<p>Hello, I want to ask a few SAT Chemistry questions. I am practicing with real SAT Chemistry papers. These questions appeared on the 2007 JAN test.</p>
<p>(A) 10 mL of 0.01M H2SO4
(B) 10 mL of 0.01M HCl
(C) 10mL of 0.001M HCl
(D) 10mL of 0.001M NH4Cl
(E) 10mL of 0.01M NH3</p>
<ol>
<li>Has a pH greater than 7</li>
<li>After 10mL of 0.001M NaOH(aq) is added, has a pH of 7</li>
<li><p>After 10mL of 0.01M NaOH(aq) is added, has a pH less than 6</p></li>
<li><p>2NO2(g) = N2O4(g) + energy
NO2(g) is sealed in a glass tube and allowed to reach equilibrium at room temperature
according to the equation above. Placing the tube in ice water will result in which of the following?
I. An increase in gas volume
II. An increase in gas pressure
III. A decrease in concentration of NO2(g)</p></li>
</ol>
<p>(A). I only
(B). II only
(C). III only
(D). I and II only
(E). I, II, and III</p>
<ol>
<li><p>Of the following, which is the strongest Bronsted base?
(A). NH3
(B). CO3(2-)
(C). H2O
(D). Cl-
(E). OH-</p></li>
<li><p>The mass of a piece of copper wire and the mass of a crucible are measured. The copper wire is placed in the crucible, covered with sulfur, and heated to redness. The mass of the crucible containing the product is measured after cooling. All of the following are possible experimental observations EXCEPT:
(A). The product is black.
(B). The product is solid.
(C). The product weighs more than the copper.
(D). Each copper atom has lost two electrons.
(E). The copper wire has disappeared.</p></li>
<li><p>Under normal conditions, air contains the smallest amount of which of the following gases?
(A). N2(g)
(B). O2(g)
(C). CO2(g)
(D). H2O(g)
(E). H2(g)</p></li>
<li><p>2SO2 + O2 = 2SO3
If all measurements are made under the same conditions, what volume of SO3(g) is produced when 40mL of SO2(g) and 20mL of O2(g) react completely according to the equation above?
(A). 10mL
(B). 20mL
(C). 40mL
(D). 60mL
(E). 80mL</p></li>
<li><p>If 100. mL of each of the following aqueous solutions is mixed with 100. mL samples of 1.0M NaOH(aq), the greatest amount of heat is liberated by
(A). 0.010M H2SO4
(B). 0.010M HCl
(C). 0.10M HCl
(D). 0.10M H2SO4
(E). 1.0M HCl</p></li>
<li><p>_CH3OH(liquid) + _O2(g)(excess) =
Products of the complete conmbustion above include which of the following?
I. CO
II. CO2
III. H2O</p></li>
</ol>
<p>(A). I only
(B). I and II only
(C). I and III only
(D). II and III only
(E). I, II, and III </p>
<ol>
<li><p>Metallic copper dissolves readily in hot water BECAUSE copper is one of the most reactive of the metallic elements</p></li>
<li><p>When solid iodine dissolves in alcohol, the entropy of the system decreases BECAUSE ordered states are more probable than disordered states</p></li>
<li><p>A 0.1M solution of HCl has a greater concentration of H3O+ than a 0.1m solution of acetic acid, HC2H3O2, has BECAUSE the chlorine atom is less electronegative than the carbon atom is.</p></li>
</ol>
<p>PLEASE help me. These are actual questions from the actual 2007 Jan SAT Chemistry test. Please explain the answers as well. Thank you VERY MUCH!!!</p>
<p>Please help me. I’m taking the test in June and I really need your help. PLEASE!!!</p>
<ol>
<li>E) NH3, because NH3 is ammonia, which is a weak base and since bases have a pH greater than 7, NH3 must have a pH greater than 7.</li>
<li>C) 10 mL of .001 M HCl, because HCl and NaOH are strong acids (HCl)/ strong bases (NaOH). Since choice C gives the exact same moles of H+ in solution as the moles of OH- in 10 mL of .001 M NaOH, the resulting mixture would have a pH of 7. This is so because a mixture of a strong acid and a strong base in equal quantities gives a neutral pH (7). </li>
<li>I am not as sure but I think the answer is A since H2SO4 is a strong acid that contains two acidic protons to contribute to pH. Thus, there would be a higher concentration of H+ than OH- (from the NaOH). Accordingly, the pH would be acidic, meaning less than 7, or in this case maybe less than 6? I am not as sure about this one but I have a feeling choice A is correct for the above reasons.</li>
<li>C) Because by adding the mixture to a ice water tube, the reaction system loses heat, which would shift the equilibrium position toward the products, by Le Chatelier’s Principle. Thus, the concentration of NO2, a reactant, would decrease. The other conditions do not apply in this case.</li>
<li>E) OH-, because OH- is the strongest of the Bronsted Lowry bases. </li>
<li>D) One cannot see the loss of electrons experimentally. </li>
<li>E) H2 gas. This is simply a fact, no explanation. </li>
<li>C) 40. mL. Stoichiometry, with the fact that mole ratio also gives volume ratio.</li>
<li>I am not very confident but my guess is E). Please correct me anyone if I am wrong.</li>
<li>D) This is a combustion reaction. Combustion reactions result in the products of H2O and CO2. I don’t think CO is a viable product for a combustion reaction.</li>
<li>T T CE</li>
<li>F F, review the definition of entropy: an increase in disorder</li>
<li>T T CE
Please correct me on some of these questions because I am unsure about quite a few, especially 13.
