energy & chemistry peanuts supply sufficient energy to boil a cup of water.burning peanuts...
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Energy & ChemistryEnergy & Chemistry
• Burning peanuts supply sufficient energy to boil a cup peanuts supply sufficient energy to boil a cup of water.of water. • Burning sugar (sugar Burning sugar (sugar
reacts with KClOreacts with KClO33, a , a
strong oxidizing agent)strong oxidizing agent)
Ch. 6 Energy and Chemistry1. Vocabulary
2. Heat using specific heat and heat capacity
3. Enthalpy from single reactiona) Single reagent
b) Limiting reagent
4. Enthalpy from multiple reactionsa) Hess’ Law
5. Enthalpy of reaction from ∆Hf°
Chapter 6: Energy and Chemical Change
• Energy is the ability to do work (W = F×D or P×V) and/or supply heat
• Kinetic energy is the energy an object has because of its motion– For an object of mass m with velocity v
mghEnergyPotential
mv
221 KE Energy Kinetic
• The law of conservation of energy states that energy cannot be created or destroyed
6.1 An object has energy if it is capable of doing work 5
Kinetic Energy: The Energy Of Motion
• KE=½mv2
• Energy can be transferred by moving particles
• Collision of fast particles with slower particles causes the slow particle to speed up while the fast molecule slows– this is why hot water cools in contact with
cool air
Potential Energy
Breaking Bonds requires energy
Question:
A child on a swing has the highest potential energy when: a: the swing is at its highest point b: the swing is at its lowest point c: the swing is halfway between the highest and lowest points
6.1 An object has energy if it is capable of doing work 8
Law Of Conservation Of Energy
• Energy cannot be created or destroyed but can be transformed from one form of energy to another
• Also known as the first law of thermodynamics
• How does water falling over a waterfall demonstrate this law?
6.2 Internal energy is the total energy of an object’s molecules 9
• 1st Law of Thermodynamics: For an isolated system the internal energy (E) is constant:
Δ E = Ef - Ei = 0
Δ E = Eproduct - Ereactant = 0
• We can’t measure the internal energy of anything, so we measure the changes in energy
• E is a state function• E = work + heat
Internal Energy is Conserved
6.1 An object has energy if it is capable of doing work 10
Heat And Temperature Are Not The Same
• The temperature of an object is proportional to the average kinetic energy of its particles—the higher the average kinetic energy, the higher the temperature
• Heat is energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermal equilibrium
Units• The SI unit of energy is the joule (J)
• calorie (cal)
– The dietary Calorie (note capital), Cal, is actually 1 kilocalorie
(exactly) J 4.184 cal 1
s m kg 1 kg) (2 J 1 -222
s 1m 1
21
Energy Transfer• When a cold and hot object come into
contact, they eventually reach thermal equilibrium (the same temperature)
• Movie1/06_heat transfer
• energy transferred as heat comes from the object’s internal energy
Energy Distribution
a) Thermal equilibrium: the same average KE for molecules in both objects
Energy Transfer
Question:
As the temperature of a substance increases, what happens to the average kinetic energy of the particles? a: increases b: stays the same c: decreases
Question:
Which sample of a substance has the higher most probable molecular speed? a: the warmer sample b: the cooler sample c: both samples have the same most probable speed
• current condition: state • Internal energy is a state function• state functions: independent from the
mechanism or method by which a change occurred
• object we are interested in: system • Everything else: surroundings • A boundary • System and Surroundings together: universe
Definitions
Open, closed, isolated
Question:
How is energy transferred as heat? a: energy flows from cooler objects to warmer ones. b: energy flows from warmer objects to cooler ones. c: molecules from the warmer object enter the cooler object.
Question:
What type of system exchanges energy but not matter with the surroundings? a: open b: closed c: isolated
6.3 Heat can be determined by measuring temperature changes 22
A cast iron skillet is moved from a hot oven to a sink full of water. Which of the following is not true?
