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SALT

SALT. MODULE 2. TESTING FOR GASES. red litmus paper . milky. limewater. blue litmus paper . bleached. lighted. ‘pop’ sound. glowing. ignited. w hite fumes. Acidified potassium manganate (VII). Purple . Orange solution c hange to green solution. Acidified potassium

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SALT

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  1. SALT MODULE 2

  2. TESTING FOR GASES red litmus paper milky limewater

  3. blue litmus paper bleached lighted ‘pop’ sound

  4. glowing ignited white fumes

  5. Acidified potassium manganate(VII) Purple Orange solution change to green solution Acidified potassium dichromate(VI)

  6. Glowing wooden splinter Burning wooden splinter to blue Limewater

  7. red con.ammonia colourless red potassium manganate(VII)

  8. Action of heat on carbonate salts • 1. All carbonates decompose on heating except …………………………………… • Most carbonate decompose on heating to produce ………………….and ……… Ammonium, sodium and potassium carbonate Metal oxide Carbon dioxide

  9. Write a chemical equation for reaction of zinc carbonate when heated • How can you identify the gas that is produced when a carbonate metal is heated? • ……………………………………………………………… Bubble the gas through limewater, it will turns cloudy

  10. Yellow when Hot,white when cold white brown when hot,yellow when cold white green black

  11. white white white white white white

  12. volume various

  13. Draw a graph volume of potassium iodide against height of precipitate formed from the above table

  14. yellow b) What is the colour of the precipitate formed? ……………………………………………………………………………………………. c) Name the precipitate formed …………………………………………………………………………….. Lead (II) iodide

  15. Calculate the number of moles of lead (II) nitrate solution in 5 cm 3 No of mole = MV/1000 = 0.5 x 5 / 1000 = 0.0025 mol

  16. What is the volume of potassium iodide that exactly react with 5.00 cm 3 of lead (II) nitrate solution? (volume that exactly react is the volume when the height of precipitate start constant) Volume = 5.00 cm3

  17. Calculate the number of moles of potassium iodide that reacts with 5 cm3 lead (II) nitrate solution No of mole = MV/1000 = 1.0 x 5 / 1000 = 0.005 mol

  18. Deduce empirical formula of the precipitate formed mole Pb2+ : mole I- 0.0025 : 0.0050 1:2 PbI2

  19. Hence, construct ionic equation for the reaction above Pb2+ + 2I- PbI2

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