Anyways, I did this without really paying that much attention so excuse any mistakes.</li>
</ol>
<p>any other people willing to help?</p>
<ol>
<li>E, since only NH3 is a base (HCL, H2SO4 are strong acids, NH4CL is acidic as the crystallization of a strong acid (HCL) and a weak base (NH3))</li>
<li>C, with equal moles here the equation is HCl+NaOH=NaCl+H2O, which is neutral.</li>
<li>It comes down to A and B.
The equation with HCl and NaOH is above, that of H2SO4 is
H2SO4+2NaOH=Na2SO4 + 2H2O, so only half of the H2SO4 will neutralize the NaOH, and the other half will remain as excess acid. A</li>
<li>A negative stress on the right is a stress on the left, which moves equilibrium right i.e. more N2O4. I’m pretty sure I and II are invalid, but im not sure if that is because delta gaseous moles is -1 or why. Just intuition. Thus, I guess C but would like an educated person to check that.</li>
<li>E, OH-, since nothing is more willing to except an H+ ion.</li>
<li>It’s entirely possible that a copper-sulfur compound is formed, so A, B, and E are wrong. Also, since sulfur has mass, if nothing is lost in gas form, it is entirely possible that the mass after burning of the whole shebang is bigger than the mass of just copper. Finally, D is the answer because you cannot experimentally tell how many electrons something has lost, especially when copper can lose either 1 (Cu+) or 2 (Cu 2+)</li>
<li>H2 (g), E…gotta love facts.</li>
<li>P and T are constant, so Avogadro’s principle lets us have moles in proportion with volume. Thus, simple stoichiometry yields 40 mL, E.</li>
<li>You have all ions to start, and are exothermically forming H2O bonds. Thus, the heat of enthalpy depends not on the acid used, but the amount of “stuff” used. The answer is E since only E uses up all of the NaOH.</li>
<li>In combustion reactions with C, H, and O, you ALWAYS get only H2O and CO2. D</li>
<li>F, since Copper elementally does NOT dissolve in water (no element dissolves in water, but some elements react with it to form soluble ions) and F since Copper has a positive reduction potential i.e. wants to be in elemental form, not ionic form.</li>
<li>F, F, solution is much more disordered than solid (think of the possibilities for the positions of the molecules) and in general, higher entropy is favored, so disordered states are more probable</li>
<li>T, since HCl is a strong acid and acetic acid is weak and F, since Chlorine is a halogen and thus very electronegative</li>
</ol>
<p>Any edits/corrections/solutions from anyone who actually knows what the hell they’re talking about would be much appreciated :)</p>
<p>By placing the sealed tube in ice water you are DECREASING the temperature.</p>
<p>Temperature = Avg Kinetic Energy</p>
<p>There are 2 methods of solving this problem:</p>
<p>1) Le Chatelier’s Principle.</p>
<p>Since the temperature is decreasing, you are removing energy from the RHS of the equation. Hence, the system will try “fight” this change by favouring the RHS of the equation. In doing so, there will be less NO2(g).</p>
<p>2) PV = NRT (ideal gas equation)</p>
<p>Since the temperature is decreasing. Either the pressure of volume must be deceasing too. They cannot be increasing. This means I and II are wrong.</p>
<p>Oh wow, I completely forgot about ideal gas law and only used LeChatelier’s…<em>facepalm</em></p>
<p>Thanks loads</p>