A. The water heats
B. The skillet cools
C. The heat transfer for the skillet has a (-) sign
D. The heat transfer for the skillet is the same as the heat transfer for the water
E. None of these are untrue
The symbol for heat is “q” – two ways to calculate
Question:
Which process is exothermic? a: N2 2N b: CO2 CO + O c: BF3 + NH3 BF3NH3
Ch. 6 Energy and Chemistry1. Vocabulary
2. Heat using specific heat and heat capacity
3. Enthalpy from single reactiona) Single reagent
b) Limiting reagent
4. Enthalpy from multiple reactionsa) Hess’ Law
5. Enthalpy of reaction from ∆Hf°
Heat Capacity (C) relates the heat (q) to an objects temperature change
The heat capacity is the amount of heat needed to raise the object’s temperature by one degree Celsius and has the units J/°C
t C ) - t C (tq initialfinal
James JouleJames Joule1818-18891818-1889
t C ) - t C (tq initialfinal
James JouleJames Joule1818-18891818-1889
• If my large ceramic coffee cup has a heat capacity of 35 J/ºC, how much heat does it take to raise the temperature by 50ºC
Specific Heat (s) is an intensive property, and is unique for each substance
• q=m×Δt×s
4.18 Water
0.235 Silver
0.129 Gold
0.387 Copper
C) (25 C g J Substance
Heat Specific 1-1-
t m s q
If 25.0 g of Al (SH=.902 J/gK) cools from 310 If 25.0 g of Al (SH=.902 J/gK) cools from 310 ooC to 37 C to 37 ooC, how many joules of heat energy C, how many joules of heat energy are lost by the Al?are lost by the Al?
where where T = TT = Tfinalfinal - T - Tinitialinitial
heat gain/lost = q = (mass)(SHeat)(T)
Problem SolvingA sample of nickel weighing 425 grams was
initially at a temperature of 26.20C. It required 975 joules of heat energy to increase the temperature to 31.55 C. What is the specific heat of nickel?
Problem Solving• A 500.0 gram sample of water is initially at
25.0C. It absorbs 50.0 kJ of heat from its surroundings, what is the final temperature in C? specific heat of water = 4.184 J/gC
James JouleJames Joule1818-18891818-1889
If my large ceramic cup (heat capacity = 35 J/ºC) is at 25ºC when 250 mL of 92ºC coffee is added to it. What will be the final temperature of my cup with coffee assuming no other heat loss/gain?
Problem SolvingProblem SolvingDetermine the final temperature when 255 g
of iron is heated to 100.0°C in boiling water and it is then quickly placed into 100.0 g water at 23.4°C. SHiron=0.449 J/gK
q (water) = -q (iron)
Question:
How much energy is required to heat 50 grams of water from 24ºC to 34ºC? a: 50 cal b: 500 cal c: 500 J
Question:
How many dietary calories equals 1000 kJ? a: 240 b: 240,000 c: 4184 d: 4.184
Question:
Adding 5 calories of heat energy to 1 gram of copper raises the temperature of the copper by 54.3 ºC. Adding 5 calories of heat energy to 1 gram of water raises the temperature of the water by 5.0 ºC. Which statement must be true?
a: Copper has a larger specific heat capacity than water. b: Water has a larger specific heat capacity than copper. c: There must be some mistake in these measurements. d: Mass is being converted into energy in this process.
Question:100 grams of water cool from 14 ºC to 13 ºC.
What is the heat change for the water? a: 100 cal b: -100 cal c: 418 cal d: - 418 cal
FIRST LAW OF FIRST LAW OF THERMODYNAMICSTHERMODYNAMICS
E = q + wE = q + w
heat energy transferredheat energy transferred
energyenergychangechange
work donework doneby the by the systemsystem
first law of thermodynamics, which says that energy cannot be created or destroyed
In reactions where gases are produce or consumed qv and qp can be very different
pressure opposing theis where PVPw
The heat produced by a combustion reaction is called the heat of combustion usually in a bomb calorimeter. The reaction is run at constant volume so that ΔE = qV
Heats of reactions in solution are usually run in open containers at constant pressure, so that qP =
ΔE + PΔV = ΔH
A calorimeter has metal parts (heat capacity of 850.0 J/degree) and 1020 grams of oil (specific heat 2.248 J/gC, both at 24.50C.
Two metal slugs, one a 460.0 g piece of cobalt (specific heat 25.12 J/mol-degree), one a 360.0 g piece of cadmium (specific heat=25.34 J/mol-degree), were removed from an oven maintained at 240.0C and added to the calorimeter.
If no heat was lost to the surroundings, what would be the final temperature of inside the calorimeter?
Question:
Which equals qP? a: w b: ΔE c: ΔH d: none of these
http://www.flatrock.org.nz/topics/flying/historic_craft_flies_again.htmhttp://video.google.com/videoplay?docid=-2380118142773657669&q=hindenburg
Hindenburg, 6 May 1937
6.5 Heats of reaction are measured at constant volume or constant pressure 46
A sample of 50.00mL of 0.125M HCl at 22.36 ºC is added to a 50.00mL of 0.125M Ca(OH)2 at 22.36 ºC. The calorimeter constant was 72 J/g ºC. The temperature of the solution (s=4.184 J/g ºC, d=1.00 g/mL) climbed to 23.30 ºC. Which of the following is not true?
