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UNIT 2 NATURE’S CHEMISTRY.

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Presentation on theme: "UNIT 2 NATURE’S CHEMISTRY."— Presentation transcript:

1 UNIT 2 NATURE’S CHEMISTRY

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5 hydroxyl

6 The shortened structural formula
for an organic compound is CH3CH(CH3)CH(OH)C(CH3)3 Which of the following is another way of representing this structure?

7 A Cn Hn+4 N B Cn H2n+3 N C Cn H3n+2 N D Cn H4n+1 N CH5N C2H7N C3H9N
The first four members of the amine homologous series are: What is the general formula for this homologous series? A Cn Hn+4 N B Cn H2n+3 N C Cn H3n+2 N D Cn H4n+1 N CH5N C2H7N C3H9N C4H11N

8 C4H6 C5H8 C6H10

9 What is meant by the term isomer?
ISOMERS What is meant by the term isomer? SAME MOLECULAR FORMULA BUT DIFFERENT STRUCTURAL FORMULAE

10 ADDITION

11 OH H H OH

12

13 BROMINE DECOLOURISED Both compounds are unsaturated.
Describe the chemical test for unsaturation. BROMINE DECOLOURISED

14 Name the functional group
CARBOXYL CARBOXYL HYDROXYL

15 C3H8S H8C3S SH8C3 SAME MOLECULAR FORMULA BUT
DIFFERENT STRUCTURAL FORMULAE

16 EXOTHERMIC Eh = cmΔT 4·18 x 0·1 x = 3·344 kJ

17 6·4 g of methanol (CH3OH) raises the temperature of 100cm3 of water from 21 °C to 40 °C
Calculate the energy, in kJ, absorbed by the water. Eh = cmΔT 4·18 x 0·1 x = 7·942 kJ

18 Eh = cmΔT 4·18 x 0·2 x = 33·44

19 In one experiment the burning of ethanol
resulted in the temperature of 400cm3 of water rising from 14oC to 30oC. Calculate the energy absorbed by the water. Eh = cmΔT 4·18 x 0·4 x = 26·75 kJ

20 In one experiment the burning of methanol resulted in the temperature of 100 cm3 of water rising from 19 oC to 59 oC. Calculate the energy released. Eh = cmΔT 4·18 x 0·1 x = 16·72 kJ

21 1 mole 1 mole 12 + (1x4) 12 + (16x2) 16g 44g 11g 44 x 0·25 = 4g
Calculate the mass of carbon dioxide produced when 4g of methane burns completely. CH O CO H2O 1 mole mole 12 + (1x4) (16x2) 16g g 11g 4g 44 x 0·25 = 4/16 = 0·25

22 1 mole 1 mole 12 + (1x4) 12 + (16x2) 16g 44g 1g 44/16 = 2·75g 4g
Calculate the mass of carbon dioxide produced when 4g of methane burns completely. CH O CO H2O 1 mole mole 12 + (1x4) (16x2) 16g g 1g 44/16 = 2·75g 4g 4 x 2·75g = 11g


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