Transcript Unit 1 review

Unit 1 review
By: Makoto Bowering and Nick Ennen
Charges
 Anions (has excess electron)–
Has a negative charge
 Most common
-1 charge : H-, F-, Cl-, Br-, I-, OH-2 charge : O, O2, S
-3 charge : N
Charges
 Cations (has a less electrons)
Has a positive charge
 Most common
1+ : H, Li, Na, K, Cs, Ag
2+ : Mg, Ca, Sr, Ba, Zn, Cd
3+ : Al
Bonds
 Covalent
2 non-metals
Ex:
FO
H2O
CH4
HCl
Bonds
 Ionic
1 metal and 1 non-metal
EX: NaCl
NaF
MgO
Bonds
 Metallic
2 Metal
EX: Diamond
Bronze
Copper
Organic compounds
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1 = meth2 = eth3 = prop4 = but5+ = a normal binary scale
 Alcohols have suffix ol and are and O-H bond
Percent composition
 CH2O
C= 40%
H= 6.72%
O= 53.28%
Assume that the percent can be changed into grams
and perform stoichiometry to get it to Moles. Once in
moles divide by the lowest value to get the empirical
forulma.
Universal Gas Law
 PV=NRT
P=Pressure (1.000 atm)
V=Volume (L)
N=Moles
R=constant (0.08206 L*atm/Mol*K)
T= Temperature (K = Tc + 273.15)
Standard temperature and pressure
 STP273.15 k
1.00 atm
1 mole of gas
22.4 L
Real gas equation
 (P+a(n/V)’2)(V-bn)= nRT
 First part =
The factor that the regular PV=nRT value is off because
of the fact that IMFA was not included
 Second part =
The factor that the regular PV=nRT value is off because
it now does not include the molecules themselves
Partial Pressure
 P(molecule)=(n(a)/n(total))
Root-Means-Square Speed
 This is the speed at which a molecule is moving.
 Urms= 3RT/M (all under a square root sign)
M=Molar mass (Kg/mol)
R= constant (8.314 J/Mol*K)
T= Temperature (K)
Limiting Reagent
 The limiting reagent limits the reaction from
producing the greatest it can. The limiting reagent
runs out before the excess reagent does.
 To find the value of the limiting reagent and excess
reagent, all that is needed is stochiometery
Limiting Reagent
 EX: A 2.00g sample of ammonia is mixed with 4.00g of
oxygen. Which is the limiting reactant and how much
excess reactant remains after the reaction has
stopped?
Limiting Reagent
 Step 1: write a balanced reaction
 4NH3 + 5O2  4NO + 6H2O
Limiting Reagent
 Step 2: finding which is limiting
Use stoichiometry to calculate how much product is
produced by each reactant. You can start with either
reactant, and you can calculate for either product, but
the product must be the same for both in order for the
amounts to be compared.
Limiting Reagent
 Step 3: finding the excess
To find the amount of excess reactant, we must
calculate how much of the non-limiting reactant actually
did react with the limiting reactant.