By Steve Owen, Caroline Ahmed, Chris Martin, Roger Woodward

Chemistry for the IB degree, moment variation, covers in complete the necessities of the IB syllabus for Chemistry for first exam in 2016. the second one version of this well-received Coursebook is totally up to date for the IB Chemistry syllabus for first exam in 2016, comprehensively masking all requisites. Get the simplest insurance of the syllabus with transparent evaluate statements, and hyperlinks to concept of information, International-mindedness and Nature of technology issues. examination guidance is supported with lots of pattern examination questions, on-line attempt questions and examination assistance. Chapters overlaying the choices and Nature of technology, overview tips and solutions to questions are integrated within the extra on-line fabric to be had with the e-book.

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Examples are hydrated copper sulfate (CuSO4·5H2O) and hydrated magnesium chloride (MgCl2·6H2O). The water is necessary for the formation of the crystals and is called water of crystallisation. Substances that contain water of crystallisation are described as hydrated, whereas those that have lost their water of crystallisation are described as anhydrous. So, we talk about ‘hydrated copper sulfate’ (CuSO4·5H2O) and ‘anhydrous copper sulfate’ (CuSO4). Hydrated copper sulfate can be obtained as large blue crystals, but anhydrous copper sulfate is white and powdery.

1371 mol N2. 4114 mol magnesium. 178 mol N2 are used, which is more than enough to react. This means that N2 is in excess because there is more than enough to react with all the magnesium present. Magnesium is therefore the limiting reactant. e. 534 mol magnesium. 4114 mol magnesium are present; therefore, the magnesium will run out before all the N2 has reacted. Exam tip If the number of moles of each reactant is divided by its coefficient in the stoichiometric equation, the smallest number indicates the limiting reactant.

2P + 3I2 → 2PI3 d Consider the reaction shown below. 25 g of NaCl? 3SCl2 + 4NaF → S2Cl2 + SF4 + 4NaCl The equation for the reaction is: CaCO3 → CaO + CO2 number of moles of CaCO3 = 100 (20 + 6 + (3 × 8)) = 2 mol Two moles of CaCO3 produces two moles of CO2. The mass of two moles of CO2 is 2 × (6 + (2 × 8)) = 44 kg. Hopefully you can see some mistakes in this calculation, but the result is what we got experimentally. It is also interesting to note that if, in your IB examination, you had just written down the final answer, you would probably have got full marks!

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