endo or exothermic sign of q (+ or -) melting of ice endo+ evaporationendo+ condensationexo-...
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Endo or Endo or ExothermicExothermic
Sign of q Sign of q
(+ or -)(+ or -)
Melting of IceMelting of Ice endoendo ++
Evaporation Evaporation endoendo ++
condensationcondensation exoexo --
subliminationsublimination endoendo ++
(l) (l) (s) (s) exoexo --
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s s l l
l l g g
S S g g
l l s s
g g l l
g g s s
Melting/fusion endo
vaporization endo
sublimination endo
freezing exo
condensation exo
deposition exo
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SOLID
Freezing
Melting
condensation
vaporization
LIQUID
GAS
FUSION
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The specific heat (c) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
Heat capacityHeat capacity = the amount of energy required to raise the = the amount of energy required to raise the temperature of an object (by one degree).temperature of an object (by one degree).
Molar heat capacityMolar heat capacity = heat capacity of 1 mol of a substance = heat capacity of 1 mol of a substance..
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Heat of fusion: The amount of energy released/required at the solid/liquid phase change.
Heat of vaporization: The amount of energy released/requiredfor liquid/ gas phase change.
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Heat is the total amount of energy possessed by the molecules in a piece of matter. This energy is both kinetic energy and potential energy.
Temperature is proportional to the average kinetic energies of the molecules
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moves spontaneously from moves spontaneously from matter with higher T to matter with matter with higher T to matter with lower Tlower T
What happens when two objects of different What happens when two objects of different temperatures come into contact?temperatures come into contact?
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Which of the following has the greatest heat capacity?
100 g water or 1000 g water
Which of the following has the greatest specific heat?
100 g water or 100 g copper
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1. How much heat is required to raise the temp of 205 g of waterfrom 15.2 C to 16.2 C ?
What info do we have?
m = 205 g c= 4.184 J/g C t = 16.2 – 15.2 = 1
What are we looking for ? q
What is our equation ? q = mct
q = (205 g) (4.184 J/g C) (1 C)
858 J
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2. Calculate the amount of heat released when 25 g of waterAt 25 C is cooled to 0 C ?
What info do we have?
m = 25 g c= 4.184 J/g C t = 25 – 0 = 25 C
What are we looking for ? q
What is our equation ? q = mct
q = (25 g) (4.18 J/g C) (25 C)
2615 J
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3. What mass of 67.5 C iron must be added to 235 g of 5.00 C waterTo make the final temp of both come out to be 15 C ?
What do we have?
Iron c = .444 J/g C m = ? gIron initial T = 67.5 CIron final T = 15 C
t = 67.5 – 15 = 52.5
Water c = 4.184 J/g CWater m = 235 gWater initial T = 5 CWater final T = 15 C
t= 15 – 5 = 10 C
Heat lost = Heat gained q = q
mct= mct
(?g) (.444g) (52.5) = (235g)(4.184 J/g C)(10 C)
mass = (235g)(4.184 J/g C)(10) (.444J/g C)(52.5 C)
421.8 g Fe
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4. A 195 g aluminum engine part at an initial temperature of 3.00 C absorbs 40 KJ of heat. What is the final temperature of the part?
What info do we have?
m = 195 g c= .897 J/g C T initial= 3.00 C
What are we looking for ? T final
What is our equation ? q = mct
40,000J = (195 g) (.897 J/g C) (Tfinal – 3.00 C)
q = 40 KJ 40,000J
q= mc(Tfinal-Tinit)
40000J = Tfinal – 3.00 C(195g)(.897J/g C)
40000J + 3.00 C = Tfinal (195g)(.897J/g C)
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5. When 300 J of energy is lost from 125 g object, the temperature decreases from 45 C to 40 C. What is the specific heat of this object?
What info do we have?
m = 125 g T final = 40C T initial= 45.00 C
What are we looking for ? c specific heat
What is our equation ? q = mct
300J = (125 g) (c) (5 C)
q = 300 J
300J = c(125g)( 5 C)
.48J/g C
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6. The specific heat of lead (Pb) is 0.129 J/g C. Find the amount ofheat released when 2.4g of lead are cooled from 37.2 C to 22.5 C?
What info do we have?
m = 2.4 g T final = 22.5C T initial= 37.2 C
What are we looking for ? q
What is our equation ? q = mct
q = (2.4 g) (.129 J/g C) (22.5C – 37.2 C)
c = .129J/g C
4.6 J
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7. How many kJ of energy are needed to raise the temperatureof 165 g water from 10.5 C to 47.32 C?
What info do we have?
m = 165 g T final = 47.32 C T initial= 10.5 C
What are we looking for ? q in units of kJ
What is our equation ? q = mct
q = (165 g) (4.184 J/g C) (47.32 C – 10.5 C) =
c = 4.18J/g C
25395 J 25395 J x 1 kJ = 25.4 KJ 1000 J
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8. How much heat is absorbed by 15 g of ice being melted?
H2O (s) H2O (l) Hf = 6.01 kJ/1 mol
15 g x 1 mol = .833 mol18 g
.833 mol x 6.01 kJ/mol = 5 kJ
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9. How much heat is necessary to change 5.0 g of water at 100 C to to steam at 100 C?
H2O (l) H2O (g) Hv = 2260 J/g
5g x 2260 J/g = 11300 J
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Calculate Molar Mass of
Calcium Phosphate
Ammonium Sulfate
CaPO4
(40) + (31) + 4(16) = 135 g/mol
(NH4)2SO4
(18)2 + 32+ 4(16) = 132 g/mol
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How many moles of Na2CO3 are thee in 10L of 2.0M solution
2.0 mol = X mol1 liter 10 liter
= 20 moles
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Find the molarity of a solution containing 59 g HClIn 500 ml of water.
First convert grams to moles
59 g HCl x 1 mole HCL36 g
= 1.6 mol
Molarity is moles per liter
1.6 mole x 1000 ml500 ml 1 liter
= 3.2M
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What volume (in ml) of 12.o M HCl is needed to contain 3.00 moles HCl
12.0 mol = 3.0 mol1 liter X liter
X = 3.012
X = .25 liter
X = 250ml
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How many moles of gas would be present in a gas Trapped within a 100 ml vessel at 25 C at a pressure Of 2.50 atm
PV = nRT n = PVRT
2.5atm (0.1L)(0.821)(298K)
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What volume will 1.27 moles of helium gasoccupy at STP
1.27 mol x 22.4 L mol
28.5 L