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1. B
2. D
Two long, straight, parallel conductors are 10 cm apart (see figure below).
Conductor A has a current i_A=2.0 A flowing out of the page, and conductor B has a current i_B=3.0 A flowing in the opposite direction (into the page). [10 marks]


Calculate the intensity of the magnetic field at points P1, P2 and P3, where:
 (P_1 A) ̅=2cm, (P_2 A) ̅=4cm and (P_3 B) ̅=3cm.
Consider the cross section of a current carrying wire as shown below. The current is carried by the 2 gray sections (0<r<a,and  b<r<c).  The current, i = 10 mA, is in the direction out of the page, half of it flowing in the inner wire and half in the outer shell.
a)   Assuming that the 3 sections have the same thickness, i.e. (0a) ̅=(ab) ̅=(bc) ̅=1cm, calculate the magnetic field for 0<r<a,a<r<b,b<r<c, and r>c.  
(Hint: think Ampere’s Law)     [6 marks]
b)(2/10) Sketch the variation of B vs. r in the diagram below. [4 marks]
C. Deposit contract typically implies a ‘first come, first served’ principle.
D. stomach.
D. introspection
5. −2, 220 kJ · mol−1

ΔT = Tf − Ti = 77.96 ◦C − 24.90 ◦C = 53.06 ◦C = 53.06 K
m = 1 L ·(1000 mL/L)·(1.00 g/mL) = 1000 g
n = 4.409 g propane ·(1mol/44.09 g) = 0.1 mol propane
−ΔHrxn = ΔHcal = mcΔT = 1000 g · 4.184 · 53.06 K = 220 kJ
-220 kJ / 0.1 mol = −2, 220 kJ · mol−1
2. +249 kJ

ΔH = ΣBEreac − ΣBEprod = 1868 kJ/mol − 1370 kJ/mol = 498 kJ/mol
The reaction forms 2 moles of H2, so the energy to form one mole of H2 from one mole of
water vapor is half as much: 249 kJ/mol.
1. lead

Lead has the lowest specific heat, and thus requires the least heat in order to have an
increase in temperature. Its temperature will thus increase the most for a given amount of
heat.
3. +307 kJ/mol

ΔHrxn = 1/2(498.4) + -1(142.7) + -1(200.0) = +306.5

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