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href="../19287/proteins-enzymes--1.html">Proteins & enzymes </a></li><li><a href="../19291/lipids-1.html">Lipids</a></li><li><a href="../19299/carbohydrates--1.html">Carbohydrates </a></li><li><a href="../19305/vitamins--1.html">Vitamins </a></li><li><a href="../19313/biochemistry-the-environment--1.html">Biochemistry & the environment </a></li><li><a href="../19336/proteins-enzymes-ahl--1.html">Proteins & enzymes (AHL) </a></li><li><a href="../19347/nucleic-acids--1.html">Nucleic acids </a></li><li><a href="../19361/biological-pigments--1.html">Biological pigments </a></li><li><a href="../19366/stereochemistry-in-biomolecules-1.html">Stereochemistry in biomolecules</a></li></ul></li><li><a href="../19390/option-c--1.html" class="father">Option C </a><ul class="level-2"><li><a href="../19391/energy-sources-1.html">Energy sources</a></li><li><a href="../19407/fossil-fuels-1.html">Fossil fuels</a></li><li><a href="../19411/nuclear-fusion-nuclear-fission-reactions-1.html">Nuclear fusion & 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href="../19194/antiviral-medications--1.html">Antiviral medications </a></li><li><a href="../19211/environmental-impact-of-some-medications-1.html">Environmental impact of some medications</a></li><li><a href="../19254/taxol-a-chiral-auxiliary-case-study-1.html">Taxol - a chiral auxiliary case study</a></li><li><a href="../19258/nuclear-medicine--1.html">Nuclear medicine </a></li><li><a href="../19273/drug-detection-analysis--1.html">Drug detection & analysis </a></li></ul></li></ul></li><li><a href="../16592/practical-scheme-of-work-ia--1.html">Practical scheme of work & IA </a><ul class="level-1"><li><a href="../16682/internal-assessment-1.html" class="father">Internal Assessment</a><ul class="level-2"><li><a href="../18892/scaffolding-the-investigation-1.html">'Scaffolding' the investigation</a></li><li><a href="../18896/timing-organisation--1.html">Timing & organisation </a></li><li><a href="../18905/choosing-the-research-question-1.html">Choosing the research question</a></li><li><a href="../18946/personal-engagement-1.html">Personal engagement</a></li><li><a href="../18950/exploration-1.html">Exploration</a></li><li><a href="../18957/analysis-1.html">Analysis</a></li><li><a href="../18965/evaluation-1.html">Evaluation</a></li><li><a href="../18966/communication-1.html">Communication</a></li><li><a href="../19882/internal-standardization-of-the-ia-1.html">Internal standardization of the IA</a></li><li><a href="../19901/submitting-the-samples-for-moderation-2.html">Submitting the samples for moderation</a></li><li><a href="../33754/ten-suggestions-for-ias-using-secondary-data-1.html">Ten suggestions for IAs using secondary data</a></li><li><a href="../37544/gaining-full-marks-for-a-databased-ia--1.html">Gaining full marks for a databased IA </a></li><li><a href="../19506/ia-example-marking-exercise--1.html">IA example & marking exercise </a></li><li><a href="../23929/genuine-examples-of-moderated-ia-reports-1.html">Genuine examples of moderated IA reports</a></li><li><a href="../25234/examples-of-teacher-marked-ia-reports--1.html">Examples of teacher-marked IA reports </a></li><li><a href="../37542/history-of-internal-assessment-1.html">History of Internal Assessment</a></li></ul></li><li><a href="../16598/mandatory-laboratory-components-1.html" class="father">Mandatory laboratory components</a><ul class="level-2"><li><a href="../16613/formula-of-magnesium-oxide--1.html">Formula of magnesium oxide </a></li><li><a href="../16612/determining-the-mr-of-an-unknown-gas-1.html">Determining the <i>M</i><sub>r</sub> of an unknown gas</a></li><li><a href="../16615/acid-base-titrations--1.html">Acid-base titrations </a></li><li><a href="../16616/a-green-acid-base-practical-1.html">A green acid-base practical</a></li><li><a href="../16617/analysis-of-aspirin-tablets-1.html">Analysis of aspirin tablets</a></li><li><a href="../16618/caco3-in-egg-shells-1.html">CaCO<sub>3</sub> in egg shells</a></li><li><a href="../16644/enthalpy-changes-1.html">Enthalpy changes</a></li><li><a href="../16631/reaction-rates-1.html">Reaction rates</a></li><li><a href="../16635/rate-dependent-factors-1.html">Rate-dependent factors</a></li><li><a href="../16636/determining-ea-for-a-reaction-1.html">Determining <i>E</i><sub>a</sub> for a reaction</a></li><li><a href="../16637/iodination-of-propanone-1.html">Iodination of propanone</a></li><li><a href="../18144/titrations-with-a-ph-meter-1.html">Titrations with a pH meter</a></li><li><a href="../18355/redox-titration-with-kmno4--1.html">Redox titration with KMnO<sub>4</sub> </a></li><li><a href="../16640/voltaic-cells-1.html">Voltaic cells</a></li><li><a href="../16659/3-d-molecular-modelling--1.html">3-D molecular modelling </a></li></ul></li><li><a href="../17168/other-good-practicals-1.html" class="father">Other good practicals</a><ul class="level-2"><li><a href="../17196/common-chemical-reactions-1.html">Common chemical reactions</a></li><li><a href="../227/elements-oxides-of-the-third-period--1.html">Elements & oxides of the third period </a></li><li><a href="../17239/the-halogens-1.html">The halogens</a></li><li><a href="../17162/boiling-points-of-mixtures-1.html">Boiling points of mixtures</a></li><li><a href="../17170/polarity-of-molecules-1.html">Polarity of molecules</a></li><li><a href="../18083/le-chateliers-principle--1.html">Le Chatelier's principle </a></li><li><a href="../19582/determining-kc-for-an-esterification-reaction-1.html">Determining <i>K</i><sub>c</sub> for an esterification reaction</a></li><li><a href="../18164/redox-reactions-of-vanadium--1.html">Redox reactions of vanadium </a></li><li><a href="../18351/chlorine-in-swimming-pools--1.html">Chlorine in swimming pools </a></li><li><a href="../18178/analysis-of-cuii-ions-in-solution--1.html">Analysis of Cu(II) ions in solution </a></li><li><a href="../18179/percentage-of-copper-in-brass-1.html">Percentage of copper in brass</a></li><li><a href="../18356/electrolytic-cells--1.html">Electrolytic cells </a></li><li><a href="../18598/reactions-of-organic-compounds-1.html">Reactions of organic compounds</a></li><li><a href="../18752/hydrolysis-of-halogenoalkanes-1.html">Hydrolysis of halogenoalkanes</a></li><li><a href="../18784/preparation-of-13-dinitrobenzene--1.html">Preparation of 1,3-dinitrobenzene </a></li><li><a href="../19022/determination-of-an-organic-structure-1.html">Determination of an organic structure</a></li><li><a href="../19887/preparation-of-nylon-66-1.html">Preparation of nylon 6,6</a></li><li><a href="../19343/determination-of-vitamin-c-content--1.html">Determination of vitamin C content </a></li><li><a href="../19342/hydrolysis-of-starch--1.html">Hydrolysis of starch </a></li><li><a href="../19159/preparation-purification-of-aspirin--1.html">Preparation & purification of aspirin </a></li></ul></li><li><a href="../18851/ict-in-practical-work-1.html" class="father">ICT in practical work</a><ul 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concepts</a></li><li><a href="../3507/why-tok-chemistry-1.html">Why TOK & Chemistry?