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CHEM 1120 Final ACS Review Study Guide | General Chemistry Exam Prep & Practice Questions

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CHEM 1120 Final ACS Review Study Guide | General Chemistry Exam Prep & Practice Questions

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CHEM 1120 Final ACS Review Study Guide | General Chemistry Exam Prep &
Practice Questions


When 0.10 M solutions of ammonium acetate, barium

acetate, and sodium acetate are ranked from least basic to

most basic, what is the correct ordering? - ANS ✔✔You must know the strong acids and bases in
order to use the system.



Strong acids: HCl, HBr, HI, HNO3, HClO4, H2SO4 (1st H only)

Strong bases: Group IA hydroxides (not HOH) and Ca(OH)2, Sr(OH)2, and Ba(OH)2



Here's the system:

Look at the cation and anion of the salt.

1. If the cation comes from a strong base, and the anion comes from a strong acid, the salt will
be neutral.

2. If the cation comes from a strong base, and the anion comes from a weak acid, the salt will be
basic.

3. If the cation comes from a weak base and the anion comes from a strong acid, the salt will be
acidic

4. if the cation and ion both come from weak acids and bases, it is impossible to predict. You
must know the relative strengths of the acid and base.



Using this system:

1.sodium nitrate NaNO3 - neutral

2.ammonium iodide NH4I - acidic

3.sodium bicarbonate NaHCO3 - basic

4.sodium hypochlorite NaOCl - basic

5.potassium acetate KCH3CO2 - basic

, correct answer:

NH4C2H3O2 < NaC2H3O2 < Ba(C2H3O2)2



What is the solubility of MgF2 (Ksp = 6.8 * 10-9) in pure water? - ANS ✔✔Let x = mol/L MgF2
that dissolve to estabilish equilibrium.



This will produce x mol/L Mg2+ and 2x mol/L F- via the reaction



MgF2 (s) < => Mg2+ (aq) + 2 F- (aq)



Ksp = [Mg2+][F-]^2 = (x)(2x)^2 = 4x^3 = 6.8 x 10^-9

= x^3 = 1.7 x 10^-9

x= cube root (1.7 x 10^-9)



x = 0.00119



What is the ratio Kc /Kp for the following reaction at 723 °C?

O2(g) + 3 UO2Cl2(g) = U3O8(s) + 3 Cl2(g) - ANS ✔✔kc/kp = RT^(delta n)



kc/kp = (0.0821 * (723+273))

=81.7716



What is [H3O+] in a solution formed by dissolving 1.00 g

NH4Cl (M = 53.5) in 30.0 mL of 3.00 M NH3 (Kb = 1.8 x 10-5)? - ANS ✔✔pOH = pKb +
log([BH+]/[B])

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