Pages 21, 22, 23

Properties
Taste
Feel
Conductivity
Reacts with metals
(yes or no?) what products?
pH range
Acids
Sour
Yes
Yes- forms H
0-7
2 gas
Bases
Bitter
Slippery
Yes
No
7-14

-
-

Hydrogen + a polyatomic?
Polyatomic ends in –ate the acid ends in –ic
Polyatomic ends in –ic the acid ends in –ous
-

Hydrogen + a single element?
Hydro______ ic acid

a) HCl – Hydrochloric acid b) HF – Hydrofluroic acid c) HNO
2
Nitrous acid d) H
2
SO
4
– Sulfuric acid


If the acids ends in an “ic” then the polyatomic ends in the –ate form

If the acids ends in “ous” then the polyatomic is in the – ite form
If the acid starts in Hydro- then the formula is
Hydrogen followed by the element ending in – ide
Hydrochloric acid= HCl

a) nitric acid HNO
3 b) acetic acid – HC
2
H
3
O
2 c) hydrobromic acid HBr d) sulfurous acid – H
2
SO
3


Following ionic naming rules
Name the cation regularly, the polyatomic uses its regular name.

a)KOH Potassium hydroxide b) Ba(OH)
2
Barium hydroxide c) LiOHLithium hydroxide d) NH3 – ammonia


Use ionic rules
Write the symbols.
Identify the charges- criss cross to make subscripts.

 a) Sodium hydroxide NaOH
 b) Beryllium hydroxide Be(OH)
2
 c) Calcium hydroxide Ca(OH)
2
 d) Cesium hydroxide CsOH


Arrhenius acid- contains an H+

Bronsted- Lowry acid- Donates an H+


Arrhenius base- contains an OH-

Bronsted- Lowry base- Accepts an H+


HNO
3
+ H
2
O  H
3
O + + NO
3
-
ACID BASE ACID BASE

N
2
H
4
+ H
2
O  N
2
H
3
+ H
3
O +
ACID BASE BASE ACID

N
2
H
4
+ HCl  N
2
H
BASE ACID ACID
5
+ + Cl -
BASE


Neutralization reactions:
ACID+ BASE  SALT + WATER
KOH + H
2
CO
3
KOH + H
2
CO
3

 H
2
O + K
2
CO
Don’t forget to BALANCE!
3
2 KOH + H
2
CO
3
 2 H
2
O + K
2
CO
3
HBr + Al(OH)
3
HBr + Al(OH)
3
3 HBr + Al(OH)
3

 H
2
O + AlBr
3
 3 H
2
O + AlBr
3

[H+]
1.
2.3x10
-5 M
2.
5.0x10
-8 M
3. 1.2x10
-11 M pH
4.64
7.3
10.9
[OH-] pOH
4.3x10
-10 M 9.36
2.0x10
-7 M 6.7
8.1x10
-5 M 3.1
Acid, base, neutral?
Acid
Base
Base


Titration- method for determining concentration of a solution by reacting a known volume of solution with a solution of known concentration

Equivalence point- equal amounts of OH -
& H + ions

Pages 24

Radiation Type Alpha Beta Gamma
Charge
Description
Symbol
Mass
Penetrating Power
Shielding Needed
+2 -1 0
Helium nucleus electron EMR
4
2 a or
4
2
He 0
-1 b or
0
-1 e g
4 amu 1/1840 amu 0 low medium high
Paper, cloth, skin, etc.
Aluminum foil Lead



Beta decay = electron
191 is gold’s Atomic mass, which goes on top.

Look up Gold (Au) on your periodic table to find the atomic number which goes on the bottom.
Gold’s atomic number is 79.


191
79
Au  0
-1 e +_____
191
79
Au  0
-1 e + 191
80
Hg

 alpha decay = helium particle

90 is Rubidium’s Atomic mass, which goes on top.

Look up Rubidium (RB) on your periodic table to find the atomic number which goes on the bottom. Rubidium’s atomic number is 37.

90
37
Rb  4
2
He +_____

90
37
Rb  4
2
He + 86
35
Br

A. 4
2
He + 14
7
N  _____ + 1
1
H
4
2
He + 14
7
N  17
6
C + 1
1
H
B. 102
44
Ru + 4
2
He  1
0 n +_____
102
44
Ru + 4
2
He  1
0 n + 105
46
Pd

Page 25


Potential energy- stored energy due to position

Kinetic energy- energy of motion
*remember that Temperature is a measure of the average kinetic energy


Heat- (Q) = the process of flowing from warmer to colder temperature.

Temperature- measure of the average kinetic energy (KE) in a sample


Specific heat (c) is the amount of energy required to raise the temperature of a 1 gram sample by 1 degree

High specific heat means that the substance warms and cools slowly. It resists changes in temperature.

Low specific heat means that the substance warms and cools quickly.

