Answer:
81.04°C
Explanation:
Heat loss by water = Heat gained by Aluminum
Heat loss by water;
H = MCΔT
ΔT = 100 - T2
M = 580g
c = 4.2
H = 580 * 4.2 (100 - T2)
H = 243600 - 2436T2
Heat ganed by Aluminium
H = MCΔT
ΔT = T2 - 24
M = 900g
c = 0.9
H = 900 * 0.9 (T2 - 24)
H = 810 T2 - 19440
243600 - 2436T2 = 810 T2 - 19440
243600 + 19440 = 810 T2 + 2436T2
263040 = 3246 T2
T2 = 81.04°C
Assumption;
Assume that energy diffuses throughout the pan and water so that all parts reach the same final temperature.
Answer:
Rate = k [OCl] [I]
Explanation:
OCI+r → or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.
In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.
In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.
The rate law is given as;
Rate = k [OCl] [I]
Answer:
Explanation:
Molal freezing point depression constant of butanol Kf = 8.37⁰C /m
ΔTf = Kf x m , m is no of moles of solute per kg of solvent .
mol weight of butanol = 70 g
235.1 g of butanol = 235.1 / 70 = 3.3585 moles
3.3585 moles of butanol dissolved in 4.14 kg of water .
ΔTf = 8.37 x 3.3585 / 4.14
= 6.79⁰C
Depression in freezing point = 6.79
freezing point of solution = - 6.79⁰C .
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