A. qczl=67.9 J
B. qsolution= 393.3J
C. qrxn = 461.0 J
D. qrxn = -461.0 JE. None of these
6.5 Heats of reaction are measured at constant volume or constant pressure 47
A sample of 50.00mL of 0.125M HCl at 22.36 ºC is added to a 50.00mL of 0.125M Ca(OH)2 at 22.36 ºC. The calorimeter constant was 72 J/g ºC. The temperature of the solution (s=4.184 J/g ºC, d=1.00 g/mL) climbed to 23.30 ºC. Which of the following is not true?
A. qczl=67.9 J
B. qsolution= 393.3J
C. qrxn = 461.0 J
D. qrxn = -461.0 JE. None of these
Ch. 6 Energy and Chemistry1. Vocabulary
2. Heat using specific heat and heat capacity
3. Enthalpy from single reactiona) Single reagent
b) Limiting reagent
4. Enthalpy from multiple reactionsa) Hess’ Law
5. Enthalpy of reaction from ∆Hf°
Consider the decomposition of water:Consider the decomposition of water:
HH22O(g) + O(g) + 243 kJ243 kJ H H22(g) + 1/2 O(g) + 1/2 O22(g)(g)
Endothermic reaction — heat is a “reactant”Endothermic reaction — heat is a “reactant”
HHRXNRXN = + 243 kJ = + 243 kJ
2H2H22(g) + O(g) + O22(g) (g) 2H 2H22O(g) + O(g) + 486 kJ486 kJ
HHRXNRXN = - 243 kJ = - 243 kJ
Energy is in chemical bonds
An enthalpy change for standard conditions is denoted
• The physical states are important
• The law of conservation of energy requires
kJ 38.92
)(2NH)(3H)(N 322
H
ggg
kJ 92.38
)(H3)(N)(NH2 223
H
ggg
H
Given 2H2(g) + O2 2 H2O for which HRXN = -483.6, how much heat is evolved when one mole water is formed? How is much heat is consumed when 3 moles of water are decomposed?
How much heat is evolved when 2.500 moles of water are formed by the same reaction?
Question:Consider the following information:
N2 (g) + 3 H2 (g) 2 NH3 (g) ΔH º = - 92.38 kJ
What is ΔH º when 1.0 mole of liquid ammonia is produced? a: - 92.38 kJ b: - 46.19 kJ c: - 184.8 kJ d: more information is needed
Question:
How much heat would be released if 1.0 mole of compound X were produced?
2A + Z 2X ΔH º = - 80 kJ a: - 80 kJ b: - 40 kJ c: - 160 kJ
The thermite reaction produces a tremendous amount of heat. If you begin with 10.0 g of Al and excess Fe2O3, how much heat in kilojoules is evolved at constant pressure?
2 Al (s) + Fe2O3 (s) Al2O3 (s) + 2 Fe (s)Hrxn = -851.5 kJ
Ch. 7 Energy and Chemistry1. Vocabulary
2. Heat using specific heat and heat capacity
3. Enthalpy from single reactiona) Single reagent
b) Limiting reagent
4. Enthalpy from multiple reactionsa) Hess’ Law
5. Enthalpy of reaction from ∆Hf°
An enthalpy diagram is a graphical representation of alternate paths between initial and final states
Remember to include the physical states of reactants and products in thermochemical equations.
Making HMaking H22 from H from H22O involves two steps.O involves two steps.
HH22O(liq) + 44 kJ O(liq) + 44 kJ H H22O(g)O(g)
HH22O(g) + 242 kJ O(g) + 242 kJ H H22(g) + 1/2 O(g) + 1/2 O22(g)(g)-----------------------------------------------------------------------
HH22O(liq) + 286 kJ O(liq) + 286 kJ H H22(g) + 1/2 O(g) + 1/2 O22(g)(g)
HESS’S LAW—HESS’S LAW—
If a rxn. is the sum of 2 or more others, the net If a rxn. is the sum of 2 or more others, the net H is the sum of the H is the sum of the H’s of the other rxns.H’s of the other rxns.
USING ENTHALPYUSING ENTHALPY
Rules for Manipulating Thermochemical Equations:
1) When an equation is reversed the sign of the enthalpy change must also be reversed.
