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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 (1)</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>What is the sum of the integer coefficients when ammonia is oxidized in air in the presence of a catalyst to produce nitrogen(II) oxide and water?</p><p style="margin-left: 40px;">_ NH<sub>3</sub>(g) + ­_ O<sub>2</sub>(g) → _ NO(g) + _ H<sub>2</sub>O(l)</p></div><div class="q-answer"><p><label class="radio" data-answer="37693cfc748049e45d87b8c7d8b9aacd"><input type="radio"><span> 23</span></label></p><p><label class="radio" data-answer="d3d9446802a44259755d38e6d163e820"><input type="radio"><span> 10</span></label></p><p><label class="radio" data-answer="1f0e3dad99908345f7439f8ffabdffc4"><input type="radio"><span> 19</span></label></p><p><label class="radio" data-answer="6512bd43d9caa6e02c990b0a82652dca"><input type="radio"><span> 11</span></label></p></div><div class="q-explanation"><p>The balanced equation is 4NH<sub>3</sub>(g) + ­5O<sub>2</sub>(g) → 4NO(g) + 6H<sub>2</sub>O(l)</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 concentration (in mol dm<sup>−3</sup>) of hydrochloric acid in the solution formed by adding 150 cm<sup>3</sup> of 0.20 mol dm<sup>−3</sup> HCl(aq) to 250 cm<sup>3</sup> of 0.30 mol dm<sup>−3</sup> HCl(aq)? </p></div><div class="q-answer"><p><label class="radio" data-answer="94e0b7097bece407207267f7787e3012"><input type="radio"><span> 0.22</span></label></p><p><label class="radio" data-answer="246c0903b5a64b2a854ec1e7865f174f"><input type="radio"><span> 0.29</span></label></p><p><label class="radio" data-answer="261943f3a93b683ceeac658927f3923f"><input type="radio"><span> 0.26</span></label></p><p><label class="radio" data-answer="5d6182b8169f820c3e247e91131138ea"><input type="radio"><span> 0.24</span></label></p></div><div class="q-explanation"><p>Amount of HCl in 150 cm<sup>3</sup> of 0.20 mol dm<sup>−3</sup> HCl(aq) = 150/1000 x 0.20 mol<br>Amount of HCl in 250 cm<sup>3</sup> of 0.30 mol dm<sup>−3</sup> HCl(aq) = 250/1000 x 0.30 mol<br>Amount of HCl in 400 cm<sup>3</sup> of mixture = (150/1000 x 0.20) + (250/1000 x 0.30) mol<br>Amount of HCl in 1000 cm<sup>3</sup> (1 dm<sup>3</sup>) of mixture = [(150/1000 x 0.20) + (250/1000 x 0.30)] x 1000/400 = 0.26 mol</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 volume of gas, measured at STP, will be remaining when 1.60 g of methane is combusted completely in 12.80 g of oxygen gas? (1 mol of an ideal gas occupies 22.7 dm<sup>3</sup> at STP).</p></div><div class="q-answer"><p><label class="radio" data-answer="79f489dbcd36384ebc07f7de5dc6248f"><input type="radio"><span> 11.4 dm<sup>3</sup></span></label></p><p><label class="radio" data-answer="cdcf95bf88ef75ca089a50f63735f337"><input type="radio"><span> 6.81 dm<sup>3</sup></span></label></p><p><label class="radio" data-answer="e318094a40a899777c1471dc1cd9fc4f"><input type="radio"><span> 2.27 dm<sup>3</sup></span></label></p><p><label class="radio" data-answer="542e236e676bc3c7ba096f8137ffb621"><input type="radio"><span> 0 dm<sup>3</sup></span></label></p></div><div class="q-explanation"><p>CH<sub>4</sub>(g) + 2O<sub>2</sub>(g) → CO<sub>2</sub>(g) + 2H<sub>2</sub>O(l) so 1.60 g (0.1 mol) of methane requires 0.2 mol of oxygen (6.40 g) to combust completely. This means there will be 6.40 g (0.2 mol) of excess oxygen remaining and 0.1 mol of carbon dioxide product present (water is a liquid at STP) which will occupy 3 x 2.27 = 6.81 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>Which is correct for <img align="middle" alt="begin mathsize 10px style table row 56 row 26 end table end style" class="Wirisformula" data-mathml="«math style=¨font-family:Arial¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨10px¨»«mtable»«mtr»«mtd»«mn»56«/mn»«/mtd»«/mtr»«mtr»«mtd»«mn»26«/mn»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»" src="../../../ckeditor/plugins/wiris/integration/showimage-61.php?formula=94b1d9cf2245ca08b148d8009ef313e8.png">Fe<sup>3+</sup>? </p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/screenshot-2019-10-31-at-10.38.25.png" style="width: 400px; height: 155px;"></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 atomic number of Fe is 26 so it will contain 26 protons. Its mass number is 56 so there will be 30 neutrons in the isotope. Fe contains 26 electrons. Three electrons have been removed to form Fe<sup>3+</sup> so there will be 23 electrons remaining.</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 transition in the hydrogen emission spectrum relates directly to its ionization energy?</p></div><div class="q-answer"><p><label class="radio" data-answer="6e9593394b2a9fd6a64fba824710e5a1"><input type="radio"><span> <em>n</em> = 2 to <em>n</em> = 1</span></label></p><p><label class="radio" data-answer="73b1c675353306aa7ebccc4903430fe7"><input type="radio"><span> <em>n</em> = 1 to <em>n</em> = ∞</span></label></p><p><label class="radio" data-answer="d68af8bbb2323446a0f9e139658fbdb7"><input type="radio"><span> <em>n</em> = ∞ to <em>n</em> = 1</span></label></p><p><label class="radio" data-answer="6a7aa15bea80d64279ca656f500e2c4b"><input type="radio"><span> <em>n</em> = 1 to <em>n</em> = 2</span></label></p></div><div class="q-explanation"><p>Emission spectra involve excited electrons falling from higher to lower energy levels. An electron transition from ∞ to the <em>n</em> = 1 level will emit the same amount of energy as needs to be absorbed when the atom is ionized and the electron moves from its ground state of <em>n</em> =1 to <em>n</em>=∞.</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 element will have the highest third ionization energy?