SI Unit = J/(g·°C)

a . Conduction- heat is transferred by touch.
Ex: heating a pan on the stove b. Convection- heat is transferred through liquids or gases.
Ex: Cooking in an oven c. Radiation- heat from the sun

 a. -200 kJ exothermic
 b. 32 kJ endothermic
 c. 653.8368 kJ endothermic
Exothermic reactions release energy so they lose energy (negative sign)
Endothermic reactions absorb heat so they have energy added (positive sign)

FORMULA: Q=mc∆T
Q= heat(J); m= mass(g); c=specific heat J/(g·°C);
∆T = change in temperature (Final– initial)
Q= (1.05)(.450)(63.5)
Q= 30.0 J
∆T = 88.5-25 = 63.5
The reaction is endothermic because it absorbs heat.

Balance the equation:
1 CH4(g) + 2 O2(g)  1 CO2(g) + 2 H2O(l)
Hrxn = -890.2 kJ/mole
52.4g 1 mol
16.043 g
= 3.27 moles
3.27 mol -890.2kJ
= -2910 kJ
1 mole

Page 26, 27, 28

1. Pressure (P)- atm, torr, kpa, mmHg, psi
2. Volume- (V) Liters
3. Temperature (T)- Kelvin
4. Amount- (n) moles
R = gas constant!

1. Gases consist of molecules whose separation is much larger than the size of the molecules themselves.
2. Particles in a gas move in straight line paths and
random directions.
3. Particles in a gas collide frequently with the sides of the container and less frequently with each other. All collisions are elastic (no energy is gained or lost as a result of the collisions).
4. Particles in a gas do not attract or repel one another. They do not sense any intermolecular forces.


STP = 0 0 C and 1 atm

Temperature in Kelvin = 273K
STP can be found on the STAAR chemistry reference chart under the constants and conversions section.

LAW
BOYLE’S
CHARLES’S
GAY LUSSAC’S
INDEP/DEP
VARIABLES
V, P
T, V
P, T
V, n
AVOGADRO’S
V, P, T,
COMBINED
IDEAL P,V,n, R, T
CONTROL
VARIABLES
T, n
P, n
V, n
P, T
NA
MATH
RELATIONSHIP
↑V, ↓ P inverse, indirect
↑T ↑V direct
↑T ↑P direct
↑n ↑V direct
NA
FORMULA
V
1
P
1
=V
2
P
2
V
1
= V
2
T
1
P
1
T
2
= P
2
T
1
T
2
V
1
= V
2 n
1
P
1
V
1 n
2
= P
2
V
2
T
1
T
2
R PV=nRT


STP= Standard temperature and pressure
P= 1 atm, T= 0 o C or 273K
PV=nRT
(1atm)(V)= (1.02moles)(.0821)(273)
V= 22.8L


STP= Standard temperature and pressure
P= 1 atm, T= 0 o C or 273K
PV=nRT
(1atm)(1.5L)=(n)(.0821)(273)
1.5= 22.4n
n= .067 moles
Molar mass of helium
.067 moles 4.003g
1 mole
= .27 grams


PV=nRT
(.988atm)(1.20L)=(.0470)(.0821)(T)
1.1856= .00385T
308K=T

Convert grams to moles to plug into the ideal gas law equation!
3.58g
1 mol =.177 mol
20.180g
PV=nRT
(.900atm)(V)=(.177mol)(.0821)(287)
V= 4.63L


Comparing Volume and Pressure – Boyles’ law
V
1
P
1
=V
2
P
2
(6L)(101kPa)=(V
2
)(91kPa)
606=(V
2
)(91kPa)
6.7L=(V
2
)


Comparing Volume and pressure- Boyles’ law
V
1
P
1
=V
2
P
2
(2.25L)(164kPa)=(1.50L)( P
2
)
369= (1.50L)( P
2
)
246 kPa=(
P
2
)


Comparing Temperature, volume and
Pressure – Combined gas law
P
1
V
1
T
1
= P
2
V
2
T
2
 (10.5)(.948) = (25.0)(P
2
)
500 618
 .019908= (25.0)(P
2
)
618


12.3= (25.0)(P
2
)
P
2
= .49 atm


Comparing volume and temperature- Charles
(7.36L) = ( V
2
)
(323K) (173K)
( V
2
) = 3.94L


Dalton’s law of partial pressure
*Be sure that all of the pressure values have the same units.
P
TOT
P
TOT
P
TOT
= P
1
+ P
2
+P
3
= 1.2atm +.75atm +.41atm
= 2.36 atm


Dalton’s law of partial pressure
*Be sure that all of the pressure values have the same units.
Convert kPA to atm 101kPa=1atm:
199 kPa 1 atm = 1.97atm
101kPa
P
TOT
= P
1
+ P
2
+P
3
1.97= .59+.65 +P
3
.73atm =P
3