Please answer these Olympiad/Chemistry SAT II questions

Test Preparation SAT Subject Test Preparation
1

<p>What is the concentration of the nitrate ion in a solution made by mixing 100 mL of a 0.200 M HNO3 solution with 200 mL 0.100 M Mg(NO3)2?</p>

<p>a.) 0.133M
b.) 0.150M
c.) 0.167M
d.) 0.200M</p>

<p>The phase transition from a gas to solid is called</p>

<p>a.) condensation
b.) evaporation
c.) polymerization
d.) sublimation</p>

<p>When the substances, Si, KCl, CH3OH, and C2H6 are arranged in an order of increasing melting point, what is the correct order?</p>

<p>a.) Si, KCl, CH3OH, C2H6
b.) CH3OH, C2H6, Si, KCl
c.) KCl, Si, C2H6, CH3OH
d.) C2H6, CH3OH, KCl, Si</p>

<p>Which one of the following mixtures creates a buffer
solution? </p>

<p>(A) 10.0 mL 0.10 M NaOH, 10.0 mL 0.10 M HF
(B) 20.0 mL 0.10 M NaOH, 15.0 mL 0.10 M HF
(C) 15.0 mL 0.10 M NaOH, 20.0 mL 0.10 M HF
(D) 10.0 mL 0.10 M NaOH, 5.0 mL 0.20 M HF </p>

<p>Consider a voltaic cell in which the reaction below
occurs in two half-cells connected by a salt bridge and an
external circuit. </p>

<p>2Cr(s) + 3Sn2+(aq) → 3Sn(s) + 2Cr3+(aq) E˚ = 0.603 V </p>

<p>Which change will cause the voltage to increase? </p>

<p>(A) Increasing the amount of Sn(s) in its half-cell
(B) Increasing the amount of Cr(s) in its half-cell
(C) Diluting the solution in the anode compartment
(D) Diluting the solution in the cathode compartment</p>

<p>In which list are the ions arranged in order of decreasing
size? </p>

<p>(A) S2–, Br –, K+, Ca2+
(B) Br –, S2–, K+, Ca2+
(C) K+, Ca2+, S2–, Br –
(D) Ca2+, K+, S2–, Br –</p>

<p>Which species has the largest bond angle?
(A) NO2+
(B) NO2
(C) NO2–
(D) NO3–</p>

<p>How many isomeric compounds have the formula
C3H8O?
(A) 1 (B) 2 (C) 3 (D) 4</p>

<p>Which of the compounds below will react readily with
chlorine by addition? </p>

<p>I. C2H2 (ethyne) II. C2H4 (ethene) III. C6H6 (benzene) </p>

<p>(A) I only
(B) I and II only
(C) II and III only
(D) I, II, and III </p>

<p>Given the reactions </p>

<p>CH4(g) → C (g) + 4H(g) ∆H˚ = 1656 kJ/mol</p>

<p>HC(triple bond)CH(g) → 2C(g) + 2H(g) ∆H˚ = 1648 kJ/mol</p>

<p>What is the estimated C(triple bond)C bond energy in kJ/mol?
(A) 414
(B) 820
(C) 1234
(D) 1664</p>

<p>I'm sorry this is very long, but I'm confused on most of these questions and would just like some reasoning behind each.</p>

2

<p>C
D
D
A&D? - not sure
B
C
C
C
B
B</p>

3

<ol>
<li>d - dimensional analysis, note the two moles of NO3- in second compound, divide total moles by final volume for molarity</li>
<li>d - solid-gas transition = sublimation</li>
<li>d - the two organic are lowww melting, one with OH melts higher since hydrogen bonding, KCl and Si are similar, but looking at choices, only one fits previously discussed conditions</li>
<li>c - buffer is weak acid/base and conjugate, the amount weak acid HF must neutralize strong base and have left over for buffer</li>
<li>c - diluting anode will decrease concetration of Cr3+ so shift to right for equilibrium</li>
<li>b - Br- is lowest on periodic table, S2- is above that, but has extra e- so bigger, the others have lost e- so smaller, with more e- lost, smallest</li>
<li>a - NO2+, is just O-N-O so bond angle is highest</li>
<li>c - can be C-C-C with O on end of chain (1), middle C (2), or C-O-C-C (3)</li>
<li>b - alkene/alkynes are unsaturated, which undergo addition rxn</li>
<li>not sure, written funny…</li>
</ol>

4

<p>

</p>

<p>Yeah, but the question says “the phase transition from a gas to solid” (in which case the answer would be A for condensation, I think).</p>

5

<p>Oh, sorry for my errors. HarveyMuddLove is correct.</p>

<p>10th:

  • 1656 kJ/mol is equal to the energy of 4 C-H bonds, so one such bond is 414 kJ/mol
  • C2H2 has two C-H bonds and one triple C-C bond. to compute the triple C-C bond energy, you have to substract energy of two C-H bonds from the energy of the decomposition of C2H2
  • 1648 - 414*2 = 820 kJ/mol</p>

<p>Perferatednylon: condensation is gas -> liquid transition. gas -> solid and solid -> gas is called desublimation and sublimation, accordingly.</p>

6

<p>None of the answer choices are correct for the second question. The real answer would be deposition or desublimation. </p>

<p>Also, for the next to last question, all of the molecules are unsaturated, so this doesn’t explain the answer choice. In a fully saturated hydrocarbon there are 2n + 2 hydrogen atoms for every carbon atom. So if you calculate the degrees of unsaturation ((maximum number of hydrogens - actual number of hydrogens) divided by 2) for each molecule you’ll find that benzene is actually the most unsaturated of the three.</p>

<p>A addition reaction occurs when a pi bond (double bond) is broken to form 2 new sigma (single) bonds. Alkenes and alkynes are electron rich and are thus nucleophiles (Lewis base). Chlorine in its diatomic form doesn’t look very electrophilic (Lewis acid), but note that Cl2 is polarizable and thus when it is close to a double bond a dipole is induced, making one of the chlorines have a partial positive charge (and thus it becomes electrophilic). Thus Cl2 will react with ethyne and ethene.</p>

<p>Benzene has double bonds as well, yet will not react with Cl2 without a Lewis acid catalyst. The reason is because benzene is an aromatic compound; such compounds are unusually stable. Their stability cannot be explained by resonance. What aromatic compounds actually are and why they are so stable is a bit complicated and the actual explanation requires molecular orbital theory.
Here’s a good explanation I found online:
<a href=“http://crab.rutgers.edu/~alroche/Ch16.pdf[/url]”>http://crab.rutgers.edu/~alroche/Ch16.pdf&lt;/a&gt; (ignore everything after page 13).</p>

<p>All the rest of HarveyMuddLove’s answers look correct to me (though I don’t remember anything about electrochemistry). The answer to the last question is B, as per the explanation in the post above mine.</p>