</p></div><div class="q-answer"><p><label class="radio" data-answer="f186217753c37b9b9f958d906208506e"><input type="radio"><span> O</span></label></p><p><label class="radio" data-answer="3c9547d2fcb523a7ae5681eedde43fb6"><input type="radio"><span> Be</span></label></p><p><label class="radio" data-answer="8640d08847fe5e081f0a41c4579bf26a"><input type="radio"><span> Li</span></label></p><p><label class="radio" data-answer="4dec99baa99738721da9c9b0c1a92498"><input type="radio"><span> Ne</span></label></p></div><div class="q-explanation"><p>Once the second electron has been removed the species together with their electron configuration will be Ne<sup>2+</sup> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>4</sup>, O<sup>2+ </sup>1s<sup>2</sup>2s<sup>2</sup>2p<sup>2</sup>, Li<sup>2+ </sup>1s<sup>1 </sup>and Be<sup>2+</sup> 1s<sup>2</sup>. In the same way that He has a higher 1st ionization energy than H, Be<sup>2+</sup> will have a higher ionization energy than Li<sup>+ </sup>as the electron being removed from Be<sup>2+</sup> is spin-paired and there is an extra proton in the nucleus of Be<sup>2+</sup> compared to Li<sup>+</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 is a correct description of the oxides of period 3? </p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/table-of-oxides-of-period-3.png" style="width: 400px; height: 149px;"></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 oxides of period 3 (Na → Cl) change from basic to acidic moving from left to right across the period with aluminium oxide having both basic and acidic properties.</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 hydrated ion forms a colourless solution in water?</p></div><div class="q-answer"><p><label class="radio" data-answer="0da39aa825c8bece48db5603e33f5572"><input type="radio"><span> [Mn(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup></span></label></p><p><label class="radio" data-answer="2d9cce474d221e91197feebb65cbb36e"><input type="radio"><span> [Cu(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup></span></label></p><p><label class="radio" data-answer="95e33c667e4dfb8012ad335a7eb1bfda"><input type="radio"><span> [Ni(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup></span></label></p><p><label class="radio" data-answer="a58cbd5731ee61b4286969d7d36c5747"><input type="radio"><span> [Zn(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup></span></label></p></div><div class="q-explanation"><p>[Zn(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup> is the only one of the four ions that does not contain an incomplete d sub-level (the 3d sub-level is full in [Zn(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup>) so no transitions can occur within the split d 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><div class="exercise"><div class="q-question"><p>Which compound contains only ionic bonding? </p></div><div class="q-answer"><p><label class="radio" data-answer="57a78cb883afa7bca304ad6dc90ace10"><input type="radio"><span> lithium sulfate</span></label></p><p><label class="radio" data-answer="17ab7640ce5dbcb2ff8d08c65cc205aa"><input type="radio"><span> sodium nitrate</span></label></p><p><label class="radio" data-answer="c434e31ed05115d85c3f7315fb5f7833"><input type="radio"><span> potassium nitride</span></label></p><p><label class="radio" data-answer="be00898f75130ade5c6e7a7e72d506a7"><input type="radio"><span> ammonium chloride</span></label></p></div><div class="q-explanation"><p>All four compounds are ionic but the nitrate, sulfate and ammonium ions, (NO<sub>3</sub><sup>−</sup>, SO<sub>4</sub><sup>2−</sup> and NH<sub>4</sub><sup>+</sup>) are complex ions which also contain covalent bonding. Potassium nitride, K<sub>3</sub>N only contains K<sup>+</sup> and N<sup>3−</sup> ions.</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 bond angle of 120<sup>o</sup>?</p></div><div class="q-answer"><p><label class="radio" data-answer="2e152897644f00558971926132234c63"><input type="radio"><span> phosphorus trifluoride, PF<sub>3</sub></span></label></p><p><label class="radio" data-answer="c69ef09f3bc3c358844a22ddfee1cbc0"><input type="radio"><span> sulfur hexafluoride, SF<sub>6</sub></span></label></p><p><label class="radio" data-answer="f05ea0d44bf4d796902ce12eae531bb0"><input type="radio"><span> tetrafluoromethane, CF<sub>4</sub></span></label></p><p><label class="radio" data-answer="d7c5390eea2024dfc2187a28287bc43d"><input type="radio"><span> phosphorus pentafluoride, PF<sub>5</sub></span></label></p></div><div class="q-explanation"><p>CF<sub>4</sub> and BF<sub>3</sub> have four electron domains around the central atom. CF<sub>4</sub> has a regular tetrahedral shape with bond angles of 109.5<sup>o</sup>, PF<sub>3</sub> has one non-bonding pair of electrons so has a trigonal pyramid shape with bond angles less than 109.5<sup>o</sup> as the non-bonding pair of electrons exerts a greater repulsion than the bonding pairs. SF<sub>6</sub> has six electron domains giving a regular octahedral shape with bond angles of 90<sup>o</sup> and 180<sup>o</sup>. PF<sub>5</sub> has five bonding electron domains so its shape is trigonal bipyramid with bond angles of 90<sup>o</sup>, 120<sup>o</sup> and 180<sup>o</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 factor significantly increases the strength of metallic bonding?