2) Formulas canceled from both sides of an equation must be for substances in identical physical states.
3) If all the coefficients of an equation are multiplied or divided by the same factor, the value of the enthalpy change must likewise be multiplied or divided by that factor.
Using Standard Enthalpy ValuesUsing Standard Enthalpy Values
HH22O(g) + C(graphite) O(g) + C(graphite) H H22(g) + CO(g)(g) + CO(g)
From reference books we findFrom reference books we find
• HH22(g) + 1/2 O(g) + 1/2 O22(g) (g) H H22O(g) O(g)
HHff of H of H22O vapor = - 242 kJ/molO vapor = - 242 kJ/mol
• C(s) + 1/2 OC(s) + 1/2 O22(g) (g) CO(g) CO(g)
HH ff of CO = - 111 kJ/mol of CO = - 111 kJ/mol
6.7 Thermochemical equations can be combined because enthalpy is a state function 66
What is the energy of the following process:6A + 9B + 3D + 1 F→2 GGiven that: • C → A + 2B ∆H=20.2 kJ/mol• 2C + D →E + B ∆H=30.1 kJ/mol• 3E + F →2G ∆H=-80.1 kJ/molA. 70.6 kJB. -29.8 kJC. -111.0 kJD. None of these
Problem solving:
Given C(s) + O2(g) --> CO2(g) ΔHrxn(-393.5 kJ)
and CO(g) + 1/2 O2(g) --> CO2(g) ΔHrxn(-283 kJ)
calculate enthalpy change for reaction: C(s) + 1/2 O2(g) --> CO(g)
Question:Consider the following information:A + F 2G ΔH º = +100 kJA G + E ΔH º = +200 kJWhat is ΔH º for 2E + F A?
a: +300 kJ b: +100 kJ c: -100 kJ d: -300 kJ e: +500 kJ
Problem solving:Problem solving:Calculate the enthalpy of combustion of
ethylene C2H4 to CO2 and H2O
2C(s) + 2H2(g) C2H4 ΔHrxn(52.3 kJ)C(s) + O2(g) CO2(g) ΔHrxn(-393.5 kJ)H2(g) + 1/2O2(g) H2O(l) ΔHrxn(-285.8 kJ)
Problem Solving• Two oxides of copper can be made from
copper by the following reactions:2 Cu(s) + O2(g) 2 CuO(s) H= -310 kJ 2 Cu(s) + ½O2(g) Cu2O(s) H= -169 kJConstruct an enthalpy diagram and find H for the following
• Cu2O(s) + ½O2 2CuO(s)
Ch. 6 Energy and Chemistry1. Vocabulary
2. Heat using specific heat and heat capacity
3. Enthalpy from single reactiona) Single reagent
b) Limiting reagent
4. Enthalpy from multiple reactionsa) Hess’ Law
5. Enthalpy of reaction from ∆Hf°
HHooff, standard molar enthalpy , standard molar enthalpy
of formationof formation
HH22(g) + 1/2 O(g) + 1/2 O22(g) (g) H H22O(g)O(g)
HHooff = -241.8 kJ/mol = -241.8 kJ/mol
By definition, By definition,
HHoof f = 0 for elements in their standard = 0 for elements in their standard
states.states.
Question:
For which equation is ΔH º equal to ΔH º F for SO2 (g)? a: S8 (s) + 8 O2 (g) 8SO2 (g) b: S8 (s) + 8 O2 (g) SO2 (g) c: 1/8 S8 (s) + O2 (g) SO2 (g)
standard states have been defined for reporting heats of reactions
1 bar pressure for all gases and
1 M concentration for aqueous solutions
25 °C (298 K) is often specified as wellMeasurements made under standard state conditions
have the ° mark: eg ΔH°
The standard heat of reaction is the value of the enthalpy change occurring under standard conditions involving the actual number of moles specified the the equation coefficients
HHoorxnrxn = = HHoo
f f (products) - (products) - HHoof f (reactants)(reactants)
In general, when In general, when ALLALL enthalpies of enthalpies of
formation are known, formation are known,
Calculate Calculate ΔΔH H of reaction?of reaction?
