The AP Chemistry Study Thread

Test Preparation AP Test Preparation
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<p>ok for stanfordream's question.</p>

<p>you need to neutralize 6E-5 mol H+</p>

<p>henderson-hasselbach
Ph = PKa + log(A-/HA)</p>

<p>Ka for Acetic Acid = 1.8E-5
PKa=-log(Ka)
PKa=-log(1.8E-5)=4.745</p>

<p>5=4.745+log(A-/HA)
.255=log(A-/HA)
1.8=(A-/HA)
1.8 moles of A- to 1 mole of HA to have a 5.00 Ph buffer.</p>

<p>since there is 6E-5 moles of acid to neutralize, you need 6E-5 moles of NaC2H3O2.</p>

<p>HC2H3O2 isn't a powder.... it's like HCl. you can't have a powder, you can only have a stock solution....... so I'll assume you have .1M HC2H3O2.... if you have something else, dilute it using M1V1=M2V2.</p>

<p>HC2H3O2 + NaOH --> NaC2H3O2 + H2O</p>

<p>you need 6E-5 moles of NaOH and HC2H3O2.
Molarity = mol/L
L = moles/Molarity
L = 6E-5/.1 = 6E-4L of NaOH and HC2H3O2</p>

<p>that will make 6E-5 moles of C2H3O2 and the NaOH will fully neutralize the HC2H3O2. </p>

<p>you need 1.8 moles of A- to 1 mole of HA, so solve for moles of HA that you need.
x/6E-5 = 1/1.8
x=3.3333333333E-5 moles HA.
Molarity = moles/L
L = moles/Molarity = 3.33333333E-5/.1 = 3.3333333E-4 Liters additional Acetic Acid added.</p>

122

<p>Bump, and:</p>

<p>An unknown compound contains only the three elements C,H, and O. A pure sample of the compound is analyzed and found to be 65.60 percent C and 9.44 percent H by mass. Determine the empirical formula of the compound.</p>

123

<ol>
<li>Assume 100 g
(65.6 g)(1.0 mol / 12 g) = 5.5 mol C
(9.44 g)(1.0 mol / 1.0 g) = 9.4 mol H</li>
</ol>

<p>100-(65.6+9.44) = 24.96% O</p>

<p>(24.96 g)(1.0 mol / 16 g) = 1.6 mol O</p>

<ol>
<li>Divide by the lowest mole total
C - 5.5 mol C / 1.6 = ~3.5</li>
</ol>

<p>H - 9.4 mol H / 1.6 = ~6</p>

<p>O - 1.6 mol O / 1.6 = 1</p>

<ol>
<li> Since you don't have all integers...multiply
C - 3.5(2) = 7
H - 6(2) = 12
O - 1(2) = 2</li>
</ol>

<p>Empirical formula is C7H12O2</p>

<p>EDIT: You probably think I'm crazy for answering each post in a matter of seconds...I have the whole week off (spring break)...and am pretty bored...the only thing i have to do this week is prepare for the may 14th ap chem exam + the may 6th sat2 chem test.</p>

124

<p>Here's another easy electrochem question:</p>

<p>Complete, balance, and identify the oxidizing and reducing agents.</p>

<p>A) As (s) + ClO3- (aq) -> H2AsO3 (aq) + HClO (aq) (acidic solution)
B) H2O2 (aq) + ClO2 (aq) -> ClO2- (aq) + O2 (g) (basic solution)</p>

125

<p>damnit I HATE redox.</p>

<p>As --> H2AsO3
As + 3H2O --> H2AsO3 + H+ + e-</p>

<p>ClO3- --> HClO
5H+ ClO3- +4e- --> HClO + 2H2O</p>

<p>multiply top equation by four and put together. 2H2O on right cancel w/ 2 of the 12 on the left, leaving you w/ 10 H2O. 5H+ cancel w/ 4H+ , leaving you w/ 1 H+
4As + 10H2O + H+ + ClO3- -->4H2AsO3 + HClO + 2H2O</p>

