ICSE X | CHEMISTRY Board Paper – 2015 ICSE Board Class X Chemistry Board Paper – 2015 Time: 2 hrs Max. Marks: 80 Answers to this Paper must be written on the paper provided separately. You will not be allowed 10 write during the first 15 minutes. This time is to be spent in reading the Question Paper. The time given at the head of this paper is the time allowed for writing the answers. Section I is compulsory. Attempt any four questions from section II. The intended marks for questions or parts
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A DILATOMETER The decomposition of diacetone alcohol into two molecules of acetone is catalyzed by hydroxide ions and is an example of an aldol condensation in reverse. O OH OHO 2CH3-C-CH3 CH3-C-CH2-C(CH3)2 The rate of decomposition is first-order with respect to the concentrations of both diacetone alcohol and hydroxide ion: Rate = k[OH-][diacetone alcohol] (1) However‚ since hydroxide ion is a catalyst its concentration remains constant during the reaction. The overall reaction appears
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food acid burette clamp 0.1M sodium hydroxide magnetic flea magnetic stirrer glass funnel 1 x 200mL volumetric flask phenolphthalein 4 x 50mL beakers Method Part A: Making the food acid Place distilled water into a 25mL beaker Rinse the 20mL pipette with distilled water then with the 0.3M food acid Once the pipette has been cleaned‚ pipette 20mL of the food acid and place it
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K+ + 2 NO3- 2. Solutions of sodium sulfate and calcium bromide are mixed. Na2SO4(aq) + CaBr2(aq) CaSO4(s) + 2 NaBr(aq) 2 Na+ + SO42- + Ca2+ + 2 Br- CaSO4(s) + 2 Na+ + 2 Br- 3. Solutions of aluminum acetate and lithium hydroxide are mixed. Al(C2H3O2)3(aq) + 3 LiOH(aq) Al(OH)3(s) + 3 LiC2H3O2(aq) Al3+ + 3 C2H3O2- + 3 Li+ + 3 OH- Al(OH)3(s) + 3 Li+ + 3 C2H3O2- 4. Solutions of iron(III) sulfate and sodium sulfide are mixed. Fe2(SO4)3(aq)
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Equipment: Concentrated nitric acid in a glass dropping bottle Small pieces of Copper 250ml Beaker 2 M Lead nitrate in a dropping bottle 2 M Potassium iodide in a dropping bottle 2 M Copper sulfate in a dropping bottle 2 M Sodium hydroxide in a dropping bottle 2 M Hydrochloric acid in a dropping bottle 4 Pyrex test tubes Test tube rack Spatula Bunsen burner‚ gauze mat Sandpaper Magnesium Ribbon Tongs Safety glasses Procedure: 1. This is a teacher-only
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Aim: To separate a mixture of carboxylic acid and a neutral substance by treatment with aqueous sodium hydroxide and purifying the carboxylic acid‚ measuring the melting points of the neutral and the acid components. METHOD: A mixture of( 5g) Carboxylic acid and neutral compound were separated by adding 2 mol dm-3 sodium hydroxide (25cm3) to the mixture‚ which separated the water soluble sodium from the water insoluble neutral component‚ by filtration. The solid at the filter was washed with
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dividing. Materials: Plastic Spoon 2 Blocks of agar phenolphthalein Beaker Dissecting tray Paper towel Sodium hydroxide solution (wear eye shields) Plastic ruler Procedure: 1. Use the sharp end of the ruler to cut the agar blocks into a 3cm‚ 2cm‚ and 1cm cube (use paper towel and dissecting tray to work with blocks). 2. Place all three cubes in the empty beaker. 3. Have teacher pour sodium hydroxide solution into beaker. 4. Wait for 5 minutes without interrupting the blocks. 5. Use formulas to calculate
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experiment were concentrated hydrochloric acid‚ sodium hyrdoxide‚ sodium carbonate‚ potassium acid phthalate and phenolphthalein as indicator. The pieces of glassware that were used to perform this experiment were volumetric flasks‚ Erlenmeyer flasks‚ beakers‚ volumetric pipette‚ burette‚ spatula and droppers. Also‚ the pieces of equipment that were used were analytical balance‚ top-loading balance and hot plate. Preparation of 250 mL 1.0 M sodium hydroxide solution (from solid) The amount of NaOH
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Results : Calculation : Part A Molecular weight of 1M of NaOH = 23g/mol + 16g/mol + 1g/mol = 40g/mol 1M = 40g/mol dissolved in 1L and 20g dissolved in 500ml 20g of NaOH was used to prepare 500ml of 1M NaOH. Part B Molecular weight of 1M of HCl = 35.5g/mol + 1g/mol = 36.5g/mol Specific gravity = 1.19kg/L 37% HC1 × 1.19kg/L = 0.44kg/L Convert w/v to mol/v = = 12mol/L = (12mol/L) = (1M)250ml = 20.83ml ≈ 21ml 21ml of concentrated HC1 is used to prepare 250ml of 1M of HC1. Part C For 0.1N of NaOH
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Precipitation of Copper Hydroxide NaOH solution description: The Sodium Hydroxide liquid was transparent‚ homogenous‚ and blue in colour. Notes and observations on the reaction: When the sodium hydroxide was added‚ the solution got a lot darker and had a slight increase in viscosity. It also became heterogeneous as small blue specks were forming and sticking to the glass stirring rod. generating some heat- exothermic reaction‚ chemical reaction‚ copper hydroxide was converted to the
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