GCSE CHEMISTRY High Demand Questions THE PERIODIC TABLE QUESTIONSHEET 1 The boxes represent particles of different gases. One box shows the particles of elements in group 0 (group 8). A (a) (i) B C D What name is given to group 0 (8) elements? (ii) Name two elements from group 0. ......................................................................................................................................................................... [2] (b) (i) Which box
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When a strong acid is present‚ hypochlorite ion changes to hypochlorous acid when the strong acid donates the H+ when it dissociates in the solution. In order to calculate the oxidizing capacity‚ the mass of NaOCl present in each trial should be determined using the volume of Na2S2O3 added in each
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precipitated where in actuality it should not have. It is also good to note that some instances of a precipitate not forming visibly may be due to it being suspended in the fluid and not solid at the bottom. One instance of this may be with potassium iodide added to silver nitrate. Upon a second evaluation‚ the cloudiness and whiteness could potentially be the precipitate being suspended
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Atoms‚ Airs‚ and Chemical Bonding Rich McConnell‚ CH-201 Grantham University Atoms‚ Airs‚ and Chemical Bonding 1. Alcohol in our digestive system reacts with oxygen gas to form carbon dioxide and water. How many grams of carbon dioxide are released if a 10.00 g alcohol sample reacts with 20.85 g of oxygen gas and produces 11.77 g of water? Answer: 19.08g CO2 2. Using the periodic table‚ fill in the missing information in the following table: Protons Neutrons Electrons Isotope Symbol 27 32 27
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produce iodide ion (I-) and dehydroascorbic acid‚ as shown in the following equation: [IMAGE] + I2(aq) -----------> + 2H+(aq) + 2I-(aq) Ascorbic acid (Vit. C) Dehydroascorbic acid However‚ since iodine is only slightly soluble in water‚ ascorgic acid should not be titrated directly by a standard iodine solution‚ since the end point of titration is not o obvious. Instead‚ back titration will be employed. The titration of a reducing agent with iodine to produce iodide ion is
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of valence electron(s) between atoms. It is a type of chemical bond that generates two oppositely charged ions (Larsen‚ 2010). Element that have a positive ion are able to bond with elements with negative ions. This is because the elements share the electrons so that the ionization becomes neutral and both atoms become stable. Materials: 1. Cards with elements and positive or negative ions. 2. Reference Tables for Physical Setting/CHEMISTRY 2011 Edition: Periodic Table of the Elements 3. Bond
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the reducing solution is potassium iodate solution and the oxidizing solution is sodium thiosulphate solution. Potassium iodate solution which is an oxidizing agent is added into an excess solution of acidified potassium iodide. This reaction will release iodine. Potassium iodide is acidified with sulphuric acid and the iodine released quickly titrated with sodium thiosulphate until it become light yellow. The iodine then detected with starch solution and it turn into dark blue solution and titrated
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oxidation of ascorbic acid by iodine solution. Iodine is relatively insoluble‚ but this can be improved by complexing the iodine with iodide to form triiodide as the following formula suggests: I_(2(aq))+〖I^-〗_((aq))↔〖I^-〗_(3(aq)) This triiodide ion can be used to oxidize vitamin C to form dehydroascorbic acid. In this reaction the triiodide ion is reduced to iodide ion‚ and ascorbic acid‚ C6H8O6 is oxidized to dehydroascorbic acid‚ C6H6O6. Aim The aim of this investigation is to discover the exact
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Top of Form Bottom of Form Chemical Compound Formulas Chemistry is the study of the composition of matter and its transformation. A substance can be considered matter with definite properties that establishes its identity. The tremendous number of chemical compounds has been categorized into numerous categories. A broad classification distinguishes between inorganic and organic compounds. Organic compounds are carbon based. Inorganic compounds exclude compounds exclude compounds based on carbon
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the activation enthalpy of the reaction by finding the rate of reaction at different temperatures using the Arrhenius Equation. The experiment will go as follows: Into a conical flask put 15cm3 of distilled water and add 2cm3 of [X]moldm-3 potassium iodide (KI) solution and 1cm3 of 2moldm-3 sulphuric acid. Then add to this 2.5cm3 of 5vol (0.42moldm-3) hydrogen peroxide (H2O2). For the second part of my investigation‚ the KI solution will remain a constant 0.3moldm-3 and the H2O2 solution
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