Cliffs AP Chemistry, 3rd Edition
Step 3: Calculate the number of moles of OH–. There should be twice as many moles of OH–as moles of Ba(OH) 2. (OH) .. (OH) 12 1. ...
Step 2: Determine the number of milliliters of concentrated HNO 3 solution that is needed. . . ' .' 1 25 10 200 mole H (^100013) ...
Samples: Free-Response Questions In an experiment to determine the equivalent mass of an unknown acid, a student measured out a ...
(d) Restatement: Equivalent mass of unknown acid. GEM moles of H furnished grams of acid = + . . mole H g acid 686 10 0 250 = # ...
(a) Restatement: Moles of HA in fifty-mL aliquots. At the end of titration, moles of HA = moles of NaOH. average volume of NaOH ...
(3) NaOH was 0.150 M instead of 0.100 M. It would take less NaOH to reach the equivalence point because the NaOH is stronger. B ...
Energy and Spontaneity Key Terms Words that can be used as topics in essays: chemical thermodynamics endothermic enthalpy change ...
∆G= ∆G°+ RT⋅ln Qwhere Q= reaction quotient ∆G= ∆G°+ 2.303 RT log K ∆G°= −n F E° where F = Faradays 1 F = 96,500 J ⋅mole–^1 ⋅V–1 ...
Samples: Multiple-Choice Questions Given the following standard molar entropies measured at 25°C and 1 atm pressure, calculate ...
Answer: D This problem requires us to use the Gibbs-Helmholtz equation: ∆G°= ∆H°−T∆S° Step 1:Using the given information, calcul ...
Answer: B ∆G°= Σ∆G°products−Σ∆G°reactants =[2(−237.00) + (−394.00)] −(–51.00) = −817.00 kJ/mole Calculate the approximate stand ...
Answer: A Entropy is a measure of the randomness or disorder of a system. The greater the disorder of a system, the greater its ...
Given the following data: Fe 2 O 3 (s) + 3 CO(g) →2 Fe(s) + 3 CO 2 (g) ∆H°= −27 kJ/mole 3 Fe 2 O 3 (s) + CO(g) →2 Fe 3 O 4 (s) ...
Step 2: Solve for ∆S°. FeO(s) + CO(g) →Fe(s) + CO 2 (g) ∆S°= Σ∆S°products−Σ∆S°reactants = (27.0 + 214.0) −(61.0 + 190.0) = −10.0 ...
If ∆H°and ∆S°are both negative, then ∆G°is A. always negative B. always positive C. positive at low temperatures and negative ...
Part II: Specific Topics Samples: Free-Response Questions Given the equation N 2 O 4 (g) →2 NO 2 (g) and the following data: S ...
(d) Restatement: Calculate Kat 500°C and 1 atm. . RT T log HT T K ∆ K 2 303 T T 12 21 1 %^h- = 2 (. )(. )( )( ) , logKK JK K K J ...
then it is referred to as standard molar entropy and represents the entropy at 298 K and 1 atm of pressure; for solutions, it wo ...
Reduction and Oxidation Key Terms Words that can be used as topics in essays: anode cathode disproportionation electrolysis elec ...
At equilibrium: Q= K; Ecell= 0 ln. nE K 0 0257cell : % OIL RIG: Oxidation Is Losing; Reduction Is Gaining (electrons) AN OX ...
«
5
6
7
8
9
10
11
12
13
14
»
Free download pdf