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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 (2)</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>How many atoms of oxygen are present in 12.6 g of hydrated oxalic acid, (COOH)<sub>2</sub>.2H<sub>2</sub>O?<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="a2195f6787b24968c82efa2c0fe7b175"><input type="radio"><span> 3.6 x 10<sup>24</sup></span></label></p>
<p><label class="radio" data-answer="e95e1ca27d0e39aa03eb5a611ce4122f"><input type="radio"><span> 0.6</span></label></p>
<p><label class="radio" data-answer="587a38867c15b3a4a00c853112c8a829"><input type="radio"><span> 3.6 x 10<sup>23</sup></span></label></p>
<p><label class="radio" data-answer="1679091c5a880faf6fb5e6087eb1b2dc"><input type="radio"><span> 6</span></label></p>
</div>
<div class="q-explanation">
<p><em>M</em> of (COOH)<sub>2</sub>.2H<sub>2</sub>O = (2 x 12) + (6 x 16) + (6 x 1) = 126 g mol<sup>−1</sup>, so the amount of (COOH)<sub>2</sub>.2H<sub>2</sub>O = 0.1 mol<br>
Each (COOH)<sub>2</sub>.2H<sub>2</sub>O contains six oxygen atoms, so the number of O atoms = 6 x 0.1 x 6.02 x 10<sup>23</sup> = 3.6 x 10<sup>23</sup></p>
</div>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>What volume of carbon dioxide measured at STP will be evolved when 4.0 g of calcium carbonate is added to 20 cm<sup>3</sup> of 2.0 mol dm<sup>−3</sup> hydrochloric acid solution?<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="3de3a5a18143e66894ee0f5bff3b7d26"><input type="radio"><span> 227 cm<sup>3</sup></span></label></p>
<p><label class="radio" data-answer="6c7fb1adc7762982279d3090d8129600"><input type="radio"><span> 454 cm<sup>3</sup></span></label></p>
<p><label class="radio" data-answer="79ddfa8e18346903d5fa8d661ff6446b"><input type="radio"><span> 908 cm<sup>3</sup></span></label></p>
<p><label class="radio" data-answer="769f5794bb7e44c10b70dcdcb7103f41"><input type="radio"><span> 114 cm<sup>3</sup></span></label></p>
</div>
<div class="q-explanation">
<p>CaCO<sub>3</sub>(s) + 2HCl(aq) → CO<sub>2</sub>(g) + H<sub>2</sub>O(l) + CaCl<sub>2</sub>(aq)<br>
Amount of CaCO<sub>3</sub> = 4/(40 + 12 + 48) = 0.04 mol; Amount of HCl = 20/1000 x 2.0 = 0.04 mol so HCl is the limiting reagent as 2 mol of HCl are required to react with 1 mol of CaCO<sub>3</sub> to produce 1 mol of CO<sub>2</sub>.<br>
Amount of CO<sub>2</sub> produced = 0.02 mol, so volume of CO<sub>2 </sub>evolved = 0.02 x 22.7 = 0.454 dm<sup>3</sup> = 454 cm<sup>3</sup></p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>A gas occupies 100 cm<sup>3</sup> at a temperature of 36 <sup>o</sup>C and a pressure of 9.63 x 10<sup>4 </sup>Pa. What volume (in cm<sup>3</sup>) will the same amount of gas occupy if the pressure is increased to 9.89 x 10<sup>4</sup> Pa and the temperature is increased to 63 <sup>o</sup>C?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="5e2e45566c64005549a087516ee94f36"><input type="radio"><span> 100 x (336 ÷ 309) x (9.63 ÷ 9.89)</span></label></p>
<p><label class="radio" data-answer="24fb591cdda7d043fe94237802aa4709"><input type="radio"><span> 100 x (309 ÷ 336) x (9.63 ÷ 9.89)</span></label></p>
<p><label class="radio" data-answer="24bfac5af22dfab26be1226e1266267d"><input type="radio"><span> 100 x (63 ÷ 36) x (9.89 ÷ 9.63)</span></label></p>
<p><label class="radio" data-answer="b389e20d3835601f861fec0a7b7fd62b"><input type="radio"><span> 100 x (336 ÷ 309) x (9.89 ÷ 9.63) </span></label></p>
</div>
<div class="q-explanation">
<p>Increasing the temperature increases the volume whereas increasing the pressure decreases the volume. The temperature must be measured in Kelvin so 36 <sup>o</sup>C and 63 <sup>o</sup>C is 309 K and 336 K respectively.</p>
<p>P<sub>1</sub>V<sub>1</sub>/T<sub>1</sub> = P<sub>2</sub>V<sub>2</sub>/T<sub>2</sub> so V<sub>2</sub> = V<sub>1 </sub>x T<sub>2</sub>/T<sub>1</sub> x P<sub>1</sub>/P<sub>2</sub> = 100 x (336 ÷ 309) x (9.63 x 10<sup>4</sup> ÷ 9.89 x 10<sup>4</sup>) = 100 x (336 ÷ 309) x (9.63 ÷ 9.89)</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which gives the number of neutrons and electrons present in the <sup>118</sup>Sn<sup>2+</sup> ion?</p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/sn2--data.png" style="width: 262px; height: 140px;"></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>Tin has an atomic number of 50, so contains 50 protons. This isotope has a mass number of 118 which is equal to the number of protons and neutrons, so it contains 68 neutrons. A neutral tin atom contains the same number of electrons as protons but the Sn<sup>2+</sup> ion has lost two electrons, so it contains 48 electrons.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">The wavelength for the emission from <em>n</em> = ∞ to <em>n</em> = 1 in the hydrogen spectrum is 9.12 x 10<sup>−8</sup> m. Which gives the ionization energy of hydrogen in kJ mol<sup>−1</sup>?<br>
Plank's constant = 6.63 ×10<sup>−34</sup> J s; Speed of light = 3.00 ×10<sup>8</sup> m s<sup>−1</sup>; Avogadro's constant = 6.02 ×10<sup>23</sup> mol<sup>−1</sup></div>
<div class="q-answer">
<p><label class="radio" data-answer="324904bb9107b6e52e3a489ce93deadd"><input type="radio"><span> (6.63 x 10<sup>−34</sup> x 9.12 x 10<sup>−8</sup> x 6.02 x 10<sup>23</sup>) ÷ (1 x 10<sup>3</sup> x 3.00 x 10<sup>8</sup>)</span></label></p>
<p><label class="radio" data-answer="8380357477060656d398025d916645e1"><input type="radio"><span> (6.63 x 10<sup>−34</sup> x 9.12 x 10<sup>−8</sup> x 6.02 x 10<sup>23</sup>) ÷ (3.00 x 10<sup>8</sup>)</span></label></p>
<p><label class="radio" data-answer="48b4dba9b4eeeecd65974d6fb1b0f24b"><input type="radio"><span> (6.63 x 10<sup>−34</sup> x 3.00 x 10<sup>8</sup> x 6.02 x 10<sup>23</sup>) ÷ (9.12 x 10<sup>−8</sup>)</span></label></p>
<p><label class="radio" data-answer="12a06d6ff65b9f7e6c9ae4ada5762c80"><input type="radio"><span> (6.63 x 10<sup>−34</sup> x 3.00 x 10<sup>8</sup> x 6.02 x 10<sup>23</sup>) ÷ (1 x 10<sup>3</sup> x 9.12 x 10<sup>−8</sup>)</span></label></p>
</div>
<div class="q-explanation">
<p><em>E</em> = <em>hν</em> = <em>hc</em>/<em>λ</em> for one electron = <em>hcL</em>/<em>λ</em> in Joules for one mole of electrons =<em> hcL</em>/1000<em>λ</em> kJ mol<sup>−1</sup></p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which electron configuration is possessed by an element in group 6 and period 4 in the periodic table?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="d066f9c8e663e95036eaca20a47fad84"><input type="radio"><span> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>6</sup>3d<sup>6</sup></span></label></p>