</a></li><li><a href="../3505/a-modern-paradigm-1.html">A modern paradigm</a></li></ul></li></ul></li><li><a href="../17783/fast-track-to-tests-questions-1.html">Fast track to tests & questions</a><ul class="level-1"><li><a href="../17784/sl-multiple-choice-tests-on-individual-topics-1.html">SL Multiple choice tests on individual topics</a></li><li><a href="../24351/sl-multiple-choice-quizzes-on-sub-topics-1.html">SL Multiple choice 'quizzes' on sub-topics</a></li><li><a href="../31519/sl-practice-paper-1-exams-1.html">SL Practice Paper 1 exams</a></li><li><a href="../17786/sl-questions-on-each-sub-topic-1.html">SL Questions on each sub-topic</a></li><li><a href="../20431/sl-paper-3-section-a-questions-1.html">SL Paper 3 Section A questions</a></li><li><a href="../17792/sl-questions-on-the-options-1.html">SL Questions on the options</a></li><li><a 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</div><section id="main-content"><h1><img alt="" class="noborder" src="../../../ib/chemistry/images/Exams/Taking%20exams.jpg" style="width: 1px; height: 1px;"><img alt="" height="21" src="../../../img/tib-icons/higher-level-32-1.png" title="" width="21"> 'Mock' Paper 1 multiple choice exam (3)</h1><section class="tib-teacher-only" readonly="true" title="This box is not visible to students"><div class="header" readonly="true"><img class="icon" src="../../../img/icons/teacher-only.svg"> Teacher only box</div><div class="content" readonly="false"><p>To give your students access to this page you will need to change the access from "Filtered student access" to "Direct student access".</p></div></section><hr class="hidden"><div class="greenBg"><h2>Instructions</h2><ul><li>Time allowed: 1 hour</li><li>Answer all the questions.</li><li>For each question choose the answer you consider to be the best.</li><li>Use the periodic table from Section 6 of the data booklet as your <strong>only</strong> 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data-structure="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" data-score-answers="6d4a526a457a48584f763758344a7649472f76324c53614b78524c393439595570496d5a5971374b38686f3d"><div class="exercise"><div class="q-question"><p>Which compound has a different empirical formula to the other three compounds?</p></div><div class="q-answer"><p><label class="radio" data-answer="756697adc797f57c76fed9723662c914"><input type="radio"><span> ethanoic acid <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/ethanoic-(1).png" style="width: 90px; height: 55px;"> </span></label></p><p><label class="radio" data-answer="02cf7fecedb524fa7ee4abf6aa9b4486"><input type="radio"><span> glucose <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/glucose.png" style="width: 150px; height: 130px;"></span></label></p><p><label class="radio" data-answer="698ef6f46a33537da1e3e3ac9171de7d"><input type="radio"><span> glycerol (propan-1,2,3-triol) <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/glycerol.png" style="width: 130px; height: 73px;"></span></label></p><p><label class="radio" data-answer="144cf4ddaa76a368d2ab0a4d913471bc"><input type="radio"><span> glyceraldehyde (2,3-dihydroxypropanal) <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/glyceraldehyde.png" style="width: 90px; height: 105px;"></span></label></p><p> </p></div><div class="q-explanation"><p>Glucose, C<sub>6</sub>H<sub>12</sub>O<sub>6</sub>, glyceraldehyde, C<sub>3</sub>H<sub>6</sub>O<sub>3</sub> and ethanoic acid, C<sub>2</sub>H<sub>4</sub>O<sub>2</sub> all have the empirical formula CH<sub>2</sub>O. Glycerol, C<sub>3</sub>H<sub>8</sub>O<sub>3</sub> has the empirical formula C<sub>3</sub>H<sub>8</sub>O<sub>3</sub>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which contains 3 x 10<sup>24</sup> hydrogen atoms?<br>Avogadro's constant, <em>L</em> or <em>N</em><sub>A</sub> = 6.02 x 10<sup>23</sup> mol<sup>−1</sup></p></div><div class="q-answer"><p><label class="radio" data-answer="34d4e71aeea93e2b432b6b070de6a5fb"><input type="radio"><span> 125 g of hydrated copper(II) sulfate, CuSO<sub>4</sub>.5H<sub>2</sub>O(s)</span></label></p><p><label class="radio" data-answer="2cf2599f706e68494fdaee235728598a"><input type="radio"><span> 23 g of ethanol, C<sub>2</sub>H<sub>5</sub>OH(l)</span></label></p><p><label class="radio" data-answer="dd57498465be08002cf0d34a0fb4af3b"><input type="radio"><span> 6 g of hydrogen gas, H<sub>2</sub>(g)</span></label></p><p><label class="radio" data-answer="c4231026a665f52d237bb551eca1add3"><input type="radio"><span> 5 dm<sup>3</sup> of 1.0 mol dm<sup>−3</sup> hydrochloric acid solution, HCl(aq)</span></label></p></div><div class="q-explanation"><p><em>M</em><sub>r </sub>CuSO<sub>4</sub>.5H<sub>2</sub>O = (64 + 32 + 64) + (5 x 18) = 250 so 125 g is 0.5 mol and contains 0.5 x 10 mol of H atoms = 3 x 10<sup>24</sup> H atoms.<br>6 g of H<sub>2</sub> contains 6 mol of H atoms = 3.6 x 10<sup>24</sup> H atoms; 23 g of ethanol (<em>M</em><sub>r </sub>= 46) is 0.5 mol so contains 0.5 x 6 mol of H atoms = 1.8 x 10<sup>24 </sup>H atoms. There are many moles of water in 5 dm<sup>3</sup> of the HCl(aq) solution so the acid solution contains many more than 3 x 10<sup>24</sup> H atoms.</p></div><p> </p><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>What volume (in dm<sup>3</sup>) of hydrogen measured at 37 <sup>o</sup>C and at a pressure of 1.0 x 10<sup>5</sup> Pa will be produced when 6.5 g of zinc metal reacts completely with 2.0 mol dm<sup>−3</sup> sulfuric acid, H<sub>2</sub>SO<sub>4</sub>(aq)?<br>Molar volume of a gas at STP = 22.7 dm<sup>3</sup> mol<sup>−1</sup></p></div><div class="q-answer"><p><label class="radio" data-answer="c4e261f0fa58132eb80569945ad42e31"><input type="radio"><span> 0.1 x 22.7 x (310 ÷ 298)</span></label></p><p><label class="radio" data-answer="08f0d40ac1221c9e6ef0e501518c002a"><input type="radio"><span> 0.1 x 22.7 x (298 ÷ 310) </span></label></p><p><label class="radio" data-answer="5bd9b33a12fe6edd4cd816cbf175799d"><input type="radio"><span> 0.1 x 22.7</span></label></p><p><label class="radio" data-answer="7cf61e42bcad2ffcbb7ff3fea0c6ee59"><input type="radio"><span> 0.1 x 22.7 x (310 ÷ 273)</span></label></p></div><div class="q-explanation"><p>Amount of Zn = 6.5/65 = 0.1 mol. Zn(s) + H<sub>2</sub>SO<sub>4</sub>(aq) → ZnSO<sub>4</sub>(aq) + H<sub>2</sub>(g) so amount of hydrogen produced = 0.1 mol. 1.0 mol of gas produced at STP (273 K, 1 x 10<sup>5</sup> Pa ) occupies 22.7 dm<sup>3</sup>, so volume of H<sub>2</sub> produced at 310 K = 0.1 x 22.7 x 310/273 dm<sup>3</sup>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>The diagram represents the transitions of excited electrons from higher energy levels to lower energy levels to produce the hydrogen emission spectrum.</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/hydrogen-energy-levels.png" style="width: 350px; height: 221px;"></p><p>Which emission corresponds to the third line in the visible spectrum?</p></div><div class="q-answer"><p><label class="radio" data-answer="7fc56270e7a70fa81a5935b72eacbe29"><input type="radio"><span> A</span></label></p><p><label class="radio" data-answer="9d5ed678fe57bcca610140957afab571"><input type="radio"><span> B</span></label></p><p><label class="radio" data-answer="0d61f8370cad1d412f80b84d143e1257"><input type="radio"><span> C</span></label></p><p><label class="radio" data-answer="f623e75af30e62bbd73d6df5b50bb7b5"><input type="radio"><span> D</span></label></p></div><div class="q-explanation"><p>The visible series is produced by electrons transitioning from higher levels to the <em>n</em> = 2 level. The first line in the visible series is due to <em>n </em>= 3 to <em>n</em> = 2, the second line <em>n</em> = 4 to <em>n</em> = 2 and the third line from <em>n</em> = 5 to <em>n</em> = 2.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question">Which element will have the highest fourth ionization energy?</div><div class="q-answer"><p><label class="radio" data-answer="2d612a8a156e7fe967c3b971975d067d"><input type="radio"><span> silicon</span></label></p><p><label class="radio" data-answer="057f2c3be5a2ab1d3e942e9433cefeac"><input type="radio"><span> boron</span></label></p><p><label class="radio" data-answer="da9ef8fa9403f7987d521d53d16b42fa"><input type="radio"><span> aluminium</span></label></p><p><label class="radio" data-answer="285de3d457b25471ad7b899e6ebbbbda"><input type="radio"><span> scandium</span></label></p></div><div class="q-explanation"><p>After the third electron has been removed the electron configuration of the ions will be B<sup>3+ </sup>1s<sup>2</sup>, Al<sup>3+</sup> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>, Si<sup>3+</sup> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>1</sup> and Sc<sup>3+</sup> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>6</sup>. Si will have the lowest fourth ionization energy as it has one electron in an outer energy level. The remaining three elements will all be high as they have completely full outer energy levels but boron will be the highest as the electron being removed is in the lowest energy level.