aq) + OH (aq) Assume that the specific heat of the solution is the same as that of pure water.
Express your answer with the appropriate units.
ANSWER:
­44.4 kJ
mol
Correct
Sample Exercise 5.9 Practice Exercise 1 with feedback
Part A ­ Using Hess’s Law to Calculate ΔH
Calculate ΔH for 2 NO(g) + O2 (g) → N2 O4 (g) using the following information:
N 2 O4 (g)
→
2 NO2 (g)
ΔH
=
+57.9 kJ
2 NO(g) + O2 (g)
→
2 NO2 (g)
ΔH
=
−114.1 kJ
ANSWER:
2.7 kJ
­55.2 kJ
­85.5 kJ
­171.0 kJ
+55.2 kJ
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Correct
Using Hess's Law, reversing the first equation requires that the sign of the associated ΔH value to also be changed.
Once the two equation can be added together to properly obtain the net equation, adding the ΔH of the two reactions
results in a ΔH for 2 NO(g) + O2 (g) → N2 O4 (g) of ­171.0 kJ.
Problem 5.64
Part A
From the enthalpies of reaction
2C(s) + O2 (g)→2CO(g)
ΔH = −221.0kJ
2C(s) + O2 (g) + 4H 2 (g)→2CH 3 OH(g)
ΔH = −402.4kJ
calculate ΔH for the reaction
CO(g) + 2H 2 (g)→CH 3 OH(g)
Express your answer with the appropriate units.
ANSWER:
­90.7 kJ
Correct
Formation Reactions
∘
The standard heat of formation, ΔHf , is defined as the enthalpy change for the formation of one mole of substance from its
∘
constituent elements in their standard states. Thus, elements in their standard states have ΔHf = 0. Heat of formation
values can be used to calculate the enthalpy change of any reaction.
Consider, for example, the reaction
2NO(g) + O2 (g) ⇌ 2NO2 (g)
with heat of formation values given by the following table:
∘
Substance
(kJ/mol)
NO(g)
90.2
O2 (g)
0
NO2 (g)
33.2
ΔH
f
Then the standard heat of reaction for the overall reaction is
ΔH
∘
rxn
=
ΔH
∘
f
(products) −
=
2(33.2)
=
−114 kJ
−
ΔH
∘
f
(reactants)
[2(90.2) + 0]
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Part A
∘
∘
For which of the following reactions is ΔHrxn equal to ΔHf of the product(s)?
You do not need to look up any values to answer this question.
Check all that apply.
Hint 1. How to approach the problem
∘
An element in its standard state has a ΔHf value of zero. Therefore the reaction that defines the heat of formation
of any compound is the reaction that produces one mole of that compound from its constituent elements in their
standard states.
Hint 2. Select the reaction for the formation of sodium fluoride
Choose the reaction that defines the standard heat of formation for NaF.
ANSWER:
2Na(s) + F2 (g)→2NaF(s)
Na(s) +
1
Na(s) +
1
2
2
F2 (l)→NaF(s)
F2 (g)→NaF(s)
Hint 3. Select the reaction for the formation of carbon dioxide
Choose the reaction that defines the standard heat of formation for CO2 .
ANSWER:
CO(g) +
1
2
O 2 (g)→CO 2 (g)
C(s, graphite) + O2 (g)→CO2 (g)
CaCO3 (g)→CaO + CO2 (g)
ANSWER:
CO(g) +
Na(s) +
1
2
1
2
O 2 (g)→CO 2 (g)
F2 (g)→NaF(s)
C(s, graphite) + O2 (g)→CO2 (g)
2Na(s) + F2 (g)→2NaF(s)
Na(s) +
1
2
F2 (l)→NaF(s)
CaCO3 (g)→CaO + CO2 (g)
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Correct
∘
For both of these reactions, the calculation of ΔHrxn would look something like
ΔH
∘
= [1 × ΔH
rxn
∘
f
(product)] − [0 + 0]
If the coefficient for the product were not 1, or if the reactants were not standard­state elements, then the enthalpy
∘
change of the reaction would not be equal to ΔHf for the product.
Part B
The combustion of pentane, C5 H12 , occurs via the reaction
C5 H 12 (g) + 8O2 (g)→5CO2 (g) + 6H 2 O(g)
with heat of formation values given by the following table:
Substance
ΔH
∘
f
(kJ/mol)
­119.9
C5 H 12 (g)
393.5
CO2 (g)
−
H 2 O(g)
−
241.8
Calculate the enthalpy for the combustion of 1 mole of pentane.
Express your answer to four significant figures and include the appropriate units.
Hint 1. How to approach the problem
The standard enthalpy of a reaction is calculated by finding the standard enthalpies of formation for the substances
involved in the reaction, tabulating them, and then using the formula
ΔH
∘
rxn
= ΔH
∘
f
(products) − ΔH
∘
f
(reactants)
Note that O2 (g) is a pure element in its standard state. Thus, the standard heat of formation for oxygen gas is
equal to zero.
Hint 2. Select the correct formula for calculating enthalpy of the reaction
Choose the correct formula for calculating the enthalpy of the reaction.
ANSWER:
∘
ΔHrxn
= [ΔH
∘
f
ΔHrxn
= [5ΔH
∘
ΔHrxn
= [ΔH
∘
ΔHrxn
∘
∘
f
= [(ΔH
(CO2 ) + ΔH
∘
f
∘
f
(H 2 O)] − [ΔH
(CO2 ) + 6ΔH
∘
f
(C5 H 12 ) + 8ΔH
∘
f
∘
(C5 H 12 ) + ΔH (O2 )]
f
(H 2 O)] − [ΔH
∘
f
∘
f
(O2 )] − [5ΔH
∘
f
∘
f
(C5 H 12 ) + 8ΔH
(CO2 ) + 6ΔH
∘
∘
∘
f
f
f
∘
f
∘
f
(O2 )]
(H 2 O)]
(C5 H 12 ) + ΔH (O2 )] − [ΔH (CO2 ) + ΔH (H 2 O)]
Hint 3. Calculate the sum of the products
∘
Taking coefficients into account, what is the sum of the ΔHf values of the products, 5CO2 (g) + 6H2 O(g)?