<p>ill let sum1 else do the other one.</p>

126

<p>This is the second one:
H2O2(aq) + Cl2O7(aq) → ClO2-(aq) + O2(g)</p>

<p>One half-reaction: H2O2(aq) → O2(g)</p>

<p>H2O2(aq) → O2(g) + 2H+(aq)</p>

<p>H2O2(aq) → O2(g) + 2H+(aq) + 2e-</p>

<p>H2O2(aq) + 2OH-(aq) → O2(g) + 2H+(aq) + 2OH-(aq) + 2e-</p>

<p>H2O2(aq) + 2OH-(aq) → O2(g) + 2H2O(l) + 2e-</p>

<p>Other half-reaction: Cl2O7(aq) → ClO2-(aq)</p>

<p>Cl2O7(aq) → 2ClO2-(aq) + 3H2O(l)</p>

<p>Cl2O7(aq) + 6H+(aq) → 2ClO2-(aq) + 3H2O(l)</p>

<p>Cl2O7(aq) + 6H+(aq) + 8e- → 2ClO2-(aq) + 3H2O(l)</p>

<p>Cl2O7(aq) + 6H+(aq) + 6OH-(aq) + 8e- → 2ClO2-(aq) + 3H2O(l) + 6OH-(aq)</p>

<p>Cl2O7(aq) + 6H2O(l) + 8e- → 2ClO2-(aq) + 3H2O(l) + 6OH-(aq)</p>

<p>Cl2O7(aq) + 3H2O(l) + 8e- → 2ClO2-(aq) + 6OH-(aq)</p>

<p>Least common multiple to balance electrons</p>

<p>4H2O2(aq) + 8OH-(aq) → 4O2(g) + 8H2O(l) + 8e- (oxidation half-reaction)</p>

<p>Cl2O7(aq) + 3H2O(l) + 8e- → 2ClO2-(aq) + 6OH-(aq) (reduction half-reaction)</p>

<p>Final balanced redox reaction</p>

<p>4H2O2(aq) + Cl2O7(aq) + 2OH-(aq) → 4O2(g) + 2ClO2-(aq) + 5H2O(l)</p>

127

<p>I will post a question.
What would happen if the electrodes were put in saturated solution of glucose dissolved in water?
A) Light Bulb would glow
B) Light bulb would remain dark
C) Apparatus would combust
D) Glucose molecules would dissassociate
E) None of the above.
This is a SAT II question but lets see how you guys do. Hopefully its not to easy.</p>

128

<p>I think the answer is B because glucose does not dissociate. As a result, the electrons will not be transferred.</p>

<p>It's definetly not D b/c glucose is a covalent bond and not E b/c A and B are opposites.</p>

129

<p>you are right scareya. want to give a question</p>

130

<p>Can I give another electrochemistry question?</p>

131

<p>Electriochemistry~I avoid them. makes my head hurt. so I won't answer usually, that is..</p>

<p>Another Q:
What is the formal charge on HNO3?</p>

<p>Write formulas for the following; no need to balance, just net ionic equations:
Calcium oxide powder is added to distrilled water.
Methylamine gas is passed over hot, solid sodium oxide.</p>

132

<p>H = +1
N = +5
O = -2</p>

<p>CaO + H2O -> Ca+ + OH-
_____ + Na2O -> ____ + _____</p>

<p>I'm not so good at the equation part...I took a educated guess</p>

133

<p>Just a quick q, but all metals are reducing agents right? And all nonmetals are oxiding agents?</p>

134

<p>H2O + Na2O -> Na+ + OH-<br>
??? I have no clue</p>

135

<p>does anyone know how to solve this problem?</p>

<p>How many grams of calcium nitrate Ca(NO3)2 contains 24 g of O atoms?</p>

136

<p>you figure out how many moles are 24 g of O atoms --
24g/16g/mol =1.5 mol of O
and there are 6 O's in each mole of calcium nitrate so you can have 1.5/6= .25 moles of calcium nitrate with the amount of O that you have
so you multiply the molecular weight of calcium nitrate by .25 moles</p>

137

<p>24g of O is 1.5 moles of O. Ca(NO3)2 contains 6 moles of O, so to find the ratio of how many moles Ca(NO3)2 you'll have, divide 1.5 by 6. You get .25 moles of Ca(NO3)2, so multiply .25 by the molar mass of Ca(NO3)2.</p>

<p>Edit: whylion got to it first</p>

138

<p>thanks guys..one more question?
whihc of the following doesnt behave as an electrolyte when dissolved in water?</p>

<p>CH3OH
K2CO3
NH4Br
HI
CH3COONa</p>

139

<p>Isn't it NH4Br because it won't dissolve into ions?</p>

140

<p>NH4Br. I think. Ammonium doesn't form insoluble compounds with anything. Also CH3COONa seems like a plausible guess. I though Alkali metal don't form insoluble compounds. Not sure though</p>