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which increases upon descending the halogens group (F → I) but decreases upon descending the alkali metals group (Li → Cs)?</p></div><div class="q-answer"><p><label class="radio" data-answer="0898d2400a88c5f27f7ece8a21df9c4e"><input type="radio"><span> first ionization energy</span></label></p><p><label class="radio" data-answer="f2458e6a4957d9e45b8764e0d6d724ed"><input type="radio"><span> electronegativity</span></label></p><p><label class="radio" data-answer="828c88adeddf2eef700d66162ef494f4"><input type="radio"><span> atomic radius</span></label></p><p><label class="radio" data-answer="653fa5e1b6dc2bab54d338d97016363d"><input type="radio"><span> melting point</span></label></p></div><div class="q-explanation"><p>The atomic radius increases upon descending both groups. The first ionization energy and electronegativity both decrease upon descending both groups. Melting points decreases from Li to Cs but increase from F to I.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which ion shows diamagnetism?</p></div><div class="q-answer"><p><label class="radio" data-answer="18ca74745021a2ec84cda7f26e425963"><input type="radio"><span> Cu<sup>+</sup></span></label></p><p><label class="radio" data-answer="f511408df0627c0ac610cf1a98642383"><input type="radio"><span> Sc<sup>2+</sup></span></label></p><p><label class="radio" data-answer="8b88e415423e363bb5852f15ff1bed36"><input type="radio"><span> Cr<sup>3+</sup></span></label></p><p><label class="radio" data-answer="54ee65a5fd1eaec5f99ea995034101e4"><input type="radio"><span> Mn<sup>2+</sup></span></label></p></div><div class="q-explanation"><p>The electron configuration of Cu<sup>+</sup> is [Ar]3d<sup>10</sup>. As it does not contain any unpaired electrons it will be diamagnetic. The other three ions all contain unpaired electrons so are paramagnetic.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>[Mn(H<sub>2</sub>O)<sub>6</sub>}<sup>2+</sup> is pink and [Cu(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup> is blue.</p><p>Which statement is correct?</p></div><div class="q-answer"><p><label class="radio" data-answer="ca61042c8bd5f75ce239d5ca70595d23"><input type="radio"><span> The absorption of light occurring in [Mn(H<sub>2</sub>O)<sub>6</sub>}<sup>2+</sup> is at a longer wavelength than in [Cu(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup>.</span></label></p><p><label class="radio" data-answer="3a7f91fb049d50143ac97ef40bee45ea"><input type="radio"><span> Both colours are due to excited electrons dropping from a higher energy level to a lower energy level.</span></label></p><p><label class="radio" data-answer="4dcadb9db1b371c1980e9bc8d83a0156"><input type="radio"><span> The movement of nine 3d electrons in [Cu(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup> gives out more energy than the movement of five 3d electrons in [Mn(H<sub>2</sub>O)<sub>6</sub>}<sup>2+</sup>. </span></label></p><p><label class="radio" data-answer="1503c779f20336bda127ac1ac87f1686"><input type="radio"><span> The energy difference between the split d sub-levels is greater for [Mn(H<sub>2</sub>O)<sub>6</sub>}<sup>2+</sup> than [Cu(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup>.</span></label></p></div><div class="q-explanation"><p>The colour of transition metal complexes is due to light being absorbed as electrons are excited from a lower to a higher level in the split d sub-level. The colour that is observed is the complementary colour to the colour absorbed. Pink is the complementary colour to blue so the absorption occurring in [Mn(H<sub>2</sub>O)<sub>6</sub>}<sup>2+</sup> is of higher energy/shorter wavelength than the absorption occurring in the red region of the spectrum in [Cu(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>What is the formula of lead(II) sulfate?</p></div><div class="q-answer"><p><label class="radio" data-answer="ee155c382d5026ecb2f2b7610a865008"><input type="radio"><span> Pb(SO<sub>4</sub>)<sub>2</sub></span></label></p><p><label class="radio" data-answer="d664c003a3cde79cd108d84f1973250e"><input type="radio"><span> Pb<sub>2</sub>SO<sub>4</sub></span></label></p><p><label class="radio" data-answer="9813877f57717e776a5cdb35e3e356a8"><input type="radio"><span> Pb<sub>3</sub>(SO<sub>4</sub>)<sub>2</sub></span></label></p><p><label class="radio" data-answer="d9110b558d838cd883b7a2f1282a5abe"><input type="radio"><span> PbSO<sub>4</sub></span></label></p></div><div class="q-explanation"><p>The lead(II) ion is Pb<sup>2+</sup> and the sulfate ion is SO<sub>4</sub><sup>2−</sup>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which compound has molecules in which the central atom has an incomplete octet of electrons in its outer shell?</p></div><div class="q-answer"><p><label class="radio" data-answer="125081a32c1476d3a4cdca995be6371f"><input type="radio"><span> BCl<sub>3</sub></span></label></p><p><label class="radio" data-answer="8c1dffb33ba9f5ec1a0f46bcad6b1917"><input type="radio"><span> CHCl<sub>3</sub></span></label></p><p><label class="radio" data-answer="04e6e538ed6d0921670aa371c6fe132d"><input type="radio"><span> PCl<sub>3</sub></span></label></p><p><label class="radio" data-answer="ad51ea5f116507bb07f50eccfb6dc155"><input type="radio"><span> Al<sub>2</sub>Cl<sub>6</sub></span></label></p></div><div class="q-explanation"><p>There are only six electrons (three bonding pairs) around the boron atom in BCl<sub>3</sub>. AlCl<sub>3</sub> also has an incomplete octet but when it dimerises to form Al<sub>2</sub>Cl<sub>6</sub> coordinate bonds are formed between Cl and Al to complete the octet. The carbon atom in CHCl<sub>3</sub> has four bonding pairs of electrons and the phosphorus atom in PCl<sub>3</sub> contains three bonding and one non-bonding pairs of electrons.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>In which compound can the bonding between the atoms be explained fully with just one Lewis (electron dot) structure rather than by two or more?</p></div><div class="q-answer"><p><label class="radio" data-answer="ca47bac5c64861f3420429564ce85049"><input type="radio"><span> nitrate ion, NO<sub>3</sub><sup>−</sup></span></label></p><p><label class="radio" data-answer="4dfafa2f58c448056383b16d97f9964e"><input type="radio"><span> aluminium trichloride dimer, Al<sub>2</sub>Cl<sub>6</sub></span></label></p><p><label class="radio" data-answer="531c576f49e8b06f60292194be14b320"><input type="radio"><span> carbonate ion, CO<sub>3</sub><sup>2−</sup></span></label></p><p><label class="radio" data-answer="3fa541d9b2e2fcab21ea7cce7018b006"><input type="radio"><span> ozone, O<sub>3</sub></span></label></p></div><div class="q-explanation"><p>Ozone is described by two resonance structure and the carbonate and nitrate ions are each described by three resonance structures. In aluminium chloride dimer all the bonds are single bonds so can be described by a single structure, <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/al2cl6.png" style="width: 120px; height: 70px;">.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which compound contains sp<sup>2</sup> and sp<sup>3</sup> hybridized carbon atoms?