<p><label class="radio" data-answer="acb9f2639c789243a6bf7e7e43e33a1f"><input type="radio"><span> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>6</sup>3d<sup>10</sup>4s<sup>2</sup>4p<sup>4</sup></span></label></p>
<p><label class="radio" data-answer="572d8267c3d65a38d7a0b02a6817dc1e"><input type="radio"><span> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>6</sup>3d<sup>5</sup>4s<sup>1</sup></span></label></p>
<p><label class="radio" data-answer="7135a6e7e5161a2238ab98d55363f5c2"><input type="radio"><span> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>6</sup>3d<sup>4</sup>4s<sup>2</sup></span></label><span></span></p>
</div>
<div class="q-explanation">
<p>If it is in the fourth period the outer level of electrons must be occupying the 4s, 3d or 4p sub-levels and as it is in group 6 there must be six electrons in the outer level. This might be expected to be [Ar]3d<sup>4</sup>4s<sup>2</sup> but it is more energetically favourable for the configuration to be [Ar]3d<sup>5</sup>4s<sup>1</sup> as all the outer electrons are unpaired which lowers the repulsion between them. The element is chromium.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>What is the oxidation state of metal M in the complex ion [M(H<sub>2</sub>O)<sub>4</sub>(OH)<sub>2</sub>]?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="2a24213f4f964092cabeb9859056f970"><input type="radio"><span> +4</span></label></p>
<p><label class="radio" data-answer="38beb697680e0b7945aac76b12ad4ab2"><input type="radio"><span> +2</span></label></p>
<p><label class="radio" data-answer="02e3c1cfd03c87011d3b438dee3b08c4"><input type="radio"><span> −2</span></label></p>
<p><label class="radio" data-answer="13c2f11de38484856c4a519b3405e5cf"><input type="radio"><span> +6</span></label></p>
</div>
<div class="q-explanation">
<p>Water is a neutral ligand and each hydroxide ion has a charge of minus 1. Since the overall charge on the complex is zero the metal must have an oxidation state of +2.</p>
</div>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>The spectrochemical series follows the order</p>
<p>I<sup>−</sup> < Br<sup>− </sup>< S<sup>2−</sup> < Cl<sup>−</sup> < F<sup>−</sup> < OH<sup>−</sup> < H<sub>2</sub>O < SCN<sup>−</sup> < NH<sub>3</sub> < CN<sup>−</sup> ≈ CO</p>
Which is a correct statement?</div>
<div class="q-answer">
<p><label class="radio" data-answer="05d84e118e39549146a0e923ac01d4ea"><input type="radio"><span> All transition metal complexes containing cyanide ions will be the same colour.<sub> </sub></span></label></p>
<p><label class="radio" data-answer="db8b6f8f4e52e31dbd7ad9c45a208cb5"><input type="radio"><span> All the ligands are monodentate ligands except S<sup>2−</sup>, which is a bidentate ligand.</span></label></p>
<p><label class="radio" data-answer="e600858a6079985ec900f556df373c52"><input type="radio"><span> Iodide ions cause the d orbitals to split the most.</span></label></p>
<p><label class="radio" data-answer="56a93acf1cbdf78d6e4d8d6e1ab53e3f"><input type="radio"><span> The energy difference between the split d orbitals will increase when ammonia replaces water as a ligand. </span></label></p>
</div>
<div class="q-explanation">
<p>The spectrochemical series lists ligands according to the energy difference they produce between the two sets of d-orbitals in an octahedral complex, so I<sup>−</sup> causes the d orbitals to split the least and ammonia will split the d orbitals more than water. They are all monodentate ligands as they only form one coordinate bond to the metal. Bidentate ligands, such as H<sub>2</sub>N-CH<sub>2</sub>-CH<sub>2</sub>-NH<sub>2</sub> can form two coordinate bonds to the metal. The colour of a transition metal complex also depends on the identity of the metal and its oxidation state as well as the nature of the ligand(s).</p>
</div>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which contains at least one coordinate covalent bond?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="51e1b92f02db8ff8d2d8b2e38490b8ff"><input type="radio"><span> methanoic acid, HCOOH</span></label></p>
<p><label class="radio" data-answer="02927405a2ffacb5edbe5e4465346b98"><input type="radio"><span> carbon dioxide, CO<sub>2</sub></span></label></p>
<p><label class="radio" data-answer="67b2079e2cae80b6d80c63e9a6d4959e"><input type="radio"><span> the carbonate ion, CO<sub>3</sub><sup>2−</sup></span></label></p>
<p><label class="radio" data-answer="f84ddecf38831bd4ca904b3252687198"><input type="radio"><span> carbon monoxide, CO</span></label></p>
</div>
<div class="q-explanation">
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/lewis-structure-of-co.png" style="width: 300px; height: 27px;"><br>
In carbon monoxide the triple bond between the C and O atom is formed by the carbon atom sharing two electrons with two electrons from the O and the O donating a further pair of electrons. In the other three species all the bonds are normal covalent single or double bonds, even in the carbonate ion which forms a resonance hybrid.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>What is the shape of<sup> </sup>the bromine trifluoride molecule, BrF<sub>3</sub>?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="fc0c2213bdb53de34335dcc2a380ba40"><input type="radio"><span> T-shaped</span></label></p>
<p><label class="radio" data-answer="06832db27dd4af4a8dbed8cd4e69f8b5"><input type="radio"><span> tetrahedral</span></label></p>
<p><label class="radio" data-answer="861089d08daaaaa77488005f795d9e58"><input type="radio"><span> trigonal pyramidal</span></label></p>
<p><label class="radio" data-answer="26f878dc79489afc096a57165c1696d1"><input type="radio"><span> trigonal planar</span></label></p>
</div>
<div class="q-explanation">
<p>There are five electron domains around the central bromine atom consisting of three bonding pairs and two non-bonding pairs. This results in a T-shaped molecule.</p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/brf3.png" style="width: 90px; height: 89px;"></p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which describes the strongest type of intermolecular attractive forces between molecules in the compound?</p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/hydrogen-bonding-etc-(1).png" style="width: 500px; height: 154px;"></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>Both propan-2-ol and propanoic acid have a hydrogen atom bonded directly to an electronegative oxygen atom so form hydrogen bonds between molecules whereas in propanal the hydrogen atom is bonded to a carbon atom and so only dipole-dipole attractions occur.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>What is the hybridization of the carbon atoms in ethenylbenzene (styrene), C<sub>6</sub>H<sub>5</sub>CHCH<sub>2</sub>?</p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/styrene-(1).png" style="width: 150px; height: 100px;"></p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="8665fc7d84815659987c7f12966d28ad"><input type="radio"><span> sp and sp<sup>2</sup> </span></label></p>
