Grade 11 Physical Science– June 2015 MEMO
Ex: GR
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BISHOPS
GRADE 11 - PHYSICAL SCIENCE MEMO
June2015
EXAMINER: GRMARKS:150
MODERATOR:PWTIME:3 HOURS
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SECTION A
QUESTION 1ONE – WORD / TERMITEMS
1.1Dative(1)
1.2Valence shell electron pair repulsion theory(1)
1.3Resultant(1)
1.4London forces(1)
1.5Water of crystallisation(1)
[5]
QUESTION 2MULTIPLE-CHOICE QUESTIONS
2.1 D2.2B2.3C2.4C2.5C
2.6A2.7D2.8A2.9A2.10C
[2 x 10 = 20]
SECTION B
QUESTION 3
3.1
3.1.1 (2)
3.1.2Easterly comp = 360 cos 30° = 311,77 N
Southerly comp = 360 sin 30° = 180,00 N(4)
3.1.3x-axis = 311,77 + 550 = 861,77 N
y-axis = (250 + 180) – 800 = -370 ie 370 N(3)
3.1.4R =;(pythag) tan ϴ = 861,77/370 ϴ = 66,76° (trig)
R = 937,84 N on a bearing of 66,76°(4)
3.2
3.2.1decrease
3.2.2increase
3.2.3decrease
3.2.4increase(4)
[17]
QUESTION 4
4.1
4.1.1Whenonebodyexertsaforceonasecondbody, thesecondbodyexertsaforceofequalmagnitudeintheoppositedirectiononthefirst body. (2)
4.1.2Force of block on surface (if force of surface on block)(2)
4.1.3µs = Fmax/FN 0,48 =Fcos22°/(11x9,8 – Fsin22°)(Fmax is parallel comp)
0,48(107,8 – 0,375F) = 0,927F
51,74 – 0,18F = 0,927F
F = 51,74/1,107 = 46,74 N(5)
4.2
4.2.1 (3)
4.2.2Let up be positive
Fnet = T + W
150 x 2 = T + (150 x (-9,8)) max 2/4 if no direction, or logic showing understanding of direction
T = 1470 + 300
T = 1770 N up(4)
4.2.3
(a)Inertia of the block(1)
(b)Newtons first law of motion.(1)
(c)Continues moving upwards but slows down due to downward acceleration from gravity to a stop then accelerates downwards till it hits the ground (3)
4.3
4.3.1Eachbodyintheuniverseattractsevery otherbodywithaforcethatisdirectlyproportionaltothe productoftheirmassesand inverselyproportionaltothesquareof thedistancebetween their centres. (2)
4.3.2F = Gm1m2/r2
5000 = 6,67x10-11 x 615 x 5,97x1024/r2
r = 7x106 m
height above the earth = 628,46 km (accept 630 km)(5)
[28]
QUESTION 5
5.1
5.1.1In a collision between two objects linear momentum is conserved on both magnitude and direction. (2)
5.1.2Total momentum before collision = Total momentum after collision
Let right be positive
(mAi + mBi)vABi = mAfvAf + mBfvBf
(1 + 2) x 3 = 1 vAf + 2 x 4,7
vAf = 9 – 9,4 = -0,4 m.s-1ie 0,4 m.s-1 to the left(4)
5.1.3F∆t = m∆v F = 1(-0,4 -3)/0,2 = -17 N ie B to A was 17 N left therefore A on B was 17 N right.
ORF∆t = m∆v F = 2(4.7 -3)/0,2 = 17 N ie A to B was 17 N right.(4)
5.2Let forward direction be positive
∆p = m∆v = 0,057(-32 – 25) = -3,25 kg.m.s-1ie 3,25 kg.m.s-1 in opposite direction to original (4)
[14]
QUESTION 6
6.1
6.1.1NaSO
36,5/2325,4/3238,1/16
= 1,59mol= 0,79 mol= 2,38 mol
1,59/0,790,79/0,792,38/0,79
= 2,01= 1= 3,01
therefore Na2SO3(4)
6.1.2c = m/MxVol
0,2= m/(46+32+48) x 0,25 carry error from 6.1.1
m = 6,3 g(4)
6.2
6.2.1n (H3PO4) = m/M = 49000/(3+31+64) = 500 moles
therefore 125 moles Ca3(PO4)2 needed.
m (Ca3(PO4)2) = nM = 125 x (120+2(31+64)) = 38750 g = 38,75 kg
Therefore 3 x 125 molesCa(H2PO4)2 expected.
ie m = 375 x (40+2(2+31+64)) = 87,75 kg(max 3/6 if incorrect limiting agent calculated – gives 200 mol Ca3(PO4)2 and a mass of 140,40 kg Ca(H2PO4)2) (6)
6.2.280/87,75 x 100 = 91,17%(carry error from 6.2.1 56,98%)(2)
[16]
QUESTION 7
7.1
7.1.1 (2)
7.1.2CO2 is non-polar. There are polar bonds but due to there being no lone pairs, the bonding pairs distribute themselves symmetrically around the carbon giving a linear molecule, which results in the vector nature of the bonds cancelling each other out. (3)
7.1.3109,5° and tetrahedral. The H’s distribute themselves symmetrically around the C in three dimensions, hence cannot be a bond angle of 90°. (3)
7.2
7.2.1n = m/M = 30/100 x 32x1015/(12+32) = 2,18 x 1014 moles(4)
7.2.22,18 x 1014 moles water reacts
m = nM= 2,18 x 1014x 18 = 3,93 x 1015 g = 3,93 x 1012 kg water
V = m/D = 3,93 x 1012/1029 = 3,82 x 109 m3 water(4)
7.2.3
(3)
7.3H is 2,1 and S is 2,5. Therefore there is a difference in enegativitywhich results in polar bonds. In the outer shell of S there are two lone pairs of electrons and the interaction of these with the bonding pairs in an asymmetrical molecule which is therefore polar. (4)
[23]
QUESTION 8
8.1
8.1.1Fluorine is the smallest atom in Group 7 and has the highest electronegativity. The H is therefore attracted more strongly to F than it is to any of the other elements in that group. The dipole that is formed is therefore way stronger than the dipoles of the other group 7 hydrides and so it is more difficult to melt HF because the forces holding the molecules together are so strong – due to hydrogen bonding. (4)
8.1.2Van der Waals forces increase with increasing molecular mass therefore melting points increase going down the group. (2)
8.1.3Dipole dipole(1)
8.2
8.2.1C14H30(1)
8.2.2CH3Br(1)
8.2.3He(1)
8.2.4CH3CH2OH(1)
8.3
8.3.1Bond length of H to H is shorter, therefore more difficult to separate atoms. OR Overlapping pairs of electrons of H2 are closer to nuclei than the pair from Cl2, therefore more difficult to separate. (2)
8.3.2The more bonds there are, the more difficult it is to separate the atoms, therefore bond length is shorter. (2)
8.3.3
(a)n = m/M =92/(24+6+16) = 2 moles(3)
(b)energy taken in = 1xC-C + 1xC-O + 5xC-H + 1xO-H + 3xO=O
= 348 + 358 + 5x 415 + 460 + 3x498
= 4735 kJ.mol-1 or kJ
energy given out = 4xC=O + 6xO-H
= 4 x 800 + 6 x 460
= 5960 kJ.mol-1or kJ
Net energy output = 4735 – 5960 = -1225 kJ.mol-1or kJ
Therefore 2450 kJ given out (x2)(6)
(c)2 moles ethanol produces 4 moles CO2 (mole ratio of 1:2)
V = nVo= 4 x 22,4 = 89,6 dm3(3)
[27]
TOTAL 150
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