</p></div><div class="q-answer"><p><label class="radio" data-answer="6bf53433153055d38513d04d9f62a561"><input type="radio"><span> ethanenitrile, CH<sub>3</sub>CN</span></label></p><p><label class="radio" data-answer="58f634c81ee840fbc178be4e0766575c"><input type="radio"><span> methylbenzene, C<sub>6</sub>H<sub>5</sub>CH<sub>3</sub></span></label></p><p><label class="radio" data-answer="afd0413c2568cf46e18a225fd0bf6dbd"><input type="radio"><span> graphite, C(s)</span></label></p><p><label class="radio" data-answer="24802140d0b9c117f6b61a4f0204e8a5"><input type="radio"><span> diamond, C(s)</span></label></p></div><div class="q-explanation"><p>In diamond all the carbon atoms are sp<sup>3</sup> hybridized to give a tetrahedral structure and in graphite they are all sp<sup>2</sup> hybridized to give layers made of hexagonal rings with bond angles of 120<sup>o</sup>. In ethanenitrile the carbon atom bonded to the nitrogen atom is sp hybridized. In methylbenzene the six carbon atoms in the phenyl group are sp<sup>2</sup> hybridized and the methyl carbon atom is sp<sup>3</sup> hybridized.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which compound contains six times as many sigma (σ) bonds as it does pi (π) bonds?</p></div><div class="q-answer"><p><label class="radio" data-answer="1221a6286b6024c619cb99a0eb6dcb7a"><input type="radio"><span> ethene, C<sub>2</sub>H<sub>4</sub></span></label></p><p><label class="radio" data-answer="e31d712890cc3c907a33a7335165c230"><input type="radio"><span> ethanoic acid, CH<sub>3</sub>COOH</span></label></p><p><label class="radio" data-answer="7da68204537b07c5075616547d73e096"><input type="radio"><span> ethanal, CH<sub>3</sub>CHO</span></label></p><p><label class="radio" data-answer="f973a4802e4eec36185447acd413d4a1"><input type="radio"><span> propyne, CH<sub>3</sub>CCH</span></label></p></div><div class="q-explanation"><p>Ethanal contains six σ and one π bond. Propyne contains six σ and two π bonds. Ethene contains five σ and one π bond. Ethanoic acid contains seven σ and one π bond.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>What will be the temperature rise, in <sup>o</sup>C, when 50.0 cm<sup>3</sup> of 1.00 mol dm<sup>−3</sup> hydrochloric acid, HCl(aq) at 25 <sup>o</sup>C is mixed with 50.0 cm<sup>3</sup> of 1.00 mol dm<sup>−3</sup> sodium hydroxide solution, NaOH(aq) at 25 <sup>o</sup>C?</p><p>H<sup>+</sup>(aq) + OH<sup>−</sup>(aq) → H<sub>2</sub>O(l) Δ<em>H</em> = − 57.6 kJ mol<sup>−1</sup></p><p>Specific heat capacity of water = 4.18 kJ kg<sup>−1</sup> K<sup>−1</sup><br>Assume the density of water is 1.00 kg dm<sup>−3</sup> and that no heat loss or gain from elsewhere occurs.</p></div><div class="q-answer"><p><label class="radio" data-answer="1815000120a3e1e062f89f9613b5f1a7"><input type="radio"><span> 28.8 ÷ 4.18</span></label></p><p><label class="radio" data-answer="4dd31b58475c54e2bbff0f663f3c4f10"><input type="radio"><span> 5.76 ÷ 4.18</span></label></p><p><label class="radio" data-answer="29f5c993cd56450d161929dd6f5b60f5"><input type="radio"><span> 57.6 ÷ 4.18</span></label></p><p><label class="radio" data-answer="22570d804aa4ffa81ead30c611f023fd"><input type="radio"><span> 2.88 ÷ 4.18</span></label></p></div><div class="q-explanation"><p>50.0 cm<sup>3</sup> of 1.00 mol dm<sup>−3</sup> hydrochloric acid, HCl(aq) contains 50.0/1000 = 0.05 mol so amount of heat evolved = 57.6 x 0.05 = 2.88 kJ. Total volume of solution = 100 cm<sup>3</sup>, so mass of solution = 0.100 kg.<br>Δ<em>T</em> = 2.88 ÷ (4.18 x 0.1) = 28.8 ÷ 4.18 <sup>o</sup>C </p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Propene undergoes an addition reaction with hydrogen bromide to form 2-bromopropane.</p><p>CH<sub>3</sub>CH=CH<sub>2</sub>(g) + HBr(g) → CH<sub>3</sub>CHBrCH<sub>3</sub>(l)</p><p>What is the value for enthalpy change Δ<em>H</em>, in kJ, for this reaction using the data below?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/average-bond-enthalpies.png" style="width: 350px; height: 108px;"></p><p>Enthalpy of vaporization of 2-bromopropane = + 30 kJ mol<sup>−1</sup></p></div><div class="q-answer"><p><label class="radio" data-answer="4403222ec7849592e58048fe428a8c48"><input type="radio"><span> (285 + 346+ 414 ) – (614 + 366) – 30</span></label></p><p><label class="radio" data-answer="b379669f69941a9ca500a0e904be0e1c"><input type="radio"><span> − [(285 + 346 + 414) – (614 + 366) − 30]</span></label></p><p><label class="radio" data-answer="448b1635d8dd69cc7b03669b8debe322"><input type="radio"><span> − [(285 + 346+ 414) – (614 + 366) + 30]</span></label></p><p><label class="radio" data-answer="303324b0fb27e1f3b6733e1acde6672b"><input type="radio"><span> (285 + 346+ 414) – (614 + 366) + 30</span></label></p></div><div class="q-explanation"><p>Average bond enthalpies only refer to the gaseous state. To form gaseous prop-2-ene the bonds broken are C=C and H−Br so the energy in = 614 + 366 kJ. The bonds formed are C−Br, C−C and C−H so the energy out = − (285 + 346 + 414) kJ. The change in enthalpy = − [(285 + 346 + 414) − (614 + 366)] kJ. However the product is in the liquid state so a further 30 kJ of energy will be given out as the gas condenses to a liquid so Δ<em>H</em> = − [(285 + 346 + 414) − (614 + 366) + 30] kJ.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which equation represents the hydration energy of sodium chloride?</p></div><div class="q-answer"><p><label class="radio" data-answer="02161386b9b9a707da02b969923ef6a4"><input type="radio"><span> Na<sup>+</sup>(g) + Cl<sup>−</sup>(g) → NaCl(aq)</span></label></p><p><label class="radio" data-answer="e189fa88fb24418e17f7cb6a40eceacf"><input type="radio"><span> NaCl(s) → NaCl(aq)</span></label></p><p><label class="radio" data-answer="32a7bb222b2002deae8e742900865b9a"><input type="radio"><span> Na(s) + ½Cl<sub>2</sub>(g) → NaCl(aq)</span></label></p><p><label class="radio" data-answer="9ad90a86b06cc7a7b128c15b66dfb74b"><input type="radio"><span> Na<sup>+</sup>(aq) + Cl<sup>−</sup>(aq) → NaCl(aq)</span></label></p></div><div class="q-explanation"><p>Hydration energy is the enthalpy change when molar quantities of gaseous ions dissolve in sufficient water to give an infinitely dilute solution. The process is highly exothermic. Generally the smaller and more highly charged the ions the greater the hydration energy.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which reaction will be spontaneous at low temperatures but non-spontaneous at high temperatures?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/four-reactions-for-spontaneity.png" style="width: 500px; height: 143px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="e1354cc842cc323c307b3424ed3dfa81"><input type="radio"><span> A.</span></label></p><p><label class="radio" data-answer="b06c26aed02d969d0f2315ba11b5432b"><input type="radio"><span> B.</span></label></p><p><label class="radio" data-answer="9bce147872014965a531500da2666847"><input type="radio"><span> C.</span></label></p><p><label class="radio" data-answer="d0904fd99a2cfb13f897223b9213c6f1"><input type="radio"><span> D.</span></label></p></div><div class="q-explanation"><p>B and C will have positive Δ<em>S</em> values as the amount in the gaseous state increases as the reaction proceeds whereas A and C will have negative Δ<em>S</em> values. Δ<em>G</em> = Δ<em>H</em> − TΔ<em>S </em>so to be spontaneous at low temperatures (where TΔ<em>S </em>is relatively small) Δ<em>H</em> needs to be negative and for it to be non-spontaneous at higher temperatures Δ<em>S </em>must be negative so that Δ<em>H</em> − TΔ<em>S </em>has a positive value.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which affects the activation energy of a reaction?</p></div><div class="q-answer"><p><label class="radio" data-answer="a16df8f81986245b070264b0e1545e0e"><input type="radio"><span> Altering the pressure of a gaseous reactant.</span></label></p><p><label class="radio" data-answer="e48b96ee153198cf98ae4962367a7894"><input type="radio"><span> Adding a suitable catalyst.</span></label></p><p><label class="radio" data-answer="2e793ace6407b156db4eb18dc63941b5"><input type="radio"><span> Decreasing the concentration of one of the reactants.</span></label></p><p><label class="radio" data-answer="38be63a3f176767e1f184a64a3e421ca"><input type="radio"><span> Increasing the particle size of a solid reactant.