<p><label class="radio" data-answer="cbd0e9c2ea681d83b33a0480e05a68e8"><input type="radio"><span> sp only</span></label></p>
<p><label class="radio" data-answer="0e04ea3f485ac176b80f7b9706751ca4"><input type="radio"><span> sp, sp<sup>2</sup> and sp<sup>3</sup></span></label></p>
<p><label class="radio" data-answer="09d95db2cbe47fcb7693258eee6b0000"><input type="radio"><span> sp<sup>2</sup> only</span></label></p>
</div>
<div class="q-explanation">
<p>The two carbon atoms in the −CH=CH<sub>2</sub> group are sp<sup>2</sup> hybridized as are the six carbon atoms in the phenyl group. C<sub>6</sub>H<sub>5</sub>− so all are sp<sup>2</sup> hybridized.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>How many sigma (σ) and pi (π) bonds are present in ethanenitrile (acetonitrile), CH<sub>3</sub>CN?</p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/sigma-pi-for-ch3cn.png" style="width: 250px; height: 144px;"></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 nitrile carbon atom is sp hybridized so forms one sigma and two pi bonds with the nitrogen atom and the four bonds formed by the sp<sup>3</sup> hybridized carbon atom on the methyl group are all sigma bonds.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>The superscript symbol, <sup>⦵</sup> placed after Δ<em>H</em> means that the enthalpy change for the reaction is the standard enthalpy change.<br>
Which statement must be true for the standard enthalpy change of any chemical reaction?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="7232d0155fd17965dcfbc8fadd4d658a"><input type="radio"><span> The enthalpy change must be measured at 298 K</span></label></p>
<p><label class="radio" data-answer="3afccf4ba1e8e0e3b006597067997606"><input type="radio"><span> Δ<em>H</em><sup>⦵ </sup>reaction = Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> reactants) − Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> products)</span></label></p>
<p><label class="radio" data-answer="994f4ae927c23cfd9003f82a66fb95ab"><input type="radio"><span> The value should always be given in kJ mol<sup>−1</sup></span></label></p>
<p><label class="radio" data-answer="c2e16a093da585a9d261fb862bb1f215"><input type="radio"><span> Reactants and products must be in their normal most pure state measured at 100 kPa.</span></label></p>
</div>
<div class="q-explanation">
<p>Temperature is not a part of the definition of standard state (although 298 K is commonly given as the temperature of interest).<br>
Δ<em>H</em><sup>⦵ </sup>reaction = Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> products) − Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> reactants) not Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> reactants) − Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> products).<br>
Although Δ<em>H</em><sup>⦵</sup><sub>f </sub>and<sup> </sup>Δ<em>H</em><sup>⦵</sup><sub>c</sub> are normally given in kJ mol<sup>−1</sup> as they refer to one particular substance, Δ<em>H</em><sup>⦵</sup>reaction is normally given in kJ, not kJ mol<sup>−1</sup> as, for example in the reaction N<sub>2</sub>(g) + 3<sub>2</sub>(g) ⇌ 2NH<sub>3</sub>(g), it is not clear if the mol<sup>−1</sup> is referring to N<sub>2</sub>(g), H<sub>2</sub>(g) or NH<sub>3</sub>(g).</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>What is the enthalpy change in kJ mol<sup>−1</sup> for the combustion of propan-2-ol, CH<sub>3</sub>CH(OH)CH<sub>3</sub>?</p>
<p>2CH<sub>3</sub>CH(OH)CH<sub>3</sub>(l) + 9O<sub>2</sub>(g) → 6CO<sub>2</sub>(g) + 8H<sub>2</sub>O(l)</p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/propan-2-ol-data.png" style="width: 350px; height: 91px;"></p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="edfbbc6dc576e9dda40f1dad01aeafbc"><input type="radio"><span> [(2 x − 318) − (6 x − 394) − (8 x − 286)]</span></label></p>
<p><label class="radio" data-answer="b1e3a64370857b3c8188c4a540cf4b25"><input type="radio"><span> ½ [(6 x − 394) + (8 x − 286) − (2 x − 318)]</span></label></p>
<p><label class="radio" data-answer="d3c8eedf6160ad2fa8c047b2ccbb09bd"><input type="radio"><span> [(6 x − 394) + (8 x − 286) − (2 x − 318)]</span></label></p>
<p><label class="radio" data-answer="aa1ebca58c42aae200e6ace5da24d4fc"><input type="radio"><span> ½ [(2 x − 318) − (6 x − 394) − (8 x − 286)]</span></label></p>
</div>
<div class="q-explanation">
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/propan-2-ol-combustion-cycle.png" style="width: 350px; height: 142px;"></p>
<p><em>x</em> = [(6 x −394) + (8 x − 286) − (2 x − 318)] kJ</p>
<p>But <em>x</em> is the enthalpy change for the combustion of 2 mol of propan-2-ol</p>
<p>so Δ<em>H</em> = ½ [(6 x −394) + (8 x − 286) − (2 x − 318)] kJ mol<sup>−1</sup></p>
</div>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which will be an exothermic step in the Born-Haber cycle for the formation of MgO(s)?</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="854735bdbdfd36efb4771f61aa9b407b"><input type="radio"><span> Mg(g) → Mg<sup>+</sup>(g) + e<sup>−</sup></span></label></p>
<p><label class="radio" data-answer="79108d82eb830ec04a23117151034dda"><input type="radio"><span> Mg<sup>+</sup>(g) → Mg<sup>2+</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>
</div>
<div class="q-explanation">
<p>Removing electrons is always endothermic. Adding one electron to an atom is always exothermic but adding another electron to a negative ion is endothermic.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which reaction will result in the smallest change in entropy?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="39e9495e0d28c7ef3c22d62390795cc2"><input type="radio"><span> 3H<sub>2</sub>(g) + N<sub>2</sub>(g) → 2NH<sub>3</sub>(g)</span></label></p>
<p><label class="radio" data-answer="d778d0078c1db74f79648c0521e5f3f0"><input type="radio"><span> N<sub>2</sub>O<sub>4</sub>(g) → 2NO<sub>2</sub>(g)</span></label></p>
<p><label class="radio" data-answer="ed5ccb2fa0fbf951c2131cdbc63cd380"><input type="radio"><span> 2SO<sub>2</sub>(g) + O<sub>2</sub>(g) → 2SO<sub>3</sub>(g)</span></label></p>