Using Standard Enthalpy ValuesUsing Standard Enthalpy Values
Calculate the heat of combustion of methanol, Calculate the heat of combustion of methanol,
CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) (g) CO CO22(g) + 2 H(g) + 2 H22O(g)O(g)
HHoorxnrxn = = HHoo
f f (prod) - (prod) - HHoof f (react)(react)
TABLE 7.2 Standard Enthalpies of Formation of Typical Substances
Substance (kJ mol−1)
Substance (kJ mol−1)
Substance (kJ mol−1)
Ag(s) 0 CaO(s) −635.5 KCl(s) −435.89
AgBr(s) −100.4 Ca(OH)2(s) −986.59 K2SO4(s) −1433.7
AgCl(s) −127.0 CaSO4(s) −1432.7 N2(g) 0
Al(s) 0 CaSO4·½H2
O −1575.2 NH3(g) −46.19
Al2O3(s) −1669.8 CaSO4·2H2O(
s) −2021.1 NH4Cl(s) −315.4
C(s, graphite) 0 Cl2(g) 0 NO(g) 90.37
CO(g) −110.5 Fe(s) 0 NO2(g) 33.8
CO2(g) −393.5 Fe2O3(s) −822.2 N2O(g) 81.57
CH4(g) −74.848 H2O(g) −241.8 N2O4(g) 9.67
CH3Cl(g) −82.0 H2O(l) −285.9 N2O5(g) 11
CH3I(g) 14.2 H2(g) 0 Na(s) 0
CH3OH(l) −238.6 H2O2(l) −187.6 NaHCO3(s)
−947.7
CO(NH2)2(s) urea) −333.19 HBr(g) −36 Na2CO3(s) −1131
CO(NH2)2(aq) −391.2 HCl(g) −92.30 NaCl(s) −411.0
C2H2(g) 226.75 HI(g) 26.6 NaOH(s) −426.8
C2H4(g) 52.284 HNO3(l) −173.2 Na2SO4(s) −1384.5
C2H6(g) −84.667 H2SO4(l) −811.32 O2(g) 0
C2H5OH(l) −277.63 HC2H3O2(l) −487.0 Pb(s) 0
Ca(s) 0 Hg(l) 0 PbO(s) −219.2
CaBr2(s) −682.8 Hg(g) 60.84 S(s) 0
CaCO3(s) −1207 I2(s) 0 SO2(g) −296.9
CaCl2(s) −795.0 K(s) 0 SO3(g) −395.2
Using Standard Enthalpy ValuesUsing Standard Enthalpy Values
CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) (g) CO CO22(g) + 2 H(g) + 2 H22O(g)O(g)
HHoorxnrxn = = HHoo
f f (prod) - (prod) - HHoof f (react)(react)
HHoorxnrxn = 1 mol = 1 mol HHoo
f f (CO(CO22) + 2 mol ) + 2 mol HHoof f (H(H22O) O)
- {3/2 mol - {3/2 mol HHoof f (O(O22) + 1 mol ) + 1 mol HHoo
f f (CH(CH33OH)} OH)}
= (-393.5 kJ) + 2 (-241.8 kJ) = (-393.5 kJ) + 2 (-241.8 kJ)
- {0 + (-201.5 kJ)}- {0 + (-201.5 kJ)}
HHoorxnrxn = -675.6 kJ per mol of methanol = -675.6 kJ per mol of methanol
Problem Solving
Calculate H for the following reactions:
2NO(g) + O2(g) 2NO2(g)
Hf (kJ/mol)NO (g) = 90.4NO2 (g) = 33.85N2O (g) = 81.56
Problem Solving: Calculate the enthalpy of combustion of
acetylene C2H2 to produce carbon dioxide and water, given the molar enthalpy of formations for the following compounds:
C2H2 = +226.7 kJ/mol
CO2 = -393.5 kJ/mol
H2O = -285.8 kJ/mol
Question:For which reaction is ΔH º equal to
ΔH ºCombustion for C2H2?
a: C2H2 (g) + 5/2 O2 (g) 2 CO2 (g) + H2O (g)
b: C2H2 (g) + 5/2 O2 (g) 2 CO2 (g) + H2O (l)
c: 2 C2H2 (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (g)
d: 2 C2H2 (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (l)
Question:
Which has ΔH º F = 0? a: O (g) b: H2 (l) c: Hg (l) d: F2 (s)
Question:When a reaction is reversed and multiplied by
2, what happens to its ΔH º? a: sign changed; multiplied by 1/2 b: sign changed; multiplied by 2 c: squared; then the sign changed d: sign kept; multiplied by ½
Question:
What does the superscript º mean in notation such as ΔH º? a: measurements at absolute zero b: degrees Celsius c: measurements at standard conditions d: measurements taken in dietary calories
Problem Solving • Calculate the standard molar enthalpy of combustion for ethanol C2H5OH using the standard molar enthalpies of formation
Question:
How much heat would be released if 1.0 mole of compound X were produced?