</span></label></p></div><div class="q-explanation"><p>Although increasing the concentration of the reactants, increasing the pressure of gaseous reactants, increasing the temperature and decreasing the particle size (i.e. increasing the surface area) of solid reactants all increase the rate of the reaction they do this by either increasing the number of collisions or by increasing the energy of the particles so that more possess the minimum energy required to react. The addition of a suitable catalyst changes the reaction pathway to one with a lower activation energy.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which method could <strong>not </strong>be used to measure the rate of the exothermic reaction between hydrochloric acid and calcium carbonate?</p><p>2HCl(aq) + CaCO<sub>3</sub>(s) → CaCl<sub>2</sub>(aq) + CO<sub>2</sub>(g) + H<sub>2</sub>O(l)</p></div><div class="q-answer"><p><label class="radio" data-answer="779251bbd7d491c1e7a774a2476239f4"><input type="radio"><span> Monitoring the change in temperature of the solution over time.</span></label></p><p><label class="radio" data-answer="4c75cd389be335ed4316cec6371d6b0f"><input type="radio"><span> Monitoring the change in mass in the reaction vessel over time.</span></label></p><p><label class="radio" data-answer="5a5297334784dbb916e0578fb3c435a9"><input type="radio"><span> Monitoring the change in colour over time.</span></label></p><p><label class="radio" data-answer="7c8c4b6e68b4cc00dac6d1ba17a7cbc5"><input type="radio"><span> Monitoring the change in volume of gas evolved over time.</span></label></p></div><div class="q-explanation"><p>The temperature of the solution, volume of gas evolved and mass of remaining reactants as the carbon dioxide evolves all change over time. There is no change in colour over time as the white calcium carbonate solid dissolves in colourless hydrochloric acid to give a colourless solution.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which graph could be obtained for a zero order reaction with respect to reactant A?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/order-of-reaction-graphs.png" style="width: 360px; height: 288px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="7fc56270e7a70fa81a5935b72eacbe29"><input type="radio"><span> A</span></label></p><p><label class="radio" data-answer="9d5ed678fe57bcca610140957afab571"><input type="radio"><span> B</span></label></p><p><label class="radio" data-answer="0d61f8370cad1d412f80b84d143e1257"><input type="radio"><span> C</span></label></p><p><label class="radio" data-answer="f623e75af30e62bbd73d6df5b50bb7b5"><input type="radio"><span> D</span></label></p></div><div class="q-explanation"><p>The rate at which the concentration of A decreases over time will be constant for a zero order reaction, so a graph of [A] against time will be a straight line with a negative slope. D is the graph for a second order reaction and A and B are both for first order reactions.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Hydrogen reacts with nitrogen(II) oxide according to the equation:</p><p>2NO(g) + 2H<sub>2</sub>(g) → 2H<sub>2</sub>O(g) + N<sub>2</sub>(g)</p><p>The proposed mechanism which is consistent with the experimentally determined rate equation for the reaction consists of three separate steps:</p><p>Step 1: (fast) NO(g) + NO(g) → N<sub>2</sub>O<sub>2</sub>(g)</p><p>Step 2:(slow) N<sub>2</sub>O<sub>2</sub>(g) + H<sub>2</sub>(g) → N<sub>2</sub>O(g) + H<sub>2</sub>O(g)</p><p>Step 3: (fast) N<sub>2</sub>O(g) + H<sub>2</sub>(g) → N<sub>2</sub>(g) + H<sub>2</sub>O(g)</p><p>Which statement is correct?</p></div><div class="q-answer"><p><label class="radio" data-answer="934c6e0f1c6f8a14007db65ea5a86f03"><input type="radio"><span> The units of the rate constant for the reaction are dm<sup>3</sup> mol<sup>−1</sup> s<sup>−1</sup></span></label></p><p><label class="radio" data-answer="d86bec3bb2ffba07ad8f7d4f86a44a3c"><input type="radio"><span> The rate equation for the reaction is: rate = <em>k</em>[N<sub>2</sub>O(g)][H<sub>2</sub>(g)]</span></label></p><p><label class="radio" data-answer="49c14ff2ac7ce63b82318cf72acf8ac2"><input type="radio"><span> The molecularity of the rate determining step is four</span></label></p><p><label class="radio" data-answer="1e034879a5f3e267e27cee6b0b2b5099"><input type="radio"><span> The overall order of the reaction is three</span></label></p></div><div class="q-explanation"><p>For the mechanism to be consistent with the rate equation the rate equation must be rate = <em>k</em>[NO(g)]<sup>2</sup>[H<sub>2</sub>(g)] so the overall order is three. The units of the rate constant will be dm<sup>6</sup> mol<sup>−2</sup> s<sup>−1</sup> and the molecularity of the rate determining step (the second step) will be two..</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which statement shows what will happen when one of the conditions is altered and the rest remain the same for the reaction between nitrogen and hydrogen?</p><p>N<sub>2</sub>(g) + 3H<sub>2</sub>(g) ⇌ 2NH<sub>3</sub>(g) Δ<em>H</em> = − 92.4 kJ</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/equilibrium-changes(2).png" style="width: 500px; height: 199px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="7fc56270e7a70fa81a5935b72eacbe29"><input type="radio"><span> A</span></label></p><p><label class="radio" data-answer="9d5ed678fe57bcca610140957afab571"><input type="radio"><span> B</span></label></p><p><label class="radio" data-answer="0d61f8370cad1d412f80b84d143e1257"><input type="radio"><span> C</span></label></p><p><label class="radio" data-answer="f623e75af30e62bbd73d6df5b50bb7b5"><input type="radio"><span> D</span></label></p></div><div class="q-explanation"><p><em>K</em><sub>c</sub> is a constant at a specified temperature so altering the pressure has no effect on the value of <em>K</em><sub>c</sub>. The reaction is exothermic as Δ<em>H </em>has a negative value so lowering the temperature will shift the position of equilibrium towards the products and increase the value of the equilibrium constant.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>What does the position of equilibrium correspond to?</p></div><div class="q-answer"><p><label class="radio" data-answer="fc2aa212d4cbf9b02b68e50837d8dfe9"><input type="radio"><span> A maximum value of entropy and a minimum in the value of the Gibbs free energy. </span></label></p><p><label class="radio" data-answer="e5f9dee70fd664c80c9a3b8afa566ee6"><input type="radio"><span> A maximum value of entropy and a maximum in the value of the Gibbs free energy.</span></label></p><p><label class="radio" data-answer="31bd1cca730220989a0adaa58a20e76e"><input type="radio"><span> A minimum value of entropy and a minimum in the value of the Gibbs free energy.</span></label></p><p><label class="radio" data-answer="2d3a0f4dab2c400bc8ae81ce60268979"><input type="radio"><span> A minimum value of entropy and a maximum in the value of the Gibbs free energy.</span></label></p></div><div class="q-explanation"><p>The entropy will have a maximum value at equilibrium so that the available energy is distributed in as many ways as possible. At equilibrium Δ<em>G</em> = 0. This is because the rate of the forward reaction is equal to the rate of the reverse reaction so no net reaction occurs hence Δ<em>G</em> will be zero.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which is the conjugate base of the hydroxide ion?