<p><label class="radio" data-answer="71f1909517cca042c125b18c1824e192"><input type="radio"><span> H<sub>2</sub>(g) + I<sub>2</sub>(g) → 2HI(g)</span></label></p>
<p> </p>
</div>
<div class="q-explanation">
<p>There is no change in the amount (in mol) of gas in the reaction between hydrogen and iodine whereas in the other three reactions the number of moles of gas either increases or decreases.<br>
The actual values in J K<sup>−1</sup> are:<br>
H<sub>2</sub>(g) + I<sub>2</sub>(g) → 2HI(g) Δ<em>S</em> = +21.5<br>
N<sub>2</sub>O<sub>4</sub>(g) →2NO<sub>2</sub>(g) Δ<em>S</em> = +177<br>
2SO<sub>2</sub>(g) + O<sub>2</sub>(g) → 2SO<sub>3</sub>(g) Δ<em>S</em> = −188<br>
3H<sub>2</sub>(g) + N<sub>2</sub>(g) → 2NH<sub>3</sub>(g) Δ<em>S</em> = −199</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>The graph shows the fraction of reacting particles with a particular energy for an uncatalysed reaction at two temperatures, T<sub>1</sub> and T<sub>2</sub>. The activation energy for the reaction is shown as <em>E</em><sub>a</sub>.</p>
<p> </p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/maxwell-boltzman-graph.png" style="width: 400px; height: 281px;"></p>
<p> </p>
<p>Which is statement is correct?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="c22b3709afd5ba0fac592dfd1442183a"><input type="radio"><span> Adding a catalyst moves the peak of T<sub>1</sub> and the peak of T<sub>2</sub> further to the right.</span></label></p>
<p><label class="radio" data-answer="ff950bd7a1320a09a53ee64d231771af"><input type="radio"><span> T<sub>1</sub> > T<sub>2</sub></span></label></p>
<p><label class="radio" data-answer="f98979065c8df0fad23fd48910d216ea"><input type="radio"><span> The area under T<sub>2</sub> > the area under T<sub>1</sub>.</span></label></p>
<p><label class="radio" data-answer="3dadd4316eb6315e5c4257a23f3dcc34"><input type="radio"><span> The activation energy for the catalysed reaction will be less than <em>E</em><sub>a</sub>.</span></label></p>
</div>
<div class="q-explanation">
<p>Increasing the temperature increases the average energy of the reacting particles so T<sub>2</sub> > T<sub>1</sub> but does not change the number of reacting particles so the area under the graph remains constant. Adding a catalyst does not change the energy of the reacting particles but does provide a different pathway for the reaction with a lower activation energy.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>100 cm<sup>3</sup> of 2.0 mol dm<sup>−3</sup> hydrochloric acid was added to 1.0 g of a strip of magnesium metal. A plot of volume of hydrogen evolved against time gave graph A.</p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/mg-hcl-graph(1).png" style="width: 300px; height: 224px;"></p>
<p>Assuming all the other conditions were kept the same which changes would have given graphs B and C?</p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/data-for-mg-hcl(2).png" style="width: 600px; height: 211px;"></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>100 cm<sup>3</sup> of 2.0 mol dm<sup>−3</sup> HCl(aq) = 0.20 mol, 1.0 g of Mg = 1.0/24.3 = < 0.05 mol so the acid is in excess and Mg is the limiting reagent. If only 0.5 g of magnesium is used half the volume of hydrogen will be evolved so C and D cannot be correct. Using powdered Mg instead of a single piece of Mg will increase the rate whereas using 1.0 mol dm<sup>−3</sup> HCl(aq) instead of 2.0 mol dm<sup>−3</sup> HCl(aq) will decrease the rate but the same volume of hydrogen will eventually be evolved.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>The initial rate of a reaction between A and B to form products C and D was determined using different concentrations of A and B at a constant temperature.</p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/initial-rate-data(1).png" style="width: 500px; height: 113px;"></p>
<p>What will be the rate equation for this reaction?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="8e9be15ba4a0c17f0a555eaec868611b"><input type="radio"><span> rate = <em>k</em>[A]<sup>2</sup>[B]</span></label></p>
<p><label class="radio" data-answer="69a8ec50a9f8aee15995504188b00592"><input type="radio"><span> rate = <em>k</em>[A][B]</span></label></p>
<p><label class="radio" data-answer="9432d7aeb29f31bc8fc86b014718b97a"><input type="radio"><span> rate = <em>k</em>[A][B]<sup>2</sup></span></label></p>
<p><label class="radio" data-answer="2ded37d85b058a85c7682033931adc4e"><input type="radio"><span> rate = <em>k</em>[B]<sup>2</sup></span></label></p>
</div>
<div class="q-explanation">
<p>From experiments 1 and 2 keeping [A] constant and doubling [B] doubles the rate so the reaction is first order with respect to [B]. From experiments 2 and 3 keeping [B] constant and doubling [A] quadruples the rate so the reaction is second order with respect to [A].</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>The rate constant for the decomposition of dinitrogen pentoxide to dinitrogen tetroxide and oxygen at 298 K is 3.38 x 10<sup>−5</sup> s<sup>−1</sup>.</p>
<p>2N<sub>2</sub>O<sub>5</sub>(g) → 2N<sub>2</sub>O<sub>4</sub> + O<sub>2</sub>(g)</p>
<p>What can be deduced from this information?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="50567f98954e43fa5cad4540885d09c7"><input type="radio"><span> The slow step involves a collision between two dinitrogen pentoxide molecules.</span></label></p>
<p><label class="radio" data-answer="9155e2a71c97a65c9cfb83793ce1b39d"><input type="radio"><span> The molecularity of the reaction is 5</span></label></p>
<p><label class="radio" data-answer="e356efa719fa4ba37cc36cc0c0ee50e3"><input type="radio"><span> The reaction mechanism consists of three separate steps.</span></label></p>
<p><label class="radio" data-answer="16f47ba0c6314b1c970909047187d801"><input type="radio"><span> The rate of the reaction is first order with respect to dinitrogen pentoxide.</span></label></p>
</div>
<div class="q-explanation">
<p>The units of the rate constant are given as s<sup>−1</sup>. This means that the reaction is first order as the units of rate are mol dm<sup>−3</sup> s<sup>−1</sup> and the units of concentration to the power of 1 are mol dm<sup>−3</sup>. Since the only reactant is dinitrogen pentoxide the reaction must be first order with respect to N<sub>2</sub>O<sub>5</sub>. The slow step must involve the breakdown of a molecule of N<sub>2</sub>O<sub>5</sub> rather than a collision between two N<sub>2</sub>O<sub>5</sub> molecules but the total number of separate steps in the mechanism cannot be deduced from the information provided. The molecularity of an elementary step is the number of reactant particles taking part in that step.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which change will increase the value of the equilibrium constant for the reaction of sulfur dioxide with oxygen to produce sulfur trioxide?</p>