2A + Z 2X ΔH º = - 80 kJ a: - 80 kJ b: - 40 kJ c: - 160 kJ
Problem SolvingUse the following thermodynamic equations to find the
enthalpy change for the following reaction:
M2O3(s) + 3CO(g) 3CO2 + 2M(s)
M(s) + CO2(g) CO(g) + MO(s) H = -140.0 kJ
3MO(s) + CO2(g) CO(g) + M3O4(s) H = +1.20 kJ
2M3O4(s) + CO2(g) CO(g) + 3M2O3(s) H =-380.0 kJ
M2O3(s) + CO2(g) CO(g) + 2MO2(s) H = -121.8 kJ
2 MO2(s) + CO2(g) CO(g) + M2O5(s) H = +344.5 kJ
Problem SolvingProblem SolvingWhite phosphorus, P4, ignites in air to
produce heat, light and P4O10. If 3.56 g of P4 is burned, 37.4 kJ of heat is evolved at constant pressure. What is the molar enthalpy of reaction for the burning of white phosphorus?
Question:
When two atoms come together and form a bond, what happens to the energy? a: it disappears into space. b: it is stored in the form of potential energy.
Question:
An aqueous chemical reaction in a coffee cup calorimeter causes the water temperature in the cup to increase. Is the reaction endothermic or exothermic? a: endothermic b: exothermic c: neither d: it’s impossible to tell
Question:
Which is not a state function? a: internal energy b: enthalpy c: heat d: temperature
A 0.3000 g sample of tin ore was dissolved in acid solution converting all the tin to tin(II). In a titration, 8.08 mL of 0.0500 M KMnO4 was required to oxidize the tin(II) to tin(IV). What was the percentage tin in the original sample? The reaction also produced MnO2
O4H2MnO3Sn8H2MnO3Sn 224-
42
Problem Solving:
Problem SolvingWhen 250.00 mL of 0.200M of sodium sulfite is
reacted with 3.50 g of potassium dichromate (K2Cr2O7 FW=294.18) in an acidic solution, the products of the reaction include sulfate ion and chromium(III) ion. What is the concentration of chromium(III) ion if the total volume is now 251.25 mL?
How would the problem be different with 5.50 g of potassium dichromate?
Problem SolvingA solution of sodium thiosulfate (Na2S2O3) reacts
with chlorine gas in an acidic solution to give sulfate ion and chloride ion. How many mL of 0.100 M sodium thiosulfate would be needed to react with 4.25 g of chlorine gas?
Problem SolvingAll the iron in a 2.000 g sample of iron ore is
dissolved in an acidic solution and converted to iron(II) ion. When titrated with 0.100 M potassium permanganate the iron was oxidized to iron(III) and the permanganate to manganese dioxide. The titration required 27.45 ml of permanganate solution to reach the endpoint.
Problem SolvingA student, using the same calorimeter, then reacted
100.0 mL of 1.00 M NaOH with 100.0 mL of 1.00 M HCl. The Tinitial of the base was 24.70 °C, and of the acid 24.74 °C, while the Tfinal was 30.53 °C.
Calculate the molar enthalpy of neutralization for this reaction. Density of solution 1.02 g/mL, specific heat of solution 4.016 J/gK
Problem SolvingHow much heat is required to raise the
temperature of 300. g of copper from 21°C to 75°C? SH(copper)=0.385 J/gK
If the same amount of heat is added to each of the following metals, which will experience the greatest temperature change? Specific Heat in J/gC: Cu(0.385) Mg(1.02) Hg(0.138) Pb(0.129)
Bonus: Problem SolvingHow much heat energy is needed to raise the
temperature of 500 gallon of water in a whirlpool bath by 5C. Assume the specific heat of the water is 1.00 cal/gC
The specific heat of lead is 0.129 J/gC. If 74.0 J of heat is added to a 125 piece of lead at 23 C, what is the final temperature of the lead?
Bonus: Problem SolvingWhat is the specific heat of ethanol in J/gC if 560.0
J of heat is required to raise the temperature from 22.0C to 39.0C for a 50.0 g sample?
A lake that is 2.0 miles square and has an average depth of 10. ft contains 3.2 x1010 liters of water. How many joules of energy must be transferred from the lake when its temperature decreases by 10.°C. Assume lake density 1.0 g/mL SHH20 = 4.184 J/gK