</p></div><div class="q-answer"><p><label class="radio" data-answer="7bcfbacd98cfa6ce29d0ad8201b14f09"><input type="radio"><span> H<sup>+</sup></span></label></p><p><label class="radio" data-answer="b89e91ccf14bfd7f485dd7be7d789b0a"><input type="radio"><span> H<sub>2</sub>O</span></label></p><p><label class="radio" data-answer="f46de30a1dfccaa06e04ff49121bff09"><input type="radio"><span> O<sup>2−</sup></span></label></p><p><label class="radio" data-answer="357ac9de262ea126d648229817bd89bd"><input type="radio"><span> H<sub>3</sub>O<sup>+</sup></span></label></p></div><div class="q-explanation"><p>The conjugate base is formed when a proton is donated, OH<sup>−</sup>(aq) → H<sup>+</sup>(aq) + O<sup>2−</sup>(aq)</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which describes the relative properties when 0.50 mol dm<sup>−3</sup> aqueous solutions of hydrochloric acid and ethanoic acid are compared? </p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/acid-comparisons(3).png" style="width: 500px; height: 120px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="7fc56270e7a70fa81a5935b72eacbe29"><input type="radio"><span> A</span></label></p><p><label class="radio" data-answer="9d5ed678fe57bcca610140957afab571"><input type="radio"><span> B</span></label></p><p><label class="radio" data-answer="0d61f8370cad1d412f80b84d143e1257"><input type="radio"><span> C</span></label></p><p><label class="radio" data-answer="f623e75af30e62bbd73d6df5b50bb7b5"><input type="radio"><span> D</span></label></p></div><div class="q-explanation"><p>Strong acids are completely dissociated into ions so have a higher electrical conductivity than weak acids which are only partially dissociated. [H+(aq)] will be higher for strong acids so the pH will be lower.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which is the increasing order of acid strength (weakest acid first) of the following four acids?</p><p>(CH<sub>3</sub>)<sub>3</sub>CCOOH p<em>K</em><sub>a</sub> = 5.03</p><p>C<sub>6</sub>H<sub>5</sub>COOH <em>K</em><sub>a</sub> = 6.3 x 10<sup>−5</sup></p><p>CH<sub>2</sub>ClCOOH <em>K</em><sub>a</sub> = 1.3 x 10<sup>−3</sup></p><p>HCOOH <em>pK</em><sub>a</sub> = 3.75</p></div><div class="q-answer"><p><label class="radio" data-answer="ce264776e23c3820cfc10c5257b076ce"><input type="radio"><span> CH<sub>2</sub>ClCOOH < HCOOH < (CH<sub>3</sub>)<sub>3</sub> CCOOH < C<sub>6</sub>H<sub>5</sub>COOH</span></label></p><p><label class="radio" data-answer="695b7d61efd3a4bf1f9c3c170cf764b4"><input type="radio"><span> HCOOH < CH<sub>2</sub>ClCOOH < (CH<sub>3</sub>)<sub>3</sub>CCOOH < C<sub>6</sub>H<sub>5</sub>COOH</span></label></p><p><label class="radio" data-answer="a9e88942f00ee828c509c417a6536287"><input type="radio"><span> C<sub>6</sub>H<sub>5</sub>COOH < (CH<sub>3</sub>)<sub>3</sub>CCOOH < CH<sub>2</sub>ClCOOH < HCOOH</span></label></p><p><label class="radio" data-answer="714285a2f91eb1cb5fb1823393b9d0ad"><input type="radio"><span> (CH<sub>3</sub>)<sub>3</sub>CCOOH < C<sub>6</sub>H<sub>5</sub>COOH < HCOOH < CH<sub>2</sub>ClCOOH</span></label></p></div><div class="q-explanation"><p>p<em>K</em><sub>a </sub>= − log<sub>10</sub><em>K</em><sub>a</sub>. The smallest value for <em>K</em><sub>a</sub> (in the order of 10<sup>−6</sup> so smaller than C<sub>6</sub>H<sub>5</sub>COOH which is in the order of 10<sup>−5</sup>) is for (CH<sub>3</sub>)<sub>3</sub>CCOOH. The <em>K</em><sub>a</sub> of HCOOH is in the order of 10<sup>−4</sup> so the largest value for<em> K</em><sub>a</sub> (in the order of 10<sup>−3</sup>) is for CH<sub>2</sub>ClCOOH.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which salt will dissolve in water to give a solution with the highest pH value ?</p></div><div class="q-answer"><p><label class="radio" data-answer="0ecd62db1795f841a355bb92f94ad47f"><input type="radio"><span> NaNO<sub>3</sub></span></label></p><p><label class="radio" data-answer="3cab4a31a4903a9b82026dcb54ff961d"><input type="radio"><span> Na<sub>2</sub>CO<sub>3</sub></span></label></p><p><label class="radio" data-answer="07b03257b8b49a9217e57aa910538461"><input type="radio"><span> NH<sub>4</sub>NO<sub>3</sub></span></label></p><p><label class="radio" data-answer="e1743b63de15ba0bc8049c5c4b8a8ecd"><input type="radio"><span> NH<sub>4</sub>CH<sub>3</sub>COO</span></label></p></div><div class="q-explanation"><p>NaNO<sub>3</sub> is the salt of a strong acid and a strong base so will give a neutral solution. NH<sub>4</sub>CH<sub>3</sub>COO is the salt of a weak acid and a weak base so will also give a solution with a pH of approximately 7 (the exact pH depends upon which of the two dissociates the most). NH<sub>4</sub>NO<sub>3</sub> is the salt of a strong acid and a weak base so will give an acidic solution with a pH less than 7, whereas Na<sub>2</sub>CO<sub>3</sub> is the salt of a strong base and a weak acid so will give an alkaline solution with a pH greater than 7.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>In which compound is the oxidation state of iodine +5?</p></div><div class="q-answer"><p><label class="radio" data-answer="b3dc61c7a66fe83ed56f852e7a0089c5"><input type="radio"><span> Iodine dioxide, IO<sub>2</sub></span></label></p><p><label class="radio" data-answer="1fcbd428d2000e9709af6339f8497e7d"><input type="radio"><span> Orthoperiodic acid, H<sub>5</sub>IO<sub>6</sub></span></label></p><p><label class="radio" data-answer="9a7bc25ee4f5049234e7656fecf6a614"><input type="radio"><span> Iodic acid, HIO<sub>3</sub></span></label></p><p><label class="radio" data-answer="d230f35de3c80f948978019ceb6d4085"><input type="radio"><span> Hypoiodous acid, HOI</span></label></p></div><div class="q-explanation"><p>H has an oxidation state of +1 (except for hydride ions, H<sup>−</sup> where it is −1 ) and O has an oxidation state of −2 (except for peroxides −O−O− where it is −1) so iodine must have an oxidation state of +5 in iodic acid to give the sum of the oxidation states equal to zero. In hypoiodous acid the oxidation state of iodine is +1, in iodine dioxide it is +4 and in orthoperiodic acid it is +7.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>In which reaction is hydrogen peroxide acting both as an oxidising agent and as a reducing agent?</p></div><div class="q-answer"><p><label class="radio" data-answer="eaa8cf777ec57f527c9f9dfde33bc1bf"><input type="radio"><span> 2KMnO<sub>4</sub>(aq) + 3H<sub>2</sub>O<sub>2</sub>(aq) → 2MnO<sub>2</sub>(s) + 2KOH(aq) + 2H<sub>2</sub>O(l) + 3O<sub>2</sub>(g)</span></label></p><p><label class="radio" data-answer="f1988c2fb33ba6e44eae71c2c29e1f83"><input type="radio"><span> NaOCl(aq) + H<sub>2</sub>O<sub>2</sub>(aq) → O<sub>2</sub>(g) + NaCl(aq) + H<sub>2</sub>O(l)</span></label></p><p><label class="radio" data-answer="db429d722068b951bc13defaaaa72c05"><input type="radio"><span> 2H<sub>2</sub>O<sub>2</sub>(aq) → 2H<sub>2</sub>O(l) + O<sub>2</sub>(g)</span></label></p><p><label class="radio" data-answer="eccd6f0013a14beeb3c4b96adad42a5d"><input type="radio"><span> FeSO<sub>4</sub>(aq) + H<sub>2</sub>O<sub>2</sub>(aq) + H<sub>2</sub>SO<sub>4</sub> (aq) → Fe<sub>2</sub>(SO<sub>4</sub>)<sub>3</sub>(aq) + 2H<sub>2</sub>O(l)</span></label></p></div><div class="q-explanation"><p>When hydrogen peroxide decomposes it simultaneously oxidises and reduces itself so that the oxidation state of oxygen in hydrogen peroxide (+1) increases to zero in oxygen and decreases to minus two in water (this is an example of disproportionation). When it acts as a reducing agent, as with KMnO<sub>4</sub> and NaOCl, oxygen is formed and when it acts as an oxidising agent, as with FeSO<sub>4</sub>, it is reduced to water.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question">A steady current of 1.0 A is passed through a dilute solution of sulfuric acid for fixed amount of time using inert electrodes. It is found that 40 cm<sup>3</sup> of gas is evolved at the cathode (negative electrode). If the experiment is repeated with a steady current of 3.0 A for half the time what volume of gas, in cm<sup>3</sup>, will be evolved at the other electrode assuming all the other conditions remain the same?