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Wednesday, April 9, 2008.
1) IRP - going thru tomorrow, 10/4
2) Physics vectors wksht - going thru tomorrow, 10/4
3) Maths page 2(6), supposingy do ALL. - *not sure when going thru*

ALL APPLIES ONLY TO THOSE WHO HAVEN COMPLETE THEIR WORK.

PLEASE ALSO REMEMBER to do RESPONSE FOR GP

**********

answers for remedial worksheet for 7/4

(page 45, Q5)

1. H2O

H = 1.0 2H 2 x 1.0 = 2.0
O = 16.0 1O 1 x 16.0 = 16.0 :. Mr = 18.0


2. CO2

C = 12.0 1C 1 x 12.0 = 12.0
O = 16.0 2O 2 x 16.0 = 32.0 :. Mr = 44.0


3. Na2CO3

Na = 23.0 2Na 2 x 23.0 = 46.0
C = 12.0 1C 1 x 12.0 = 12.0
O = 16.0 3O 3 x 16.0 = 48.0 :. Mr = 106.0


4. (NH4)2CO3

N = 14.0 2N 2 x 14.0 = 28.0
H = 1.0 8H 8 x 1.0 = 8.0
C = 12.0 1C 1 x 12.0 = 12.0
O = 16.0 3O 3 x 16.0 = 48.0 :. Mr = 96.0


5. Fe(NO3)3

Fe = 55.8 1Fe 1 x 55.8 = 55.8
N = 14.0 3N 3 x 14.0 = 42.0
O = 16.0 9O 9 x 16.0 = 144.0 :. Mr = 241.8


6. CuSO4.5H2O

Cu = 63.5 1Cu 1 x 63.5 = 63.5
S = 32.1 1S 1 x 32.1 = 32.1
O = 16.0 9O 9 x 16.0 = 144.0
H = 1.0 10H 10 x 1.0 = 10.0 :. Mr = 249.6


7. MgSO4.7H2O

Mg = 24.3 1Mg 1 x 24.3 = 24.3
S = 32.1 1S 1 x 32.1 = 32.1
O = 16.0 11O 11 x 16.0 = 176.0
H = 1.0 14H 14 x 1.0 = 14.0 :. Mr = 246.4

NOTE : ALL ANSWERS TO 1d.p!



(page 49, Q5)

(a) No. of mole of 6g of C

= 6 / 12.0
= 0.500 mol.


(b) No. of mole of 8g of S

= 8 / 32.1
= 0.249 mol.


(c) No. of mole of 112g of Fe

= 112 / 55.8
= 2.01 mol.


(d) Mr of MgO = 24.3 + 16.0 = 40.3

No. of mole of 10g of MgO

= 10 / 40.3
= 0.248 mol.


(e) Mr of Fe2O3 = (2 x 55.8) + (3 x 16.0) = 159.6

No. of mole of 16g of Fe2O3

= 16 / 159.6
= 0.100 mol.

NOTE : ALL ANSWERS TO 3s.f, write 'No. of mole' NOT 'No. of moles', write statement!



(page 50, Q7)

(a) mass of 3 moles of Na

= 3 x 23.0
= 69.0 g.


(a) Mr of H2SO4 = (2 x 1.0) + 32.1 + (4 x 16.0) = 98.1

mass of 1.5 moles of H2SO4

= 1.5 x 98.1
= 147.15
= 147 g.

NOTE : ALL ANSWERS TO 3s.f, include unit of mass in answer (i.e. grams), write statement!



(page 50, Q8)
[(a) is example in paper!]

(b) No. of mole of O = (6.02 x 10^23) / (6.02 x 10^23) = 1 mol

:. mass of O = 1 x 16.0 = 16.0 g.


(c) No. of mole of H2O = (3.01 x 10^23) / (6.02 x 10^23) = 0.5000 mol

:. mass of H2O = 0.5000 x [(2 x 1.0) + 16.0] = 9.00 g.


(d) No. of mole of CH4 = (2.01 x 10^23) / (6.02 x 10^23) = 0.3333 mol

:. mass of O = 0.3333 x [12.0 + 4(1.0)] = 5.33 g.