</p></div><div class="q-answer"><p><label class="radio" data-answer="16114e340b0ee2e06b655fef7907e5b9"><input type="radio"><span> decrease in the charge on the metal ion </span></label></p><p><label class="radio" data-answer="056d9c94c9c2f8de1e5c5558733bfdf6"><input type="radio"><span> decrease in the radius of the metal ion</span></label></p><p><label class="radio" data-answer="3884dd83c7030daf673f603aa05ef502"><input type="radio"><span> increase in the number of neutrons in the metal ion</span></label></p><p><label class="radio" data-answer="268e8e1b3e4419c01650bb231624e10f"><input type="radio"><span> increase in the number of electrons in the metal ion</span></label></p></div><div class="q-explanation"><p>The three factors which increase the strength of metallic bonding are a decrease in the radius of the metal ion, an increase in the charge on the metal ion (these two factors increase the charge density) and the number of delocalized electrons. An increase in the number of electrons is not the same as an increase in the number of delocalized electrons. Potassium, for example, has 39 electrons whereas lithium only has 3 electrons but both have only one electron in the outer s orbital that can delocalize and the much smaller lithium ion means lithium has a much higher boiling point than potassium. </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 combination of atomic orbitals will result in the formation of a pi (π) bond?</p></div><div class="q-answer"><p><label class="radio" data-answer="3239876c71208864562d9835d98ad867"><input type="radio"><span> p<sub>y</sub> + p<sub>y</sub></span></label></p><p><label class="radio" data-answer="066f2bfec5db39dcb7278f34e904c93f"><input type="radio"><span> p<sub>x</sub> + p<sub>x</sub></span></label></p><p><label class="radio" data-answer="7e97efaa47a9fbf52ff5d5da10cbe8f1"><input type="radio"><span> s + s</span></label></p><p><label class="radio" data-answer="f8f7b3c750269a63d9e4ed93b6e65481"><input type="radio"><span> s + p</span></label></p></div><div class="q-explanation"><p>A π bond is formed when two atomic orbitals combine 'sideways' on. When one of the atomic orbitals is an s orbital a sigma (σ) bond will always be formed and when two p<sub>x</sub> atomic orbitals combine they meet 'head' on so also form a σ 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>How many of the carbon atoms in acetophenone (1-phenylethan-1-one) are sp<sup>3</sup> hybridized?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/named-acetohenone.png" style="width: 250px; height: 153px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="c81e728d9d4c2f636f067f89cc14862c"><input type="radio"><span> 2</span></label></p><p><label class="radio" data-answer="1679091c5a880faf6fb5e6087eb1b2dc"><input type="radio"><span> 6</span></label></p><p><label class="radio" data-answer="cfcd208495d565ef66e7dff9f98764da"><input type="radio"><span> 0</span></label></p><p><label class="radio" data-answer="c4ca4238a0b923820dcc509a6f75849b"><input type="radio"><span> 1</span></label></p></div><div class="q-explanation"><p>The six carbon atoms in the phenyl group are sp hybridized and the carbonyl carbon atom is sp<sup>2 </sup>hybridized. Only the carbon atom in the methyl group 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>Consider the following enthalpy values:</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/combustion-values.png" style="width: 400px; height: 42px;"></p><p>Which expression gives the enthalpy value, in kJ mol<sup>−1</sup>, for the formation of propane, C<sub>3</sub>H<sub>8</sub>(g)? </p></div><div class="q-answer"><p><label class="radio" data-answer="ea0f56a3be89ac18a16b15237ea38b92"><input type="radio"><span> 3(– 394) + 4(− 286) – 2219</span></label></p><p><label class="radio" data-answer="23685e922536b70ad8f5023ef4e8734b"><input type="radio"><span> − 394 − 286 + 2219</span></label></p><p><label class="radio" data-answer="877c33a0bdfb80120b8e8b93e19bd15f"><input type="radio"><span> − 394 − 286 − 2219</span></label></p><p><label class="radio" data-answer="5696b1d5b08b7d40f52526ccc210d6c3"><input type="radio"><span> 3(−394) + 4(− 286) + 2219</span></label></p></div><div class="q-explanation"><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/hess-s-law-for-propane-formation.png" style="width: 250px; height: 154px;"></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 an exothermic process?</p></div><div class="q-answer"><p><label class="radio" data-answer="f14cd47225a5e05fbe8faaf328cf2dd6"><input type="radio"><span> O(g) + e<sup>−</sup> → O<sup>−</sup>(g) </span></label></p><p><label class="radio" data-answer="e7c43cccd8d79b370cabf01d0cd07397"><input type="radio"><span> Ca(g) → Ca<sup>+</sup>(g) + e<sup>−</sup></span></label></p><p><label class="radio" data-answer="9288f6340207dba68c2d314a5f8f5801"><input type="radio"><span> O<sup>−</sup>(g) + e<sup>−</sup> → O<sup>2−</sup>(g) </span></label></p><p><label class="radio" data-answer="9df76ba5a2fc62784b827ab5638983c4"><input type="radio"><span> Ca<sup>+</sup>(g) → Ca<sup>2+</sup>(g) + e<sup>−</sup></span></label></p></div><div class="q-explanation"><p>Removing an electron from a gaseous atom or ion is always endothermic. Adding an electron to a gaseous atom is always exothermic but adding an electron to a gaseous negative ion is always endothermic.</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 enthalpy value at SATP conditions can be determined using only average bond enthalpies? </p></div><div class="q-answer"><p><label class="radio" data-answer="1e3989853c3e568f480e6b0501761084"><input type="radio"><span> The reaction between hydrogen and iodine to produce hydrogen iodide.</span></label></p><p><label class="radio" data-answer="3a509d38803548d9b91179bb47e9aa6d"><input type="radio"><span> The reaction between oct-1-ene and hydrogen to produce octane.</span></label></p><p><label class="radio" data-answer="46255e7ef34f2b09ced50e9f8aa5697f"><input type="radio"><span> The reaction between methane and chlorine to produce chloromethane and hydrogen chloride.</span></label></p><p><label class="radio" data-answer="ad61f461613cc5aae9097c832afc14e7"><input type="radio"><span> The reaction between ethene and water to produce ethanol.