</div><div class="q-answer"><p><label class="radio" data-answer="34173cb38f07f89ddbebc2ac9128303f"><input type="radio"><span> 30</span></label></p><p><label class="radio" data-answer="d645920e395fedad7bbbed0eca3fe2e0"><input type="radio"><span> 40</span></label></p><p><label class="radio" data-answer="072b030ba126b2f4b2374f342be9ed44"><input type="radio"><span> 60</span></label></p><p><label class="radio" data-answer="98f13708210194c475687be6106a3b84"><input type="radio"><span> 20</span></label></p></div><div class="q-explanation"><p>For every two mol of H<sub>2</sub> evolved at the cathode 1 mol of O<sub>2</sub> will be evolved at the anode. By Avogadro's Law the ratio of the volumes will be the same as the molar ratio. If the current is tripled and the time is halved the amount of H<sub>2</sub> evolved will be increased 1.5 times to 60 cm<sup>3</sup>. Hence the volume of O<sub>2</sub> evolved will be 30 cm<sup>3</sup>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>A cell is made from two half-cells:</p><p>Ni<sup>2+</sup>(aq) + 2e<sup>−</sup> ⇌ Ni(s) <em>E</em><sup>⦵</sup> = − 0.26 V</p><p>Cu<sup>2+</sup>(aq) + 2e<sup>−</sup> ⇌ Cu(s) <em>E</em><sup>⦵</sup> = + 0.34 V</p><p>What is the value of Δ<em>G,</em> in J mol<sup>−1</sup>, for the reaction that occurs when the cell is operating under standard conditions?</p><p>Faraday's constant (<em>F</em>) = 9.65 x 10<sup>4</sup> C mol<sup>−1</sup></p></div><div class="q-answer"><p><label class="radio" data-answer="e68feb76dd715cc1f84014a1c4e8c03f"><input type="radio"><span> 2 x 9.65 x 10<sup>4</sup> x 0.60</span></label></p><p><label class="radio" data-answer="6c7b556735210b44bb557f088471e613"><input type="radio"><span> 2 x 9.65 x 10<sup>4</sup> x 0.08</span></label></p><p><label class="radio" data-answer="4ecb46c0815a144a7931e5d65e7bd9d4"><input type="radio"><span> − 2 x 9.65 x 10<sup>4</sup> x 0.60 </span></label></p><p><label class="radio" data-answer="0eddb8a76467551b8ceb5060c60c16f5"><input type="radio"><span> − 2 x 9.65 x 10<sup>4</sup> x 0.08</span></label></p></div><div class="q-explanation"><p>The spontaneous reaction that occurs is Ni(s) + Cu<sup>2+</sup>(aq) → Ni<sup>2+</sup>(aq) + Cu(s) with an EMF of 0.60 V (where 0.60 V is the difference between the two <em>E</em><sup>⦵</sup> values for the two half-cells). Δ<em>G</em> = − <em>nFE</em><sup>⦵</sup> = − 2 x 9.65 x 10<sup>4</sup> x 0.60 J mol<sup>−1</sup>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which compound contains a tertiary carbon atom?</p></div><div class="q-answer"><p><label class="radio" data-answer="d810cda4e76f0aa4cdb2d2198fe93c05"><input type="radio"><span> CH<sub>3</sub>CH<sub>2</sub>CBr(CH<sub>3</sub>)<sub>2</sub></span></label></p><p><label class="radio" data-answer="9677a8d56d254c370d5f4ea4f5a92c40"><input type="radio"><span> CH<sub>3</sub>CH<sub>2</sub>CHClCH<sub>3</sub></span></label></p><p><label class="radio" data-answer="42be35a3525e5b8ef8c19047a55ed70f"><input type="radio"><span> CH<sub>3</sub>CH(OH)CH<sub>3</sub></span></label></p><p><label class="radio" data-answer="e82712b6aff4a0ab1d6f08d48def8b87"><input type="radio"><span> CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>Cl</span></label></p></div><div class="q-explanation"><p>A tertiary carbon atoms is bonded to a functional group such as a halogen or -OH and to three R groups (the R groups may be the same as each other or different). A tertiary carbon atom is not bonded to a hydrogen atom.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Testosterone is a male sex hormone.</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/testosterone.png" style="width: 250px; height: 172px;"></p><p>Which functional groups are all present in a molecule of testosterone?</p></div><div class="q-answer"><p><label class="radio" data-answer="368e5615c5ffeda85aca379c4a6b3425"><input type="radio"><span> ether, alkenyl and hydroxyl</span></label></p><p><label class="radio" data-answer="db976b0f2761497e644e1acc9579de0b"><input type="radio"><span> phenyl, ketone and hydroxyl</span></label></p><p><label class="radio" data-answer="2388f06274a9a412631191cb1d8ceb5d"><input type="radio"><span> alkenyl, carbonyl and hydroxyl</span></label></p><p><label class="radio" data-answer="9d7922243e2258cef679f12c58c4b1dc"><input type="radio"><span> phenyl, alkenyl and aldehyde</span></label></p></div><div class="q-explanation"><p>>C=O, >C=C< and −OH (carbonyl, alkenyl and hydroxyl) are all present. −CHO, C<sub>6</sub>H<sub>5</sub>− and R−O−R' (aldehyde, phenyl and ether) are all absent.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which organic product is formed when propanoic acid, CH<sub>3</sub>CH<sub>2</sub>COOH is warmed with propan-2-ol, CH<sub>3</sub>CH(OH)CH<sub>3 </sub>in the presence of a few drops of concentrated sulfuric acid?</p></div><div class="q-answer"><p><label class="radio" data-answer="ecda348ecff8a1c50bb89e4beb637a76"><input type="radio"><span> CH<sub>3</sub>CH<sub>2</sub>COOCH(CH<sub>3</sub>)<sub>2</sub></span></label></p><p><label class="radio" data-answer="4f3a020bd4a951aa0a081c36bec7beb1"><input type="radio"><span> (CH<sub>3</sub>)<sub>2</sub>CHCOOCH<sub>3</sub>CH<sub>2</sub></span></label></p><p><label class="radio" data-answer="37e7bbf990dde6495f289d7486a5f10d"><input type="radio"><span> CH<sub>3</sub>CH<sub>2</sub>OCOCH(CH<sub>3</sub>)<sub>2</sub></span></label></p><p><label class="radio" data-answer="ff5862e66637ac73b9a35819e2985c48"><input type="radio"><span> CH<sub>3</sub>CH(OH)CH<sub>2</sub>COOCH<sub>2</sub>CH<sub>3</sub></span></label></p></div><div class="q-explanation"><p>In the presence of a sulfuric acid catalyst, carboxylic acids condense with alcohols to form an ester and water.<br>CH<sub>3</sub>CH<sub>2</sub>COOH + CH<sub>3</sub>CH(OH)CH<sub>3 </sub>→ CH<sub>3</sub>CH<sub>2</sub>COOCH(CH<sub>3</sub>)<sub>2</sub> + H<sub>2</sub>O</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which intermediate is formed when hydrogen bromide adds to 2-methylpropene, (CH<sub>3</sub>)<sub>2</sub>C=CH<sub>2</sub>?</p></div><div class="q-answer"><p><label class="radio" data-answer="5c2d330fc1da6f0aa78b66ed843e4804"><input type="radio"><span> <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/markovnikov-b.png" style="width: 120px; height: 75px;"></span></label></p><p><label class="radio" data-answer="c6f6c84528f8ce72d365ec1832948225"><input type="radio"><span> <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/markovnikov-c.png" style="width: 120px; height: 98px;"></span></label></p><p><label class="radio" data-answer="07bf7ba450ef6fb28717c5c6717d04fc"><input type="radio"><span> <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/markovnikov-a.png" style="width: 120px; height: 101px;"></span></label></p><p><label class="radio" data-answer="73d22288e66d31ff3ac55298505a7167"><input type="radio"><span> <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/markovnikov-d.png" style="width: 130px; height: 74px;"></span></label></p></div><div class="q-explanation"><p>Hydrogen bromide is polar,<sup> δ+</sup>H−Br<sup>δ−</sup>. The mechanism is electrophilic addition. The hydrogen atom adds to the carbon atom that already contains the most hydrogen atoms bonded to it. This is because the tertiary carbocation that is formed is more stable than the primary carbocation that would be formed if it added to the other carbon atom making up the double bond.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which can show conformational isomerism but not configurational isomerism?</p></div><div class="q-answer"><p><label class="radio" data-answer="734f6b974bdd84c7a2a16f5dd80b6586"><input type="radio"><span> butane, C<sub>4</sub>H<sub>10</sub></span></label></p><p><label class="radio" data-answer="e6ea3c4e388bd87fdd11d027ed48b7e1"><input type="radio"><span> 1,3-dichlorocyclobutane</span></label></p><p><label class="radio" data-answer="595d8a852554ed48380a5afa9182169e"><input type="radio"><span> but-2-ene, CH<sub>3</sub>CHCHCH<sub>3</sub></span></label></p><p><label class="radio" data-answer="ecf193fe2e77e777250f63b9e72e80ef"><input type="radio"><span> butan-2-ol, CH<sub>3</sub>CH<sub>2</sub>CH(OH)CH<sub>3</sub></span></label></p></div><div class="q-explanation"><p>Both conformational and configurational isomerism are forms of stereoisomerism. Conformational isomerism involves rotations about single bonds whereas the rotations in configurational isomerism involve the breaking and reforming of chemical bonds. Butane can be rotated about the single bond between the second and third carbon atoms.