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MRSM Pengkalan Chepa Quick revision

MRSM Pengkalan Chepa Quick revision. Give reasons on form 4 chemistry topics. Name……………………….. Class………………. Chapter periodic table. K is situated below Na in group 1 elements. Atom K is more reactive than atom Na. Explain ( 3 marks). Size of K atom is bigger than Na atom .

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MRSM Pengkalan Chepa Quick revision

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  1. MRSM PengkalanChepaQuick revision Give reasons on form 4 chemistry topics Name……………………….. Class………………

  2. Chapter periodic table K is situated below Na in group 1 elements. Atom K is more reactive than atom Na. Explain ( 3 marks) • Size of K atom is bigger than Na atom . • nuclei attraction ( force of attractions between nucleus and electrons) in Katom is weaker than Na atom. • Potassium atom easier to donate/ release electron to form positive ion .

  3. Chapter periodic table 2) Cl is situated above Br in group 17 elements. Atom Cl is more reactive than Atom Br . Explain ( 3 marks) • Size of Clatom is smaller than Br atom . • nuclei attraction ( force of attractions between nucleus and electrons) in Clatom is stronger than Br atom. • Chlorine atom easier to gain electron to form negative ion .

  4. Chapter periodic table 3) Ne does not form compound with other atoms. Explain ( 2 marks) • Atom neon has stable electron arrangement of octet. • So atom neon does not need to gain, release or share electron with other atoms • Note: Only helium has duplet electron arrangement. Other atoms in group 18, has octet electon arrangement.

  5. 4) Atom X and atom Y are situated in the same period in the periodic table. Which atom is smaller in size? . Explain( 3 marks) • size of atom Y is smaller than X • number of proton of atom Y is more than atom X • nuclei attraction in atom Y is stronger

  6. 5) Sodium and lithium has similar chemical properties. Why?( 1 marks) • sodium atom and lithium atom has the same number of valence electrons, that is one

  7. 6) Fluorine and chlorine has similar chemical properties. Why?( 1 marks) • Fluorine atom and chlorine atom has the same number of valence electrons, that is seven

  8. Chapter on chemical bonds 7) Covalent compound has low melting point and boiling point. Why?( 2 marks) • The molecules are held by weak Van de Waals forces of attractions • So less heat energy is needed to overcome the attraction 8) Ionic compound has high melting point and boiling point. Why?( 2 marks) • The ions are held by strong electrostatic forces of attractions • So more heat energy is needed to overcome the attraction

  9. Chapter on chemical bonds 9) Covalent compound does not conduct electricity in all states, but ionic compound conduct electricity in molten and aqueous state . Why?( 2 marks) • Covalent compound consists of molecules and do not contain mobile ions. • But ionic compound in molten or aqueous state contain mobile ions to conduct electricity

  10. Chapter electrochemistry 10)Electrolysis of copper(II) sulphate solution using carbon electrodes. After one hour, the intensity of blue solution decrease. Explain( 1 marks) Because the concentration of copper(II) ion decreases OR Copper(II) ions are discharge to copper atoms ( Cu 2+ + 2e  Cu)

  11. 11)Electrolysis of copper(II) sulphate solution using copper electrodes . After one hour, the intensity of blue solution remain the same. Explain( 3 marks) • Concentration of copper(II) ions remain constant Rate of discharge of copper(II) ions at cathode ( Cu2+ + 2e  Cu) Is equal to Rate of ionization of copper atoms to copper(II) ions at anode ( Cu  Cu2+ + 2e)

  12. 12) Explain the observation after 30 minutes experiment is carried out. ( 6 marks) Copper(II) ions are discharged to form copper atoms ( Cu2+ + 2e  Cu) a) Brown deposit on copper metal b) Zinc electrode becomes thinner Zinc atom ionize to form zinc ions ( Zn  Zn2+ + 2e) c) Intensity of blue solution decrease Concentration of copper(II) ions decrease / Copper(II) ions are discharged to form copper atoms ( Cu2+ + 2e  Cu)

  13. 13) Diagram above shows neutralization reaction between sulphuric acid and NaOH. In this experiment, 20 cm3 of sulphuric acid needed to neutralize exactly the 25 cm3 alkali , NaOH • The experiment is repeated with 1.0 moldm-3 hydrochloric acid to replace the sulphuric acid . Predict the volume of hydrochloric acid needed to neutralise 25.0 cm3 the sodium hydroxide, NaOH solution. Give your reason.( 2 marks) Replace with hydrochloric acid • 40 cm3 , conc of hydrogen ions in HCl is half than conc of hydrogen ions in sulphuric acid

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