</span></label></p></div><div class="q-explanation"><p>Average bond enthalpies refer only to the gaseous state and additional enthalpy terms involving change of state are required for all the reactions except for the reaction between methane and chlorine where all the reactants and products are in the gaseous state.</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 enthalpy change for a particular reaction is +101 kJ mol<sup>−1</sup> and the entropy change is +136 J K<sup>−1 </sup>mol−1.<br>Which is a true statement about this reaction.</p></div><div class="q-answer"><p><label class="radio" data-answer="cb63e4ec7b8a2835788003e7f94286d1"><input type="radio"><span> It will be spontaneous at 298 K but as the temperature is increased it will eventually become non-spontaneous.</span></label></p><p><label class="radio" data-answer="557dd0dadf430081f050f42a50a4050c"><input type="radio"><span> It is non-spontaneous at 298 K but as the temperature is increased it will eventually become spontaneous.</span></label></p><p><label class="radio" data-answer="691dcd47bb5821f1a5be8e888d6d3fac"><input type="radio"><span> It will be spontaneous at all temperatures.</span></label></p><p><label class="radio" data-answer="7d80c65d279d2f20643591155c388936"><input type="radio"><span> It will be non-spontaneous at all temperatures.</span></label></p></div><div class="q-explanation"><p>Δ<em>G</em> = Δ<em>H</em> − TΔ<em>S = </em>+101 − T(136/1000) kJ mol<sup>−1</sup>. At 298 K Δ<em>G </em>has a positive value so the reaction is non-spontaneous but as T increases eventually T(136/1000) will become greater than 101 and Δ<em>G </em>will have a negative value at which point the reaction becomes spontaneous.</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 graph shows the reaction pathway for an endothermic reaction with and without a catalyst.</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/activation-energy-profile.png" style="width: 300px; height: 204px;"></p><p>Which gives the correct quantitative values? </p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/values-for-ea-etc.png" style="width: 400px; height: 138px;"></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>E</em><sub>a</sub> for the forward catalysed reaction is the difference between the starting energy and the peak of the dotted red line (catalysed pathway), Δ<em>H</em> is the difference in energy between the initial and final states and <em>E</em><sub>a</sub> for the reverse catalysed reaction is the difference between the new starting energy and the peak of the catalysed pathway.</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 reaction between magnesium and hydrochloric acid was studied to see how changing the surface area of the magnesium metal, changing the temperature and changing the concentration of the acid affects the initial rate of the reaction.</p><p>Mg(s) + 2HCl(aq) → MgCl<sub>2</sub>(aq) + H<sub>2</sub>(g)</p><p>Which combination of factors gives the slowest initial rate?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/mg-hcl-data.png" style="width: 600px; height: 139px;"></p><p> </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 larger the particle size the smaller the surface area of the magnesium which will lower the number of collisions. The lower the temperature the lower the number of successful collisions as less particles will collide with the necessary activation energy. The lower the concentration of the acid the lower the number of collisions between the acid and the magnesium.</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 a correct unit for the rate constant for a second order reaction? </p></div><div class="q-answer"><p><label class="radio" data-answer="c0cd754bc13e0e56af5a088abd1b70cb"><input type="radio"><span> mol<sup>−2</sup> dm<sup>6</sup> s<sup>−1</sup></span></label></p><p><label class="radio" data-answer="0cea9e82263d0c7ad41dc382776651d7"><input type="radio"><span> mol<sup>−1</sup> dm<sup>3</sup> s<sup>−1</sup></span></label></p><p><label class="radio" data-answer="dcb8eec26dd1db8bd982e3685c6ded07"><input type="radio"><span> mol dm<sup>−3</sup> s<sup>−1</sup> </span></label></p><p><label class="radio" data-answer="d7ceaa60f30f0f87a6824ef4b9d3b849"><input type="radio"><span> s<sup>−1</sup></span></label></p></div><div class="q-explanation"><p>For a second order reaction rate = <em>k</em> x concentration<sup>2</sup>, so k = (mol dm<sup>−3</sup> s<sup>−1</sup>) ÷ (mol dm<sup>−3</sup>)<sup>2</sup> = mol<sup>−1</sup> dm<sup>3</sup> s<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>The logarithmic form of the Arrhenius equation is:</p><p> <img alt="" src="../../../ib/chemistry/images/quiz/mc-test/arrhenius-equation.png" style="width: 132px; height: 48px;"></p><p>Which is a correct statement?</p></div><div class="q-answer"><p><label class="radio" data-answer="dd79e8f5988a8bab8a31912a90f4c113"><input type="radio"><span> <em>A</em> is known as the frequency factor.</span></label></p><p><label class="radio" data-answer="46792329664230d4d37756d28fe65dd7"><input type="radio"><span> <em>k</em> is the value for the equilibrium constant for the reaction at temperature T.</span></label></p><p><label class="radio" data-answer="7c22b08d30afdf52400997de3c630381"><input type="radio"><span> When 1/T is plotted against ln<em>k</em> the gradient is equal to the activation energy. </span></label></p><p><label class="radio" data-answer="906071f764c32ee6e36468711a5ad38a"><input type="radio"><span> The value of <em>E</em><sub>a</sub> is dependent upon the temperature.