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# Chem class(1mac)

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### Chem class(1mac)

1. 1. 4S6Welcome To the Chemistry World~
2. 2. Cl- Zn2+ Cl-1 ZnCl2 = 1 Zn2+ ion 2 Cl- ions
3. 3. 3.4 THE MOLE AND THE VOLUME - the molar volume of a gas is the volume occupied by one mole of the gas or particles of gas - one mole of any gas always has the same volume under the same temperature and pressure.
4. 4.  - the molar volume of any gas at STP is 22.4 dm3mol-1 ________________. -Standard Temperature and Pressure (STP) is 0°C the conditions where temperature is ______ and 1 atm pressure is ________. Example: 22.4 dm3mol-1 a. 1 mole of oxygen gas = _____________ 22.4 dm3mol-1 b. 1 mol of carbon dioxide gas= _____________ 22.4 dm3mol-1 c. 1 mol of ammonia gas = _______________
5. 5.  - the molar volume of any gas at room temperature 24 dm3mol-1 is ________________. -At room temperature is the conditions where 25°C 1 atm temperature is ______ and pressure is ________. Example: 24 dm3mol-1 a. 1 mole of oxygen gas = _____________ 24 dm3mol-1 b. 1 mol of carbon dioxide gas= _____________ 24 dm3mol-1 c. 1 mol of ammonia gas = _______________
6. 6. STP=22.4 dm3mol-1 Volume RTP =24 dm3mol-1 Number of Molar Moles volume(More Van More Vibrate)
7. 7. EXAMPLE 1: What is the volume of 2.5mol helium gas at s.t.p? 22.4 dm3mol-1Volume of gas= No. of moles x molar volume= 2.5 mol x 22.4dm3mol-1= 56 dm3
8. 8. EXAMPLE 2: What is the number of moles in 240 cm3 oxygen gas at r.t.p? 24 dm3mol-1number of moles= volume of the gas Molar Volume= 0.24 dm3 24 dm3 mol-1= 0.01 mol 240 cm3 = 240 1000 = 0.24dm3
9. 9. EXAMPLE 3: Calculate the volume of the gases at room temperature and pressure that contains:a. 12 x 1022 C3H8 moleculesb. 2.4 x 1023 N2 molecules(1 mole of gas occupies 24 dm3at room temperature; NA= 6 x 1023 )
10. 10. CALCULATE THE VOLUME OF THE GASES AT ROOMTEMPERATURE AND PRESSURE THAT CONTAINS: 24 dm3mol-1A. 12 X 1022 C3H8 MOLECULESNumber of mole= 12 x 1022 6 x 1023= 0.2 molVolume= 0.2 x 24dm3= 4.8dm3
11. 11. CALCULATE THE VOLUME OF THE GASES AT ROOMTEMPERATURE AND PRESSURE THAT CONTAINS:B. 2.4 X 1023 N2 MOLECULESNumber of mole= 2.4 x 1023 6 x 1023= 0.4 mol Volume= 0.4 x 24dm3= 9.6 dm3
12. 12. ANSWERa) Number of mole = 12 x 1022 6 x 1023 = 0.2 mol1 mole of gas occupies 24dm3 at room temperature0.2 mole gas occupies 0.2 x 24dm3 = 4.8dm3b) Number of mole = 2.4 x 1023 6 x 1023 = 0.4 mol1 mole of gas occupies 24dm3 at room temperature0.4 mole gas occupies 0.4 x 24dm3 = 9.6 dm3
13. 13. EXAMPLE 4: a. What is the volume of 4g of oxygen gas in cm3 at STP? b. A sample of hydrogen gas contains molecules. Find the mass of the hydrogen gas. c. How many molecules are there in 7840 cm3 of ammonia gas in STP?