(e) No. of mole of CO2 = (4.01 x 10^23) / (6.02 x 10^23) = 0.6661 mol

:. mass of CO2 = 0.6661 x [12.0 + 2(16.0)] = 29.3 g.

NOTE : include units, ALL INTERMEDIATE ANSWERS TO 4s.f, ALL ANSWERS TO 3s.f!



(page 51, Q9)
[(a) is example in paper!]

*(b) No. of mole of 8g of O2(g) = 8 / 2(16.0) = 0.2500 mol

*take note O2 is equivalent to 2O atoms!*

No. of mole of O atoms = 2 x 0.2500 = 0.5000 mol

:. No. of O atoms = 0.5000 x 6.02 x 10^23 = 3.01 x 10^23 g


(c) No. of mole of 8g of O2(g) = 8 / 2(16.0) = 0.2500 mol

No. of molecules of O2(g) = 0.2500 x 6.02 x 10^23 = 1.51 x 10^22 g


(d) No. of mole of 35g of Cl2(g) = 35 / 2(35.5) = 0.4930 mol

No. of molecules of O2(g) = 0.4930 x 6.02 x 10^23 = 2.97 x 10^23 g


(e) No. of mole of 3g of H2O = 3 / [(2 x 1.0) + 16.0] = 0.1667 mol

No. of molecules of H2O = 0.1667 x 6.02 x 10^23 = 1.00 x 10^23 g

NOTE : ALL ANSWERS TO 3s.f, intermediate answers to 4s.f, write 'No. of mole' NOT 'No. of moles', write statement!



(page 51, Q10)

Hydrogen, H2 (4 mole) - volume = 4 x 24.0 = 96.0 dm3

Oxygen, O2 (8 mole) - volume = 8 x 24.0 = 192.0 dm3

Carbon dioxide, CO2 (0.5 mole) - volume = 0.5 x 24.0 = 12.0 dm3

Nitrogen, N2 (0.001 mole) - volume = 0.001 x 24.0 = 0.0240 dm3

NOTE : final answer to 3s.f, show working! WRITE UNITS!



(page 52, Q11)

Hydrogen, H2 (24 dm3)

no. of mole = 24.0 / 24.0 = 1 mol

Mr = 2.0

mass = 2.0 g


Oxygen, O2 (48 dm3)

no. of mole = 48.0 / 24.0 = 2 mol

Mr = 32.0

mass = 64.0 g


Carbon dioxide, CO2 (6 dm3)

no. of mole = 6.0 / 24.0 = 0.250 mol

Mr = 44.0

mass = 11.0 g


Nitrogen, N2 (48 cm3) *convert!*

no. of mole = (48 / 1000) / 24.0 = 2 x 10^-3 mol

Mr = 28.0

mass = 0.0560 g

NOTE : all answers to 3s.f, include UNITS!



(page 53, Q16)

(a) concentration = 0.5 / (250 / 1000)

= 2.00 mol/dm3


(b) no. of mole of NaOH = 2 / (23.0 + 16.0 + 1.0) = 0.05000

concentration = 0.05000 / (50 / 1000)

= 1.00 mol/dm3


(c) no. of mole of Na2SO4 = 35.5 / [2(23.0) + 32.1 + 4(16.0)] = 0.2498

concentration = 0.2498 / (500 / 1000)

= 0.500 mol/dm3

NOTE : final answerUNITS!



(page 53, Q17)

(a) no. of mole of HCl

= 0.5mol/dm3
= 1.00 mol


(b) no. of mole of NaOH

= 2.5 mol/dm3 x 4 dm3
= 10.0 mol


(c) no. of mole of H2SO4

= 2.5 mol/dm3 x 500 / 1000 dm3
= 1.25 mol


(d) no. of mole of HNO3

= 1.0 mol/dm3 x 400 / 1000 dm3
= 0.400 mol

NOTE : answers to 3s.f!



*********

sorry for untidyness and please double check if the answers above are right.(:


jojo