</span></label></p><p> </p></div><div class="q-explanation"><p><em>A</em> is the frequency factor although it is also sometimes known as the Arrhenius constant or pre-exponential factor. When 1/T is plotted against ln<em>k</em> the gradient will be equal to −<em>E</em><sub>a</sub>/<em>R</em>. <em>k</em> is the value of the rate constant at temperature T (in Kelvin) and <em>t</em>he value of the activation energy, <em>E</em><sub>a</sub> is independent of the temperature.</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 value of the equilibrium constant for the reaction between hydrogen and iodine to form hydrogen iodide at 500 K is 1.60 x 10<sup>2</sup>.</p><p style="margin-left: 40px;">H<sub>2</sub>(g) + I<sub>2</sub>(g) ⇌ 2HI(g) <em>K</em><sub>c</sub> = 1.60 x 10<sup>2</sup></p><p>What will be the value of the equilibrium constant for the reverse reaction?</p><p style="margin-left: 40px;">2HI(g) ⇌ H<sub>2</sub>(g) + I<sub>2</sub>(g) </p></div><div class="q-answer"><p><label class="radio" data-answer="57fd0111f7513ba146a2b8f4e875a526"><input type="radio"><span> 0.5 x √(1.60 x 10<sup>2</sup>)</span></label></p><p><label class="radio" data-answer="0577602f08cb0177cf7022615eb57879"><input type="radio"><span> √(1.60 x 10<sup>2</sup>)</span></label></p><p><label class="radio" data-answer="2375fa5e40b2dc49db082d1ca98f87ba"><input type="radio"><span> 1 ÷ √(1.60 x 10<sup>2</sup>)</span></label></p><p><label class="radio" data-answer="2d3551cbab2394f5667db73ffcb301c6"><input type="radio"><span> 1 ÷ (1.60 x 10<sup>2</sup>)</span></label></p></div><div class="q-explanation"><p>The equilibrium constant for the reverse reaction will be (<em>K</em><sub>c</sub>)<sup>−1</sup>, i.e. 1 ÷ (1.60 x 10<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> 0.9 mol of hydrogen gas and 0.7 mol of iodine gas were mixed at 477 <sup>o</sup>C and allowed to reach equilibrium in a flask with a fixed volume. At equilibrium 1.2 mol of hydrogen iodide was present.</p><p>H<sub>2</sub>(g) + I<sub>2</sub>(g) ⇌ 2HI(g)</p><p>What is the value of the equilibrium constant for this reaction at 477 <sup>o</sup>C?</p></div><div class="q-answer"><p><label class="radio" data-answer="c20ad4d76fe97759aa27a0c99bff6710"><input type="radio"><span> 12</span></label></p><p><label class="radio" data-answer="642e92efb79421734881b53e1e1b18b6"><input type="radio"><span> 48</span></label></p><p><label class="radio" data-answer="1a18da63cbbfb49cb9616e6bfd35f662"><input type="radio"><span> 2.3</span></label></p><p><label class="radio" data-answer="dee43f9ada8a3a1b6a0cf3e583392aa7"><input type="radio"><span> 0.020</span></label></p></div><div class="q-explanation"><p><em>K</em><sub>c</sub> = [HI(g)]<sup>2</sup> ÷ ([H<sub>2</sub>(g)] x [I<sub>2</sub>(g)] = (1.2/V)<sup>2</sup> ÷ [(0.9 − 0.6)/V x (0.7 − 0.6)/V] = 1.44 ÷ 0.03 = 48</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 ammonia, NH<sub>3</sub>? </p></div><div class="q-answer"><p><label class="radio" data-answer="70c7652df1614d1450ec4f641e8176a9"><input type="radio"><span> NH<sub>4</sub><sup>+</sup></span></label></p><p><label class="radio" data-answer="c127db953cffab3ede6ccb14fcdad4fb"><input type="radio"><span> NH<sub>4</sub>OH</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="686a32d507f38654b3d727c8a4151a9f"><input type="radio"><span> NH<sub>2</sub><sup>−</sup></span></label></p></div><div class="q-explanation"><p>Brønsted-Lowry acids donate a proton to form their conjugate base, so NH<sub>2</sub><sup>−</sup> is the conjugate base of NH<sub>3 </sub>whereas ammonia itself is the conjugate base of NH<sub>4</sub><sup>+</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 pH of 0.010 mol dm<sup>−3 </sup>H<sub>2</sub>SO<sub>4</sub>(aq)? </p></div><div class="q-answer"><p><label class="radio" data-answer="4e7ef8c3e4a97ef41a9be47c43a692d6"><input type="radio"><span> between 1 and 2</span></label></p><p><label class="radio" data-answer="c4ca4238a0b923820dcc509a6f75849b"><input type="radio"><span> 1</span></label></p><p><label class="radio" data-answer="c81e728d9d4c2f636f067f89cc14862c"><input type="radio"><span> 2</span></label></p><p><label class="radio" data-answer="0dc61e9dbcf9f51129d4b730d12c718a"><input type="radio"><span> between 2 and 3</span></label></p></div><div class="q-explanation"><p>Sulfuric acid is a strong diprotic acid so [H<sup>+</sup>(aq)] will be equal to 2 x 10<sup>−2</sup> mol dm<sup>−3</sup>. pH = − log<sub>10</sub>[H<sup>+</sup>(aq)] so the pH will be less than 2 but more than 1 (as [H<sup>+</sup>(aq) would need to be 10 x 10<sup>−2</sup> mol dm<sup>−3</sup> for it to be 1). This has assumed that H<sub>2</sub>SO<sub>4</sub> is a strong dibasic acid. In fact the first dissociation to HSO<sub>4</sub><sup>−</sup> is strong but the second dissociation is not completely strong. Even so, the actual concentration of H<sup>+</sup>(aq) will still be more than 1 x 10<sup>−2</sup> (but less than 2 x 10<sup>−2</sup> mol dm<sup>−3</sup>) so the pH will between 1 and 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"><p>Which 1.0 mol dm<sup>−3</sup> solution will have the highest pH?</p></div><div class="q-answer"><p><label class="radio" data-answer="29f586105216f1bf47287095659da95d"><input type="radio"><span> KCl(aq)</span></label></p><p><label class="radio" data-answer="773a4b6b5cb51253ab15775de5e4c073"><input type="radio"><span> NH<sub>4</sub>Cl(aq)</span></label></p><p><label class="radio" data-answer="21542ec7a745ae3dcef6b7d9c4811371"><input type="radio"><span> CH<sub>3</sub>COOH(aq)</span></label></p><p><label class="radio" data-answer="7d9b332c26730196f82a2610a4ef1089"><input type="radio"><span> Na<sub>2</sub>CO<sub>3</sub>(aq)</span></label></p><p> </p></div><div class="q-explanation"><p>NH<sub>4</sub>Cl is the salt of a weak base and a strong acid so like CH<sub>3</sub>COOH (which is a weak acid) will have a pH less than 7. KCl is the salt of a strong acid with a strong base so will have a pH = 7. Na<sub>2</sub>CO<sub>3</sub> is the salt of a strong base with a weak acid so will have a pH greater than 7 as the CO<sub>3</sub><sup>2−</sup> ions combine with H<sup>+ </sup>ions from the water to form H<sub>2</sub>CO<sub>3</sub> leaving an excess of OH<sup>−</sup> ions.