<p style="margin-left: 120px;">2SO<sub>2</sub>(g) + O<sub>2</sub>(g) ⇌ 2SO<sub>3</sub>(g) Δ<em>H</em><sup>⦵</sup> = − 196 kJ</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="46771f43ea41d7ca1eab3cc0c4531b0d"><input type="radio"><span> increasing the pressure</span></label></p>
<p><label class="radio" data-answer="9c451630bad73003fc529c19cafc3a67"><input type="radio"><span> lowering the temperature</span></label></p>
<p><label class="radio" data-answer="71c9b84347d182391ffbfd31bd594b96"><input type="radio"><span> removing some of the sulfur trioxide as it is formed</span></label></p>
<p><label class="radio" data-answer="bc8a272a6a8a67fa152d4b9a0f5a006b"><input type="radio"><span> decreasing the pressure</span></label></p>
</div>
<div class="q-explanation">
<p>All of the changes will alter the position of equilibrium but the equilibrium constant is a constant at a particular temperature so only a change in temperature will change the value of <em>K</em><sub>c</sub>. As Δ<em>H</em><sup>⦵</sup> has a negative value the reaction is exothermic so lowering the temperature will increase the value of <em>K</em><sub>c</sub>.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which row gives the correct combination for the composition of the equilibrium mixture and the sign of Δ<em>G</em> for the stated value of <em>K</em><sub>c</sub>?</p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/equilibrium-mixtures-(2).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>If <em>K</em><sub>c </sub>>> 1 then the equilibrium mixture will contain mainly products and if <em>K</em><sub>c</sub> <<1 then the equilibrium mixture will consist mainly of reactants. Δ<em>G</em> = − RTln<em>K</em><sub>c</sub> so if <em>K</em><sub>c</sub> is << 1 ln<em>K</em><sub>c</sub> will be negative and Δ<em>G</em> will be positive whereas if <em>K</em><sub>c</sub> is >>1 ln<em>K</em><sub>c</sub> will be positive and Δ<em>G</em> will be negative</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>What will be the pH of a solution made by adding 90 cm<sup>3</sup> of water to 10 cm<sup>3</sup> of an aqueous solution of a strong base with a pH of 12?</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="d3d9446802a44259755d38e6d163e820"><input type="radio"><span> 10</span></label></p>
<p><label class="radio" data-answer="aab3238922bcc25a6f606eb525ffdc56"><input type="radio"><span> 14</span></label></p>
<p><label class="radio" data-answer="6512bd43d9caa6e02c990b0a82652dca"><input type="radio"><span> 11</span></label></p>
</div>
<div class="q-explanation">
<p>In the original solution with a pH of 12 [H<sup>+</sup>(aq)] = 1 x 10<sup>−12</sup> mol dm<sup>−3</sup> and [OH<sup>−</sup>(aq)] = 1 x 10<sup>−2</sup> mol dm<sup>−3</sup>. Adding 90 cm<sup>3</sup> of water dilutes the solution ten times so [OH<sup>−</sup>(aq)] = 1 x 10<sup>−3</sup> mol dm<sup>−3 </sup>and [H<sup>+</sup>(aq)] = 1 x 10<sup>−11</sup> mol dm<sup>−3 </sup>so pH = 11.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Hydrogen chloride gas dissolves in water to form hydrochloric acid.</p>
<p>HCl(g) + H<sub>2</sub>O(l) → H<sub>3</sub>O<sup>+</sup>(aq) + Cl<sup>−</sup>(aq)</p>
<p>What is the role of the water?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="107a27a9680c7fcd6890750ac6954efd"><input type="radio"><span> It is acting as an acid and a base and a solvent</span></label></p>
<p><label class="radio" data-answer="ac445bdc12d86258fd1891a371d41803"><input type="radio"><span> It is only acting as a base</span></label></p>
<p><label class="radio" data-answer="c52c04ce0ee93a381f1a34a671454d05"><input type="radio"><span> It is only acting as a solvent</span></label></p>
<p><label class="radio" data-answer="737f1c88c6641981da9aed5d5264eb9c"><input type="radio"><span> It is acting as a base and a solvent</span></label></p>
</div>
<div class="q-explanation">
<p>Water is amphiprotic so depending upon the reaction can act as an acid or a base by losing or gaining a proton. In this reaction it is not acting as an acid as it is not losing a proton to form OH<sup>−</sup> but is acting as a Brønsted-Lowry base by gaining a proton to form H<sub>3</sub>O<sup>+</sup>. Since the gas is dissolving and the ions formed are hydrated it is also acting as a solvent.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which can act as a Lewis acid but not as a Brønsted-Lowry acid?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="ebaf9bd1f90f1e9597736a9a00607b72"><input type="radio"><span> F<sup>−</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>
<p><label class="radio" data-answer="7df701d2736645defca2a5b3d6330f6e"><input type="radio"><span> AlCl<sub>3</sub></span></label></p>
<p><label class="radio" data-answer="70c7652df1614d1450ec4f641e8176a9"><input type="radio"><span> NH<sub>4</sub><sup>+</sup></span></label></p>
</div>
<div class="q-explanation">
<p>Lewis acids can accept a pair of electrons. Brønsted-Lowry acids can donate a proton so H<sub>3</sub>O<sup>+</sup> and NH<sub>4</sub><sup>+</sup> are Brønsted-Lowry acids and Lewis acids. F<sup>−</sup> cannot accept a pair of electrons but can donate a pair so is a Lewis base. AlCl<sub>3</sub> can accept a pair of electrons but has no proton to donate so is a Lewis acid but not a Brønsted-Lowry acid.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which combination can be used to make a buffer solution with a pH greater than 7?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="0e43745de20714e1eee7cbfc491ae38f"><input type="radio"><span> NH<sub>3</sub>(aq) + NaCl(aq)</span></label></p>
<p><label class="radio" data-answer="bc97d2df46238241b92e5689f1122500"><input type="radio"><span> NH<sub>3</sub>(aq) + HCl(aq)</span></label></p>
<p><label class="radio" data-answer="302ff7c1d4a3aa804dae925667a613a1"><input type="radio"><span> NaOH(aq) + CH<sub>3</sub>COONa(aq)</span></label></p>
<p><label class="radio" data-answer="857f24d76fd2b02f37d26d4082e55b77"><input type="radio"><span> NaOH(aq) + CH<sub>3</sub>COOH(aq) </span></label></p>
</div>
<div class="q-explanation">
