{"id":8398,"date":"2019-01-09T09:10:41","date_gmt":"2019-01-09T09:10:41","guid":{"rendered":"https:\/\/www.mytutor.co.uk\/blog\/?p=8398"},"modified":"2021-08-18T13:15:19","modified_gmt":"2021-08-18T13:15:19","slug":"top-chemistry-gcse-topics-tutored","status":"publish","type":"post","link":"https:\/\/www.mytutor.co.uk\/blog\/parents\/gcse\/top-chemistry-gcse-topics-tutored\/","title":{"rendered":"Top Chemistry GCSE topics tutored"},"content":{"rendered":"

At MyTutor, we\u2019ve got lots of dedicated Chemistry tutors<\/a> across the UK who love helping teens achieve their best when exams come around. Since we started in 2013, we\u2019ve given more than 250,000 one-to-one lessons<\/strong>, and over 1 million school pupils<\/strong> have used our online resource centre. Over time we\u2019ve been able to get a strong understanding of exactly the topics in each subject that kids tend to need that extra help with.<\/p>\n

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Here are the top three Chemistry topics<\/b> our students struggle with most, and some example answers which can double as handy study notes for your child to get them revision-ready.<\/p>\n

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  1. \n
    Balance the Chemical Equation for the reaction of calcium carbonate with hydrochloric acid: CaCO3+ HCl -> CaCl2 + CO2 + H2O<\/a><\/h6>\n<\/li>\n
  2. \n
    Balance the following equation: C3H8 +O2 —-> CO2 + H2O (There will be no need for state symbols.)<\/a><\/h6>\n<\/li>\n
  3. \n
    Balance the equation for the reduction of MnO4- to Mn2+<\/a><\/h6>\n<\/li>\n
  4. \n
    Explain why chlorine (Cl2) is a gas at room temperature, but sodium chloride (NaCl) is a solid at room temperature.<\/a><\/h6>\n<\/li>\n
  5. \n
    What is the difference between ionic, covalent and metallic bonding?<\/a><\/h6>\n<\/li>\n<\/ol>\n

    1. Balance the Chemical Equation for the reaction of calcium carbonate with hydrochloric acid: CaCO3+ HCl -> CaCl2 + CO2 + H2O<\/h2>\n

    To balance chemical equations we need to look at each element individually on both sides of the equation.<\/p>\n

    To start with we can add up the elements to see which are unbalanced.\"\"

    There is 1 Carbon atom on the left and 1 Carbon atom on the right so this is balanced.<\/p>\n

    There are 3 Oxygen atoms on the left and 3 Oxygen atoms on the right so this is balanced.<\/p>\n

    There is 1 hydrogen atom on the left but 2 on the right so this is unbalanced. Therefore we need to balance it by using 2 molecules of HCl:\"\"

    Whenever we balance an equation you have to change the number of molecules used represented by the big numbers before a molecule e.g 2HCl. You cannot change the subscript (small) numbers as this is the number of each element in a molecule, and you would end up making up your own molecule that doesn’t make sense. For example you could not do H2Cl!<\/p>\n

    The easiest way to balance equations is by going through step by step and each time you make a change- check how it affects the rest of the equation.<\/p>\n

    2. Balance the following equation: C3H8 +O2 —-> CO2 + H2O.<\/b> (There will be no need for state symbols.)<\/h2>\n

    First of all, to balance these equations, list the number of each atom, for both sides of the equation. In this example, it would be:<\/p>\n

    Carbon: 3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Carbon: 1
    \nHydrogen: 8\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Hydrogen: 2
    \nOxygen: 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Oxygen: 3<\/p>\n

    The first step would be to balance out the carbons on the products side by multiplying CO2 by 3. The new equation will be<\/p>\n

    C3H8 + 02 —-> 3CO2 + H20<\/b><\/p>\n

    Now, the carbons are balanced, we will look at the hydrogens. We can multiply the number of waters on the product side by 4, to make 4 H20 molecules. This gives both sides of the equation 8 hydrogens. Next we can check our oxygens. There will be 10 oxygens on the products side and 2 on the reactants so to balance these out, we multiply the 02 on the reactants side by 5.<\/p>\n

    The final equation will be<\/p>\n

    C3H8 + 502 —-> 3CO2 + 4H20<\/b><\/p>\n

    3. Balance the equation for the reduction of MnO4- to Mn2+<\/h2>\n

    Balancing equations is usually fairly simple. However some of them involve several steps.<\/p>\n

    You may have come across balanced equations in data booklets that look intimidating. The balanced equation for reduction of Mn7+ to Mn2+ is one such equation.<\/p>\n

    Initially one might write:<\/p>\n