</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 species can act as a Lewis acid?</p></div><div class="q-answer"><p><label class="radio" data-answer="ab70a0e602697d4f18677797b16dac8b"><input type="radio"><span> Cl<sup>−</sup></span></label></p><p><label class="radio" data-answer="e8d977fc876aac63e6656d10c320a842"><input type="radio"><span> Cl<sub>2</sub></span></label></p><p><label class="radio" data-answer="7df701d2736645defca2a5b3d6330f6e"><input type="radio"><span> AlCl<sub>3</sub></span></label></p><p><label class="radio" data-answer="04db74781fff701c4bb17f664490f3ec"><input type="radio"><span> CCl<sub>4</sub></span></label></p></div><div class="q-explanation"><p>A Lewis acid can accept a pair of electron. AlCl<sub>3</sub> is electron deficient as there are only six electrons around the Al atom so it can accept a pair of electrons to form, for example AlCl<sub>4</sub><sup>−</sup>. In all the other species there is an octet (or a share in an octet) of electrons around each 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>Which compound contains sulfur with the highest oxidation state? </p></div><div class="q-answer"><p><label class="radio" data-answer="260405714079916446757f62e0feeca8"><input type="radio"><span> SO<sub>2</sub></span></label></p><p><label class="radio" data-answer="276717ad15b9c3ea595581bcb255a528"><input type="radio"><span> H<sub>2</sub>SO<sub>4</sub></span></label></p><p><label class="radio" data-answer="8f5dce78c729ac12555f31c051a4e2e2"><input type="radio"><span> H<sub>2</sub>S</span></label></p><p><label class="radio" data-answer="5e5c30d0c17a78e92ef2acea887d79cb"><input type="radio"><span> Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub></span></label></p></div><div class="q-explanation"><p>The oxidation state of sulfur in sulfuric acid, H<sub>2</sub>SO<sub>4</sub> is +6. In sodium thiosulfate, Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> and in sulfur dioxide, SO<sub>2</sub> it is +4 and in hydrogen sulfide, H<sub>2</sub>S it is −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"><p>Which is a correct statement about electrochemical cells? </p></div><div class="q-answer"><p><label class="radio" data-answer="8a4c838e982e3224662952b8ccad540a"><input type="radio"><span> Voltaic cells convert energy from non-spontaneous, endothermic chemical processes to electrical energy.</span></label></p><p><label class="radio" data-answer="2a477cd571b3171dd5b9dd17f6a66724"><input type="radio"><span> Oxidation occurs at the cathode (positive electrode) and reduction occurs at the anode (negative electrode) in a voltaic cell.</span></label></p><p><label class="radio" data-answer="a77b4c6afdca96b99857f5e5507a5417"><input type="radio"><span> Oxidation occurs at the cathode (negative electrode) and reduction occurs at the anode (positive electrode) in an electrolytic cell.</span></label></p><p><label class="radio" data-answer="74855866599863fa9e90eb5508e5bdfe"><input type="radio"><span> Electrolytic cells convert electrical energy to chemical energy, by bringing about non-spontaneous processes.</span></label></p></div><div class="q-explanation"><p>Oxidation always occurs at the anode. In a voltaic cell the anode is the negative electrode and in a electrolytic cell the anode is the positive electrode. For a voltaic cell to produce electrical energy the spontaneous chemical process must be exothermic, not endothermic.</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>Consider the two half-reactions:</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/half-reactions-(1).png" style="width: 300px; height: 65px;"></p><p>Which species is the strongest reducing agent?</p></div><div class="q-answer"><p><label class="radio" data-answer="6cb321460d08fef1babca1ef07a2ddec"><input type="radio"><span> Sn</span></label></p><p><label class="radio" data-answer="c9e0e1712df7181d34e5cbe4ec1a6403"><input type="radio"><span> Sn<sup>2+</sup></span></label></p><p><label class="radio" data-answer="517c5a04e9196e96f98e4b28557e2c75"><input type="radio"><span> Ni<sup>2+</sup></span></label></p><p><label class="radio" data-answer="a26e174e330476756d2601ea5368aec3"><input type="radio"><span> Ni</span></label></p></div><div class="q-explanation"><p>The spontaneous reaction between the two half-reactions is Ni(s) + Sn<sup>2+</sup>(aq) → Sn(s) + Ni<sup>2+</sup>(aq) <em>E</em><sup>⦵</sup><sub>total</sub> = 0.12 V, so Ni is the strongest reducing agent.</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 correctly describes what occurs at each electrode when an aqueous solution of copper(II) sulfate is electrolysed using different electrodes?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/electrolysis-of-cuso4.png" style="width: 565px; height: 200px;"></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>With inert electrodes (such as platinum) oxygen is evolved at the anode and copper is deposited at the cathode. During this process the solution loses its blue colour as the concentration of copper(II) ions decreases. With copper electrode the copper anode dissolves forming Cu<sup>2+</sup>(aq) ions and the blue colour remains constant as copper is simultaneously deposited at the cathode.</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 list contains only functional groups found in diamorphine (heroin)?