<p>An alkaline buffer solution contains a weak base and the salt that the base forms with a strong acid. This can be achieved by adding excess ammonia to hydrochloric acid so that all the acid has been turned into the salt and the excess weak base remains.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which is a redox reaction?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="cff1bbcde6d9028ad96cf3fa9eaabda7"><input type="radio"><span> AgNO<sub>3</sub>(aq) + NaCl(aq) → AgCl(s) + NaNO<sub>3</sub>(aq)</span></label></p>
<p><label class="radio" data-answer="602bc71889641b5ce5372970cf8763c2"><input type="radio"><span> BF<sub>3</sub>(g) + CsF(l) → CsBF<sub>4</sub>(s)</span></label></p>
<p><label class="radio" data-answer="fb32e5c9bca879c5662d9205eb79b1fc"><input type="radio"><span> CH<sub>4</sub>(g) + Cl<sub>2</sub>(g) → CH<sub>3</sub>Cl(g) + HCl(g)</span></label></p>
<p><label class="radio" data-answer="8cf04b78881304c7bc4fdde646e93ca0"><input type="radio"><span> [Cu(H<sub>2</sub>O)<sub>6</sub>]<sup>2+</sup> + 4Cl<sup>−</sup>(aq) → [CuCl<sub>4</sub>]<sup>2−</sup>(aq) + 6H<sub>2</sub>O(l)</span></label></p>
</div>
<div class="q-explanation">
<p>The only one of the reactions in which where there is a change in oxidation state is the reaction between chlorine and methane. Elemental chlorine is reduced from 0 to − 1 in hydrogen chloride and methane is oxidized to chloromethane. The reaction between silver ions and chloride ions to form silver chloride is precipitation. The reaction between boron trifluoride and fluoride ions is a Lewis acid - base reaction and the reaction with hydrated copper ions is a ligand exchange reaction.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>The reactions of manganese metal with nickel ions and nickel metal with lead ions occur spontaneously. </p>
<p>Mn(s) + Ni<sup>2+</sup>(aq) → Mn<sup>2+</sup>(aq) + Ni(s)</p>
<p>Ni(s) + Pb<sup>2+</sup>(aq) → Ni<sup>2+</sup>(aq) + Pb(s)</p>
<p>Which is the best oxidizing agent?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="c0ef8306b7dd2c167aad83133d5fe0ad"><input type="radio"><span> Pb<sup>2+</sup>(aq)</span></label></p>
<p><label class="radio" data-answer="1e96ba543b166bb6e437ace0935b0b5b"><input type="radio"><span> Mn(s)</span></label></p>
<p><label class="radio" data-answer="7bc89c6ae00019e7be38e3251e906434"><input type="radio"><span> Ni<sup>2+</sup>(aq)</span></label></p>
<p><label class="radio" data-answer="68d502c5dd8bdd76ec6725557ed219b4"><input type="radio"><span> Ni(s)</span></label></p>
</div>
<div class="q-explanation">
<p>Ni<sup>2+</sup>(aq) is a better oxidizing agent than Mn(s) as it can remove electrons from Mn(s) to form Ni(s), but Pb<sup>2+</sup>(aq) is an even stronger oxidizing agent as it can remove electrons from Ni(s). Mn(s) is the best reducing agent.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>A cell is made from an iron half-cell, Fe(s)/Fe<sup>2+</sup>(s), connected to a copper half-cell, Cu<sup>2+</sup>(aq)/Cu(s). What is the maximum EMF that can be obtained when two such cells are connected together in series under standard conditions at 298 K?</p>
<p>Fe<sup>2+</sup>(aq) + 2e<sup>−</sup> ⇌ Fe(s) <em>E</em><sup>⦵</sup> − 0.45 V</p>
<p>Cu<sup>2+</sup>(aq) + 2e<sup>−</sup> ⇌ Cu(s) <em>E</em><sup>⦵</sup> = + 0.34 V</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="b09b8163c6ca1a7760a002844a5f482f"><input type="radio"><span> 0.11 V</span></label></p>
<p><label class="radio" data-answer="b88c2ebf57976ddf620951ae344746ce"><input type="radio"><span> 0.79 V</span></label></p>
<p><label class="radio" data-answer="0cf3bf12b04347112a839d6c5af51ca2"><input type="radio"><span> 1.58 V</span></label></p>
<p><label class="radio" data-answer="f431ede893df1b72a44091b5c4adb756"><input type="radio"><span> 0.22 V</span></label></p>
</div>
<div class="q-explanation">
<p>The overall reaction in the cell is Fe(s) + Cu<sup>2+</sup>(aq) → Cu(s) + Fe<sup>2+</sup>(aq) <em>E</em><sup>⦵</sup><sub>total </sub>= 0.79 V so the EMF from each of the two cells is 0.79 V. When they are connected in series the EMF will be 2 x 0.79 V = 1.58 V.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Two electrolytic cells were connected in series. The first contained an aqueous solution of dilute sulfuric acid with platinum electrodes and the second contained an aqueous solution of copper(II) sulfate with copper electrodes. After a current had been passed for some time it was found that one of the electrodes in the second cell had decreased in mass by 3.2 g. Which is a correct statement?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="81320830a61c43b2c45547cacb84e65d"><input type="radio"><span> The blue colour of the copper(II) sulfate solution remained constant throughout.</span></label></p>
<p><label class="radio" data-answer="9c73c2068fbbcdb62b8f553d4c4450e7"><input type="radio"><span> 0.10 mol of hydrogen gas was produced at the cathode (−) in the cell containing sulfuric acid. </span></label></p>
<p><label class="radio" data-answer="e40b9a5b1ac60fb241545404ecdfa1c7"><input type="radio"><span> 0.10 mol of oxygen gas was evolved at the anode (+) in the cell containing sulfuric acid.</span></label></p>
<p><label class="radio" data-answer="c7def94123becbbc6c05c261abdb1908"><input type="radio"><span> 0.10 mol of hydrogen gas was produced at the cathode (−) in the cell containing copper(II) sulfate solution.</span></label></p>
</div>
<div class="q-explanation">
<p>In the copper(II) sulfate cell copper is deposited at the cathode so the cathode gains in mass, Cu<sup>2+</sup>(aq) + 2e<sup>−</sup> → Cu(s). An equal amount of copper is lost from the anode, Cu(s) → Cu<sup>2+</sup>(aq) + 2e<sup>−</sup> so [Cu<sup>2+</sup>(aq)] remains constant and the solution retains its blue colour. 3.2 g of copper was lost, i.e. 3.2/64 = 0.05 mol of copper. This means that the charge passed through both cells was 2 x 0.05 F. 2H<sup>+</sup>(aq) + 2e<sup>−</sup> → H<sub>2</sub>(g) so the amount of hydrogen formed at the cathode in the sulfuric acid cell will be 0.05 mol and the amount of oxygen formed at the anode will be 0.025 mol since 2H<sub>2</sub>O(l) → 4H<sup>+</sup>(q) + O<sub>2</sub>(g) + 4e<sup>−</sup>. </p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which is a tertiary alcohol?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="facd93f8edb009b7db7b095859e9830f"><input type="radio"><span> CH<sub>3</sub>−CH<sub>2</sub>−CH<sub>2</sub>OH</span></label></p>
<p><label class="radio" data-answer="36a012a29c9dc556ba6a623d9ab5ec1c"><input type="radio"><span> CH<sub>3</sub>−CH<sub>2</sub>−CH(OH)−CH<sub>3</sub></span></label></p>