<br><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/conformational-isomers.png" style="width: 181px; height: 80px;"></p><p> </p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which is the correct <em>E/Z</em> configuration for the two isomers?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/e-z-configuration.png" style="width: 400px; height: 174px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="7fc56270e7a70fa81a5935b72eacbe29"><input type="radio"><span> A</span></label></p><p><label class="radio" data-answer="9d5ed678fe57bcca610140957afab571"><input type="radio"><span> B</span></label></p><p><label class="radio" data-answer="0d61f8370cad1d412f80b84d143e1257"><input type="radio"><span> C</span></label></p><p><label class="radio" data-answer="f623e75af30e62bbd73d6df5b50bb7b5"><input type="radio"><span> D</span></label></p></div><div class="q-explanation"><p>Chlorine has a higher atomic number than carbon so has a higher priority according to Cahn-Ingold-Prelog (CIP) rules. In both cases the two highest groups are on the same side so both isomers have the <em>Z</em> configuration.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>23.20 cm<sup>3</sup> of 0.10 mol dm<sup>−3</sup> potassium hydroxide solution was required to neutralize 20.00 cm<sup>3</sup> of a solution of hydrochloric acid.<br>What was the concentration, in mol dm<sup>−3</sup>, of the hydrochloric acid solution?</p></div><div class="q-answer"><p><label class="radio" data-answer="d2c768dfd05952a899d0b2c64139195e"><input type="radio"><span> 0.1160 </span></label></p><p><label class="radio" data-answer="cb5ae17636e975f9bf71ddf5bc542075"><input type="radio"><span> 0.1 </span></label></p><p><label class="radio" data-answer="b1659515b9d1a59ebbc790e01084a8f0"><input type="radio"><span> 0.12</span></label></p><p><label class="radio" data-answer="339991b274e97fd544f2d285cd7661a6"><input type="radio"><span> 0.116 </span></label></p></div><div class="q-explanation"><p>The number of significant figures in a result is based on the figures given in the data. Since the concentration of the potassium hydroxide solution is only given to two significant figures the concentration of the hydrochloric acid solution should only be quoted to two significant figures.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Thymine is one of the four nucleobases in DNA. It has the molecular formula C<sub>5</sub>H<sub>6</sub>N<sub>2</sub>O<sub>2</sub>. What is the index of hydrogen deficiency (IHD) of thymine?</p></div><div class="q-answer"><p><label class="radio" data-answer="eccbc87e4b5ce2fe28308fd9f2a7baf3"><input type="radio"><span> 3</span></label></p><p><label class="radio" data-answer="e4da3b7fbbce2345d7772b0674a318d5"><input type="radio"><span> 5</span></label></p><p><label class="radio" data-answer="c81e728d9d4c2f636f067f89cc14862c"><input type="radio"><span> 2</span></label></p><p><label class="radio" data-answer="a87ff679a2f3e71d9181a67b7542122c"><input type="radio"><span> 4</span></label></p></div><div class="q-explanation"><p>When determining IHD, O atoms count as zero. For each N atom add one to the number of C atoms and add one to the number of H atoms. This gives the equivalent of C<sub>7</sub>H<sub>8</sub> which needs 8 more hydrogen atoms (4 units of H<sub>2</sub>) to become saturated as C<sub>7</sub>H<sub>16</sub>, hence the IHD is 4. </p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>What will be the same in the <sup>1</sup>H NMR spectra of butan-1-ol and butan-2-ol?</p></div><div class="q-answer"><p><label class="radio" data-answer="d3d3118fffcf98a4d5cadf2a4eadbbc9"><input type="radio"><span> the number of signals<span class="radio"></span></span></label></p><p><label class="radio" data-answer="40dc6ebde00dd5255212660bd422b709"><input type="radio"><span> chemical shift of each absorption</span></label></p><p><label class="radio" data-answer="a3062d12292c22409e8b6a912c85e8ee"><input type="radio"><span> the splitting patterns</span></label></p><p><label class="radio" data-answer="d5939667b79a9ffe6e319d216d5dcbc5"><input type="radio"><span> the ratio of the areas under each signal</span></label></p></div><div class="q-explanation"><p>Butan-1-ol, CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>CH<sub>2</sub>OH and butan-2-ol, CH<sub>3</sub>CH<sub>2</sub>CH(OH)CH<sub>3</sub> will both give five signals as the hydrogen atoms in both compounds are in five separate chemical environments. Butan-1-ol will have the ratio of 3:2:2:2:1 and butan-2-ol will have the ratio 3:2:1:1:3 for the areas under each signal. Because butan-2-ol contains one carbon atom bonded to only one hydrogen atom their splitting patterns will be different. Since the oxygen atom is on a different carbon atom their chemical shifts will also be different.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="totals"><span class="score">Total Score: </span><button class="btn btn-success check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr></section></article><section id="media-extras"><div class="page-actions no-print navbar inline hidden-desktop"><div class="navbar-inner"><ul class="nav"><li><a class="presentation" href="#" onclick="return false;"><i class="fa fa-desktop"></i></a></li><li><a class="print-section-blog" href="#" onclick="return false;"><i class="fa fa-print"></i></a></li><li><a class="page-bookmarker" data-ticket="IB Docs (2) Team" data-pid="32111" href="#" onclick="return false;"><i class="fa fa-star"></i></a></li><li><a class="personal-notes" href="#" onclick="return false;"><i class="fa fa-file-text"></i></a></li><li><a class="" data-toggle="modal" href="#modal-feedback" onclick="return false;"><i class="fa fa-envelope-o"></i></a></li><li class="dropdown"><a class="dropdown-toggle" data-toggle="dropdown" data-target="#" href="#"><i class="fa fa-share-alt"></i></a><ul class="dropdown-menu" role="menu"><li><a class="" target="_blank" title="Share on Twitter" href="https://twitter.com/share?text=%22HL+Practice+Paper+1+%283%29++%22+-+via+%40InThinker+%23chemistry%0A&url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32111%2Fhl-practice-paper-1-3-"><i class="fa fa-twitter-square"></i><span>Twitter</span></a></li><li><a class="" target="_blank" title="Share on Facebook" href="https://www.facebook.com/sharer.php?u=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32111%2Fhl-practice-paper-1-3-&t=HL+Practice+Paper+1+(3)++"><i class="fa fa-facebook-square"></i><span>Facebook</span></a></li><li><a class="" href="http://www.linkedin.com/shareArticle?mini=true&url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32111%2Fhl-practice-paper-1-3-"><i class="fa fa-linkedin-square"></i><span>LinkedIn</span></a></li><li><a class="" href="https://plus.google.com/share?url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32111%2Fhl-practice-paper-1-3-"><i class="fa fa-google-plus-square"></i><span>Google +</span></a></li><li><a class="" href="mailto:?subject=HL Practice Paper 1 (3) &body=To give your students access to this page you will need to change the access from " filtered="" student="" access"="" to="" "direct="" student="" access".="" %0ahttps%3a%2f%2fwww.thinkib.net%2fchemistry%2fpage%2f32111%2fhl-practice-paper-1-3-"=""><i class="fa fa-envelope"></i><span>Email</span></a></li></ul></li><li><a class="" href="chemistry/teaching-materials"><i class="fa fa-puzzle-piece colored"></i></a></li></ul></div></div><div style="border-top: solid 1px #eee;border-bottom: solid 1px #eee;padding: 4px 0 4px 0;line-height: 1em;color: #666;font-size: .8em"><small><em>All materials on this website are for the exclusive use of teachers and students at subscribing schools for the period of their subscription. 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