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/diamorphine-(heroin)-structure.png" style="width: 264px; height: 210px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="d3e8d15adb60d0a888ebfa260cd52c36"><input type="radio"><span> alkyl, amide, ester, alkenyl</span></label></p><p><label class="radio" data-answer="95187a91b46049f5f708c53ded608b20"><input type="radio"><span> alkenyl, phenyl, amine, ether</span></label></p><p><label class="radio" data-answer="9a4662b8bb35062ed074f43d4b68b816"><input type="radio"><span> amine, ester, ether, carboxyl</span></label></p><p><label class="radio" data-answer="97c9bc740d05708e5626bad3d6a39488"><input type="radio"><span> benzyl, carboxyl, amide, alkyl</span></label></p></div><div class="q-explanation"><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/functional-groups-of-diamorphine(3).png" style="width: 400px; height: 203px;"></p><p>Carboxyl (−COOH), benzyl (C<sub>6</sub>H<sub>5</sub>CH<sub>2</sub>−) and amide (−CO−NH<sub>2</sub>) functional groups are 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>What is the name of this compound using IUPAC rules? </p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/2%2C3-dimethylbutan-2-ol.png" style="width: 200px; height: 122px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="8c2d854229865527e6b5a0c60e6aa7a3"><input type="radio"><span> hexan-2-ol</span></label></p><p><label class="radio" data-answer="d8827c09f3bdcf0153f416437d666938"><input type="radio"><span> 2,3-dimethylbutan-2-ol</span></label></p><p><label class="radio" data-answer="72129403c3b338b55f1f69edfde937b3"><input type="radio"><span> 2-methylpentan-2-ol</span></label></p><p><label class="radio" data-answer="2b41d8bbf0492ec84d0d5101ef0d5a15"><input type="radio"><span> 2,2-dimethylbutan-3-ol</span></label></p></div><div class="q-explanation"><p>The compound is an isomer of hexanol, C<sub>6</sub>H<sub>13</sub>OH, but the longest chain contains four carbon atoms and there is a hydroxyl group on the second carbon atom. It is also bonded to two methyl groups, one on the second carbon atom and one on the third, so it is 2,3-dimethylbutan-2-ol.</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 general equation for the nucleophilic substitution of halogenoalkanes with hydroxide ions is: </p><p style="margin-left: 40px;">R−X(l) + NaOH(aq) → R−OH(aq) + NaX(aq) (where X is a halide ion)</p><p>Which halogenoalkane will be substituted the fastest by hydroxide ions under the same conditions of temperature and aprotic polar solvent?</p></div><div class="q-answer"><p><label class="radio" data-answer="94fab94cca9c0a26a1c0b13565cab0ac"><input type="radio"><span> C<sub>2</sub>H<sub>5</sub>Cl</span></label></p><p><label class="radio" data-answer="a98dceb7bff3be89bff197e1473a9b4d"><input type="radio"><span> C(CH<sub>3</sub>)<sub>3</sub>I</span></label></p><p><label class="radio" data-answer="9cd20a0f80102e69f3fe5910e45f0c61"><input type="radio"><span> C<sub>2</sub>H<sub>5</sub>I</span></label></p><p><label class="radio" data-answer="a57b1f7bfb37b66f9e452cf098022941"><input type="radio"><span> C(CH<sub>3</sub>)<sub>3</sub>Cl</span></label></p></div><div class="q-explanation"><p>Tertiary halogenoalkanes proceed by an S<sub>N</sub>1 mechanism that is faster than the S<sub>N</sub>2 mechanism followed by primary halogenoalkanes and iodine is a better leaving group than chlorine.</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 compound is produced when methanol reacts with methanoic acid in the presence of a sulfuric acid catalyst? </p></div><div class="q-answer"><p><label class="radio" data-answer="212f5b9364df5fd2f8326ec51da0676f"><input type="radio"><span> HCOOCH<sub>3</sub></span></label></p><p><label class="radio" data-answer="e69e27380a0146e24533c9ea5bbf1e86"><input type="radio"><span> CH<sub>3</sub>COOH</span></label></p><p><label class="radio" data-answer="8d6459b753533e51ff21c1219a969d6f"><input type="radio"><span> HCOOC<sub>2</sub>H<sub>5</sub></span></label></p><p><label class="radio" data-answer="7eeb5f65419261138323ed0cb14bb2f4"><input type="radio"><span> CH<sub>3</sub>COOC<sub>2</sub>H<sub>5</sub></span></label></p></div><div class="q-explanation"><p>The equation for this esterification reaction is HCOOH + CH<sub>3</sub>OH ⇌ HCOOCH<sub>3</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 species acts as the electrophile when benzene reacts with a mixture of concentrated nitric and sulfuric acids to form nitrobenzene?</p></div><div class="q-answer"><p><label class="radio" data-answer="411da425ac1426f3eb7bb13b893ee3b2"><input type="radio"><span> NO<sup>+</sup></span></label></p><p><label class="radio" data-answer="44abbd327c3bc3645d73af8f84966a65"><input type="radio"><span> HNO<sub>3</sub></span></label></p><p><label class="radio" data-answer="78d8b5a878a8797f69ab634e779a191b"><input type="radio"><span> NO<sub>2</sub><sup>+</sup></span></label></p><p><label class="radio" data-answer="4dd3dde6015c9e122597cf6d97fa7db8"><input type="radio"><span> NO<sub>2</sub><sup>−</sup></span></label></p></div><div class="q-explanation"><p>Sulfuric acid protonates the nitric acid to form H<sub>2</sub>NO<sub>3</sub><sup>+</sup> which then dissociates to form water and the nitronium ion, NO<sub>2</sub><sup>+</sup>. It is the NO<sub>2</sub><sup>+</sup> ion that then acts as the electrophile by being attracted to the delocalised π electrons on the benzene ring.</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>How many chiral carbon atoms are present in a molecule of the mild analgesic ibuprofen?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/ibuprofen.png" style="width: 120px; height: 195px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="c81e728d9d4c2f636f067f89cc14862c"><input type="radio"><span> 2</span></label></p><p><label class="radio" data-answer="cfcd208495d565ef66e7dff9f98764da"><input type="radio"><span> 0</span></label></p><p><label class="radio" data-answer="c4ca4238a0b923820dcc509a6f75849b"><input type="radio"><span> 1</span></label></p><p><label class="radio" data-answer="eccbc87e4b5ce2fe28308fd9f2a7baf3"><input type="radio"><span> 3</span></label></p></div><div class="q-explanation"><p>A chiral carbon atom is asymmetric, i.e. contains four different groups bonded to it.