<p><label class="radio" data-answer="9d9cd2eda036d4ecd51a57559b51e9a4"><input type="radio"><span> CH<sub>2</sub>(OH)−CH(OH)−CH<sub>2</sub>(OH)</span></label></p>
<p><label class="radio" data-answer="8a2b2849e83bad4cad9c632896040eac"><input type="radio"><span> C(CH<sub>3</sub>)<sub>3</sub>OH</span></label></p>
</div>
<div class="q-explanation">
<p>Tertiary alcohols contains three R− groups bonded to the carbon atom to which the −OH is bonded. Compounds such as propan−1,2,3−triol (also known as glycerol), which contain three −OH groups, are known as trihydric alcohols.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which compounds belong to the same homologous series?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="c524f62900b7e52477a540fcfa6d491b"><input type="radio"><span> CH<sub>3</sub>CH<sub>2</sub>OCH<sub>2</sub>CH<sub>3</sub> and CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>CH<sub>2</sub>OH</span></label></p>
<p><label class="radio" data-answer="905dfbdd8baa2394144eaa1cc80d527a"><input type="radio"><span> CH<sub>2</sub>CH<sub>2</sub> and CH<sub>2</sub>CHCH<sub>3</sub></span></label></p>
<p><label class="radio" data-answer="66050c0a12ca176106b191e1af185ed1"><input type="radio"><span> CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>CHO and CH<sub>3</sub>CH<sub>2</sub>COCH<sub>3</sub></span></label></p>
<p><label class="radio" data-answer="59831ffb8117174682745edc409b81c2"><input type="radio"><span> CHCH and CH<sub>2</sub>CH<sub>2</sub></span></label></p>
</div>
<div class="q-explanation">
<p>A homologous series is a series of compounds of the same family, with the same general formula, which differ from each other by a common structural unit. Ethene, CH<sub>2</sub>CH<sub>2</sub> and propene, CH<sub>2</sub>CHCH<sub>3</sub> are both members of the alkene homologous series with the general formula C<sub>n</sub>H<sub>2n</sub>. Ethyne, CHCH is an alkyne whereas ethene, CH<sub>2</sub>CH<sub>2</sub> is an alkene. Ethoxyethane, CH<sub>3</sub>CH<sub>2</sub>OCH<sub>2</sub>CH<sub>3</sub> and butan-1-ol, CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>CH<sub>2</sub>OH are an ether and an alcohol respectively. They are isomers as they both have the same molecular formula, C<sub>4</sub>H<sub>10</sub>O but are members of different classes of compounds. Butanal, CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>CHO and butan-2-one, CH<sub>3</sub>CH<sub>2</sub>COCH<sub>3</sub> are also isomers, one is an aldehyde and the other is a ketone.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>Which is a propagation step in the mechanism for the free radical substitution reaction between methane and chlorine in ultraviolet light?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="8b45968a3543013ace786a5605ff9418"><input type="radio"><span> CH<sub>4</sub> + Cl<strong>·</strong>→ CH<sub>3</sub>Cl + H<strong>·</strong></span></label></p>
<p><label class="radio" data-answer="00dd19d3f438cd38bf7d382e44eb4af6"><input type="radio"><span> CH<sub>3</sub><strong>· </strong>+ Cl<strong>·<sup> </sup></strong>→ CH<sub>3</sub>Cl</span></label></p>
<p><label class="radio" data-answer="fc89534a4ac0a37ac36cf6aeaa0144c7"><input type="radio"><span> Cl<sub>2</sub> → Cl<strong>·</strong> + Cl<strong>·</strong></span></label></p>
<p><label class="radio" data-answer="4e85b3ef29373475e57228b784a1088c"><input type="radio"><span> CH<sub>3</sub><strong>·</strong> + Cl<sub>2</sub> → CH<sub>3</sub>Cl + Cl<strong>·</strong></span></label></p>
</div>
<div class="q-explanation">
<p>The initiation step is the homolytic fission of the Cl−Cl bond by ultraviolet light to give chlorine free radicals. These radicals then react to produce more radicals in propagation steps such as CH<sub>4</sub> + Cl<strong>·</strong> → CH<sub>3</sub><strong>·</strong> + HCl and CH<sub>3</sub><strong>· </strong>+ Cl<sub>2</sub> → CH<sub>3</sub>Cl + Cl<strong>·</strong>. At no time are hydrogen, H<strong>·</strong> radicals formed. Termination steps such as CH<sub>3</sub><strong>·</strong> + Cl<strong>·</strong> → CH<sub>3</sub>Cl result in a product that is not a free radical.</p>
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<p> </p>
<div class="exercise">
<div class="q-question">
<p>An optical isomer of 2-bromobutane, CH<sub>3</sub>CHBrCH<sub>2</sub>CH<sub>3</sub> undergoes a substitution reaction with aqueous sodium hydroxide solution to form butan-2-ol by an S<sub>N</sub>2 mechanism.</p>
<p>CH<sub>3</sub>CHBrCH<sub>2</sub>CH<sub>3</sub> + NaOH → CH<sub>3</sub>CHOHCH<sub>2</sub>CH<sub>3</sub> + NaBr</p>
<p>Which statement is correct?</p>
</div>
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<p><label class="radio" data-answer="253b81d26a504ac9c8650deec7f2d64f"><input type="radio"><span> The organic product will rotate the plane of plane polarized light in the opposite direction to the organic reactant.</span></label></p>
<p><label class="radio" data-answer="8fa4e0808dd152b19338955800111c01"><input type="radio"><span> The first step in the reaction is CH<sub>3</sub>CHBrCH<sub>2</sub>CH<sub>3</sub> → CH<sub>3</sub>C<sup>+</sup>HCH<sub>2</sub>CH<sub>3</sub> + Br<sup>−</sup>. </span></label></p>
<p><label class="radio" data-answer="187824a4955cf2f119e13b090b3b6f72"><input type="radio"><span> The reaction will proceed slower if 2-iodobutane is used in place of 2-bromobutane.</span></label></p>
<p><label class="radio" data-answer="3a9820e71fae69635b5cac6f213d84b3"><input type="radio"><span> The organic product will be a racemic mixture.</span></label></p>
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<p>In the S<sub>N</sub>2 mechanism the OH<sup>−</sup> nucleophile bonds to the carbon atom before the bromine atom leaves as the bromide ion so the product has the opposite optical activity in terms of rotating the plane of plane polarized light to the reactant (this is known as Walden inversion). Iodine is a better leaving group than bromine so iodoalkanes react faster than bromoalkanes.</p>
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<p>Which compound can react with lithium aluminium hydride to form pentan-1-ol, CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>CH<sub>2</sub>CH<sub>2</sub>OH?</p>
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<p><label class="radio" data-answer="4e7b20532f079500df5b3a12a064f38f"><input type="radio"><span> pentan-3-one</span></label></p>
<p><label class="radio" data-answer="e0b024a847b94d5f81d54f60205d520b"><input type="radio"><span> pentan-2-ol</span></label></p>
<p><label class="radio" data-answer="274f1fd3aab0d6623fbbfbe8ec8ff18e"><input type="radio"><span> pentanoic acid</span></label></p>