Express your answer to five significant figures and include the appropriate units.
ANSWER:
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∘
= ­3418.3 kJ
ΔHf (products)
ANSWER:
∘
ΔHrxn
= ­3298 kJ
Correct
Sample Exercise 5.11 Practice Exercise 1 with feedback
Part A ­ Equations Associated with Enthalpies of Formation
If the heat of formation of H2 O(l) is ­286 kJ/mol, which of the following thermochemical equations for the formation of H 2 O is correct?
ANSWER:
H2 (g) + O(g) → H2 O(g) ΔH
H2 O(l) → H2 (g) + 1
2
= −286 kJ
O2 (g) ΔH = 286 kJ
2 H2 (g) + O2 (g) → 2 H2 O(l) ΔH
H2 (g) + 1
2
= −286 kJ
O2 (g) → H 2 O(l) ΔH = −286 kJ
2 H(g) + O(g) → H2 O(l) ΔH
= −286 kJ
Correct
Problem 5.76
Calcium carbide (CaC2 ) reacts with water to form acetylene (C2 H2 ) and Ca(OH)2 .
Part A
From the following enthalpy of reaction data and data in Appendix C, calculate ΔHf∘ for CaC2 (s) : CaC2 (s) + 2 H 2 O(l) → Ca(OH) (s) + C2 H 2 (g)
2
ΔH
∘
= −127.2kJ
ANSWER:
ΔH
∘
f
= ­60.6 kJ Correct
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± Nutritional Energy
Alexander, who weighs 183 lb , decides to climb Mt. Krumpett, which is 5980 m high. For his food supply, he decides to take
nutrition bars. The label on the bars states that each 100­g bar contains 10 g of fat, 40 g of protein, and 50 g of carbohydrates.
One gram of fat contains 9 Calories, whereas each gram of protein and carbohydrates contains 4 Calories.
Nutrition facts
Calories per gram
Fat
9
Protein
4
Carbohydrates
4
Part A
Alexander wants to know exactly how many bars to pack in his backpack for the journey. To provide a margin of safety, he
assumes that he will need as much energy for the return trip as for the uphill climb. How many bars should Alexander
pack?
Enter the exact answer to two significant figures. Do not round to an integer.
Hint 1. How to approach the problem
First calculate the energy needed for Alexander to climb the height of the mountain, which can be determined from
the Alexander's potential energy. Next, calculate how much energy is available in each nutrition bar. Finally,
calculate how many energy bars will be needed for the trip, remembering to double the number of bars needed for
the uphill climb to account for the return trip.
Hint 2. Calculate the energy needed to climb the mountain
The energy needed to climb the mountain is the difference between Alexander's potential energy at the foot of the
mountain and his potential energy at the top and this is equal to the gravitational potential energy.
How much energy is needed by Alexander to climb the mountain?
Express your answer in joules to four significant figures.
Hint 1. How to determine the gravitational potential energy
Gravitational potential energy is the energy stored in an object that is raised to a height. The gravitational
potential energy is related to an object's mass m, the height h to which it is raised, and the acceleration due
to gravity, g. The relationship is given by E = m ⋅ g ⋅ h. The value of g near Earth's surface is 9.81 m/s2 .
Hint 2. Convert the mass to metric units
In the gravitational potential energy equation, the mass of the object (in this case the climber) needs to be
entered in kg (kilograms). Recall that 1 pound is the equivalent of 454 g, or 0.454 kg. What is the mass of
the object (in this case the climber) in kilograms?
Express your answer in kilograms to three significant figures.
ANSWER:
mass = 83.1 kg ANSWER:
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energy needed = 4.874×106 J Hint 3. Calculate the energy in a nutrition bar
Convert the amount of fat, protein, and carbohydrates in each nutrition bar into joules of available energy.
Express your answer in joules to four significant figures.
Hint 1. Calculate the number of Calories in each bar
Nutritional information is usually provided in units of Calories on foods that we buy. For the nutrition bar
Alexander wishes to carry, one gram of fat contains 9 Calories, whereas one gram of protein and one gram of
carbohydrates each contain 4 Calories. How many Calories are in one nutrition bar?
Express your answer in Calories to three significant figures.
ANSWER:
450
Calories
Hint 2. Calorie to joule conversion
There are 1000 calories in a Calorie (the capital letter matters!). There are exactly 4.184 J (joules) in one
calorie:
1 Cal = 1000 cal = 4.184 × 1000 J
ANSWER:
energy in one bar = 1.883×106 J ANSWER:
5.2 bars Correct
Most people would not pack a fraction of a nutrition bar. For example, if your calculation told you to take 4.3 bars, you
would realistically pack 5 whole bars.
Sample Exercise 5.14 Practice Exercise 1 with feedback
Part A ­ Estimating the Fuel Value of a Food from Its Composition
A stalk of celery has a caloric content (fuel value) of 9.0 kcal. If 1.0 kcal is provided by fat and there is very little protein,
estimate the number of grams of carbohydrate and fat in the celery.
ANSWER:
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11/10/2016
Homework 5 Chapter 05
2.2 g carbohydrate and 0.1 g fat
2 g carbohydrate and 1 g fat
32 g carbohydrate and 10 g fat
1 g carbohydrate and 2 g of fat
2 g carbohydrate and 0.1 g fat
Correct
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