14. 14. WHAT IS THE VOLUME OF 4G OF OXYGEN GAS INCM3 AT STP?Number of mole= 4g 32gmol-1= 0.125 mol Volume= 0.125mol x 22.4 dm3 mol-1= 2.8dm3= 2800cm3
15. 15. A SAMPLE OF HYDROGEN GAS CONTAINS 1.806 X1023 MOLECULES. FIND THE MASS OF THEHYDROGEN GAS.Number of mole = 1.806 x 1023 6.02 x 1023 = 0.3 molmass of the hygrogen gas= 0.3 mol x 2gmol-1= 0.6g
16. 16. C.HOW MANY MOLECULES ARE THERE IN 7840CM3 OF AMMONIA GAS IN STP?7840cm3 = 7840÷ 1000= 7.84dm3Number of mole= 7.84dm3 22.4dm3mol-1= 0.35 mol Number of molecules= 0.35 mol x 6.02 x 1023= 2.107 x 1023 molecules
17. 17. a. Number of mole = 4g 32gmol-1 = 0.125 mol 0.125 mol of oxygen gas = o.125mol x 22.4 dm3mol-1 = 2.8dm3 = 2800cm3b) Number of mole = 1.806 x 1023 6.02 x 1023 = 0.3 mol 0.3 mol of hydrogen gas = 0.3 mol x 2gmol-1 = 0.6gc) 7840cm3 = 7840÷ 1000= 7.84dm3 Number of mole = 7.84dm3 22.4dm3mol-1 = 0.35 mol0.35 mol of ammonia gas = 0.35 mol x 6.02 x 1023 = 2.107 x 1023 molecules
18. 18. CHEMISTRY FORMULAE A CHEMICAL FORMULA is a representation of a Chemical substance _______________________ using symbols of the elements and subscript numbers to show the numbers of each type of atoms that are present in the substance.
19. 19. The overall chargeCHEMICAL FORMULA must be 0sodium chloride + - Na Cl NaCl
20. 20. The overall chargeCALSIUM FLOURIDE must be 0 2+ - Ca F CaF2
21. 21. The overall chargeMAGNESIUM CHLORIDE must be 0 2+ - Mg Cl MgCl2
22. 22. The overall chargeMAGNESIUM SULPHATE must be 0 Put in 2+ 2- simplest ratio Mg SO4 MgSO4
23. 23. The overall chargeALUMINIUM OXIDE must be 0 3+ 2- Al O Al2O3
24. 24. The overall chargeAMMONIUM SULFATE must be 0 + 2- NH4 SO4 (NH4 )2 SO4
25. 25. The overall chargeLEAD (II) HYDROXIDE must be 0 2+ - Pb OH Pb ( OH )2
26. 26. NAMING Naming of cations of fixed oxidation state Naming of cations of variable oxidation state Naming of anions – ide ate- (Fat) – more oxygen Ite – (Fit )– less oxygen
27. 27. CATIONS OF FIXED OXIDATION STATE Sodium ion (Na+) Eg: Sodium chloride (NaCl) Magnesium ion (Mg2+) Eg: Magnesium oxide (MgO) Calcium ion(Ca2+) Eg: Calcium carbonate (CaCO3) Aluminium ion (Al3+) Eg: Aluminium oxide (Al2O3)
28. 28. CATIONS OF VARIABLE OXIDATION STATE Iron(ll) ion (Fe2+)Eg: Iron(II) chloride (FeCl2) Iron(lll) ion (Fe3+)Eg: Iron(III) chloride (FeCl3) Copper (l) ion (Cu+)Eg: Copper(I) oxide (Cu2O) Copper (ll) ion(Cu2+)Eg: Copper(II) oxide (CuO) Lead(ll) ion(Pb2+)Eg: Lead(II) oxide (PbO) Lead(lV) ion (Pb4+)Eg: Lead(IV) oxide (PbO2)
29. 29. PREFIXES Mono 1 Di 2 Tri 3 Tetra 4 Penta 5
30. 30. NAMING OF CHEMICAL COMPOUNDFe2O3 O 2- Fe3+ Iron(III) oxide
31. 31. NAMING OF CHEMICAL COMPOUND PbO2 O 2- Pb4+ Lead (IV) oxide
32. 32. NAMING OF CHEMICAL COMPOUND ZnO Zinc oxide NaNO3 Sodium nitrate SO3 Sulphur trioxide Fe2O3 Iron(III) oxide CCl4 Carbon tetrachloride N2O4 Dinitro tetraoxide CaCl2 Calcium chloride PbI2 Lead(ll) Iodide PbO2 Lead (IV) oxide K2CO3 Potassium carbonate PCl5 Phosphorus pentachloride
33. 33. PASS UP ON MONDAY(4/3/2013) Past year paper Spotlight access 3.4 Spotlight access 3.5, question8 and 9