<p><label class="radio" data-answer="098d3dcb2e529153c6d54def998f504c"><input type="radio"><span> pentan-2-one</span></label></p>
</div>
<div class="q-explanation">
<p>Carboxylic acids are reduced by lithium aluminium hydride firstly to the aldehyde then to the primary alcohol. Ketones can also be reduced, but to form secondary alcohols, not primary alcohols.</p>
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<p>Which will be the major product when 2-methylbut-2-ene, (CH<sub>3</sub>)<sub>2</sub>C=CH(CH<sub>3</sub>) reacts with iodine monochloride, ICl?</p>
</div>
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<p><label class="radio" data-answer="62ec02e98284a63108383851084b5d18"><input type="radio"><span> <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/anti-markovnikov-(3).png" style="width: 150px; height: 94px;"></span></label></p>
<p><label class="radio" data-answer="2b79f5f8e4de17b4d922a02ec778c04a"><input type="radio"><span> <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/anti-markovnikov-(2).png" style="width: 150px; height: 95px;"></span></label></p>
<p><label class="radio" data-answer="a094e7aadc516cc3032525a074ea635d"><input type="radio"><span> <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/anti-markovnikov-(1).png" style="width: 150px; height: 93px;"></span></label></p>
<p><label class="radio" data-answer="18336242f14cf8fa2221ec55c9d5ab3d"><input type="radio"><span> <img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/anti-markovnikov-(4).png" style="width: 150px; height: 89px;"></span></label></p>
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<div class="q-explanation">
<p>Iodine monochloride is polar, <sup>δ+</sup>I−Cl<sup>δ−</sup> as chlorine is more electronegative than iodine. When iodine monochloride adds by electrophilic addition the electrophile adds to the carbon atom already bonded to the most hydrogen atoms to form the more stable tertiary carbocation intermediate which then bonds to Cl<sup>−</sup> to give 2-methyl-2-chloro-3-iodobutane.</p>
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<p>A piece of magnesium of known mass was combusted in air in a crucible and the product was weighed afterwards to determine the empirical formula of magnesium oxide. Which could explain why the result obtained was Mg<sub>4</sub>O<sub>5</sub> rather than the expected result of MgO?</p>
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<p><label class="radio" data-answer="5b158521690cda6034fb0031d0becc20"><input type="radio"><span> The piece of magnesium used already had a coating of the oxide layer. </span></label></p>
<p><label class="radio" data-answer="c4c770ef902daf057bf7dad659580806"><input type="radio"><span> The crucible was not inert and also reacted with the magnesium.</span></label></p>
<p><label class="radio" data-answer="ce1059ea41f6813967f511ebb7b1106d"><input type="radio"><span> Some of the product escaped before it could be weighed.</span></label></p>
<p><label class="radio" data-answer="ed71309f2351a35495885a3c9dc392ab"><input type="radio"><span> Not all of the magnesium combusted.</span></label></p>
</div>
<div class="q-explanation">
<p>If some of the product was lost, not all the magnesium burned or if some of the magnesium burned was already magnesium oxide then the mass of the weighed product would be less than expected, so the amount of oxygen combining with the magnesium would be less and the empirical formula would be richer in Mg compared to O. If the crucible reacted with some of the magnesium then the product might have a mass more than if it had just combined with oxygen so the formula would be richer in O compared to Mg.</p>
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<p>What is the index of hydrogen deficiency of the amino acid phenylalanine, C<sub>9</sub>H<sub>11</sub>NO<sub>2</sub>?</p>
<p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/phenylalanine.png" style="width: 130px; height: 147px;"></p>
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<p><label class="radio" data-answer="e4da3b7fbbce2345d7772b0674a318d5"><input type="radio"><span> 5</span></label></p>
<p><label class="radio" data-answer="eccbc87e4b5ce2fe28308fd9f2a7baf3"><input type="radio"><span> 3</span></label></p>
<p><label class="radio" data-answer="1679091c5a880faf6fb5e6087eb1b2dc"><input type="radio"><span> 6</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 and for a 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>10</sub>H<sub>12</sub> which needs 10 more hydrogen atoms (5 units of H<sub>2</sub>) to become saturated as C<sub>10</sub>H<sub>22</sub>, hence the IHD is 5. This can also be seen from the phenyl group which has an IHD of 4 due to the 'three double bonds' and the ring plus the IHD of 1 for the C=O in the carboxyl functional group to give a total IHD of 5.</p>
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<p>Which compound will give a singlet, triplet and quartet in its <sup>1</sup>H NMR spectrum with integration traces in the ratio of 3:3:2?</p>
</div>
<div class="q-answer">
<p><label class="radio" data-answer="910f13f0a7172ffdeb91e0d00de67413"><input type="radio"><span> CH<sub>3</sub>−CO−CH<sub>2</sub>−CH<sub>3</sub></span></label></p>
<p><label class="radio" data-answer="110aa0fed54f1ec330a5540de0421a48"><input type="radio"><span> C<sub>2</sub>H<sub>5</sub>−CH<sub>2</sub>−CO−CH<sub>3</sub></span></label></p>
<p><label class="radio" data-answer="bb07d76740952e3e24ddaf8ef6877db0"><input type="radio"><span> CH<sub>3</sub>−CH(OH)−CH<sub>3</sub></span></label></p>
<p><label class="radio" data-answer="9d331a147e38e1df456bb471c6e5783e"><input type="radio"><span> CH<sub>3</sub>−CH<sub>2</sub>−COOH</span></label></p>
</div>
<div class="q-explanation">
<p>C<sub>2</sub>H<sub>5</sub>− is actually an ethyl group, CH<sub>3</sub>−CH<sub>2</sub>−, so pentan-2-one, C<sub>2</sub>H<sub>5</sub>−CH<sub>2</sub>−CO−CH<sub>3</sub><sub> </sub>will give 4 signals. The other three compounds will each give three signals as their hydrogen atoms are in three different chemical environments. The ratio of the integration traces will be 6:1:1 for propan-2-ol, CH<sub>3</sub>CH(OH)CH<sub>3</sub>. Both propanoic acid CH<sub>3</sub>CH<sub>2</sub>COOH and but-2-one, CH<sub>3</sub>COCH<sub>2</sub>CH<sub>3 </sub>will give a singlet, triplet and quartet but their integration traces will be different. For propanoic acid it will be 1:3:2 and for but-2-one it will be 3:3:2.</p>
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