<br>The chiral carbon atom is shown below with a red asterisk, <strong><span style="color:#FF0000;">*</span>.</strong></p><p><strong><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/starred-ibuprofen.png" style="width: 120px; height: 174px;"></strong></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>An experiment was performed to measure the volume, <em>V</em> (in m<sup>3</sup>) of exactly 1.0 g of oxygen gas at different temperatures, <em>T</em> (in Kelvin) at a constant pressure of 1.0 x 10<sup>5</sup> Pa.</p><p>Which graph should be plotted from the data obtained to give a straight line with a gradient equal to the value of the gas constant, <em>R</em> (in J K<sup>−1</sup> mol<sup>−1</sup>)? Assume that oxygen behaves as an ideal gas.</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/gradient-of-graph.png" style="width: 400px; height: 125px;"></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 ideal gas equation is <em>pV </em>= <em>nRT</em>. Since <em>M</em><sub>r</sub> of O<sub>2</sub> is 32, <em>n</em> = 1/32 mol.<em> P</em> = 1.0 x 10<sup>5</sup> Pa = 1.0 x 10<sup>5</sup> Nm<sup>−2</sup> and I J = 1 Nm</p><p>32 <span style="color:#FF0000;">(mol<sup>−1</sup>)</span> x 1.0 x 10<sup>5</sup> <span style="color:#FF0000;">(Nm<sup>−2</sup>)</span> x V <span style="color:#FF0000;">(m<sup>3</sup>) </span>= <em>R</em> <span style="color:#B22222;">(J K<sup>−1</sup> mol<sup>−1</sup>)</span> x T <span style="color:#FF0000;">(K)</span></p><p>a plot of (3.2 x 10<sup>6</sup>)<em>V</em> against <em>T </em>will give a gradient of <em>R</em> in J K−1 mol−1)</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 index of hydrogen deficiency (IHD) of 2,3,7,8-tetrachlorodibenzodioxin (molecular formula: C<sub>12</sub>H<sub>4</sub>Cl<sub>4</sub>O<sub>2</sub>)?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/2%2C3%2C7%2C8-tetrachlorodibenzodioxin.png" style="width: 230px; height: 107px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="c81e728d9d4c2f636f067f89cc14862c"><input type="radio"><span> 2</span></label></p><p><label class="radio" data-answer="c9f0f895fb98ab9159f51fd0297e236d"><input type="radio"><span> 8</span></label></p><p><label class="radio" data-answer="a87ff679a2f3e71d9181a67b7542122c"><input type="radio"><span> 4</span></label></p><p><label class="radio" data-answer="45c48cce2e2d7fbdea1afc51c7c6ad26"><input type="radio"><span> 9</span></label></p></div><div class="q-explanation"><p>When calculating IHD O atoms count as zero and Cl atoms can be considered to behave as H atoms so the formula can be reduced to C<sub>12</sub>H<sub>8</sub>. If it was fully saturated it would be C<sub>12</sub>H<sub>26</sub> so it is 18 H atoms (or 9 H<sub>2</sub>) short which means that the IHD Is 9. This can also be seen from the structural formula as the two benzene rings each have an IHD of 4 (three for the 'double bonds' and one for the ring) and the additional ring makes a total of 9.</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 will give three signals, showing a splitting pattern of one singlet, one triplet and one quartet in its <sup>1</sup>H NMR spectrum?</p></div><div class="q-answer"><p><label class="radio" data-answer="e5850a54743a5fd68c76d39797fab92a"><input type="radio"><span> HCOCH<sub>2</sub>CH<sub>2</sub>CH<sub>3</sub></span></label></p><p><label class="radio" data-answer="bdf2c0c29526dbc2a1860f730bea157c"><input type="radio"><span> CH<sub>3</sub>CH(OH)CH<sub>2</sub>CH<sub>3</sub></span></label></p><p><label class="radio" data-answer="3073d8ef690d807e890bdf59130326c8"><input type="radio"><span> C(CH<sub>3</sub>)<sub>3</sub>COCH<sub>2</sub>CH<sub>3</sub></span></label></p><p><label class="radio" data-answer="08171b95af45bf3834a62bedf3e9eda8"><input type="radio"><span> CH<sub>3</sub>CH<sub>2</sub>CH<sub>3</sub></span></label></p></div><div class="q-explanation"><p>The triplet and quartet will be shown by an ethyl group −CH<sub>2</sub>CH<sub>3</sub> with no other hydrogen atoms on the neighbouring carbon atom it is bonded to as the −CH<sub>3</sub> protons are split into a triplet by the −CH<sub>2</sub> protons and the −CH<sub>2</sub> protons are split into a quartet by the −CH<sub>3</sub> protons. A singlet will be shown by all the other hydrogen atoms in the molecule being in the same chemical environment with no hydrogen atoms on neighbouring carbon atoms. C(CH<sub>3</sub>)<sub>3</sub>COCH<sub>2</sub>CH<sub>3</sub> will give this splitting pattern with an integration trace of 9 for the singlet, 2 for the quartet and 3 for the triplet.</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="31518" 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+%281%29+%22+-+via+%40InThinker+%23chemistry%0A&url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F31518%2Fhl-practice-paper-1-1-"><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%2F31518%2Fhl-practice-paper-1-1-&t=HL+Practice+Paper+1+(1)+"><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%2F31518%2Fhl-practice-paper-1-1-"><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%2F31518%2Fhl-practice-paper-1-1-"><i class="fa fa-google-plus-square"></i><span>Google +</span></a></li><li><a class="" href="mailto:?subject=HL Practice Paper 1 (1) &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%2f31518%2fhl-practice-paper-1-1-"=""><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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