Soap is the sodium or potassium salt of long chain of fatty acid. Fatty acids when treated with NaOH or KOH forms Soap. This process is called as Saponification. Examples of Soap are as follow,
1. Sodium Stearate C₁₇H₃₅COONa
2. Potassium Oleate C₁₇H₃₃COOK
Reaction of Soap with MgCl₂;
When Soap is treated with MgCl₂ or CaCl₂ it forms insoluble precipitate called S.C.U.M. The reactions with MgCl₂ are as follow,
2C₁₇H₃₅COONa + MgCl₂ --------> 2C₁₇H₃₅COOMg + 2 NaCl
2C₁₇H₃₃COOK + MgCl₂ --------> 2C₁₇H₃₅COOMg + 2 KCl
These reaction are often found in hard water. And this reaction decreases the effectiveness of soap.
<span>1 trial : you have nothing to compare the result with - you don't know if it's a mistake.
2 trials : you can compare results - if very different, one may have gone wrong, but which one?
3 trials : if 2 results are close and 3rd far away, 3rd probably unreliable and can be rejected.
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First calculate the enthalpy of fusion. M, C and m,c = mass and
specific heat of calorimeter and water; n, L = mass and heat of fusion
of ice; T = temperature fall.
L = (mc+MC)T/n.
c=4.18 J/gK. I assume calorimeter was copper, so C=0.385 J/gK.
1. M = 409g, m = 45g. T = 22c, n = 14g
L = (45*4.18+409*0.385)*22/14 = 543.0 J/g.
2. M = 409g, m = 49g, T = 20c, n = 13g
L = (49*4.18+409*0.385)*20/13 = 557.4 J/g.
3. M = 409g, m = 54g, T = 20c, n = 14g
L = (54*4.18+409*0.385)*20/14 = 547.4 J/g.
(i) Estimate error in L from spread of 3 results.
Average L = 549.3 J/g.
average of squared differences (variance) = (6.236^2+8.095^2+1.859^2)/3 = 35.96
standard deviation = 5.9964
standard error = SD/(N-1) = 5.9964/2 = 3 J/g approx.
% error = 3/547 x 100% = 0.5%.
(ii) Estimate error in L from accuracy of measurements:
error in masses = +/-0.5g
error in T = +/-0.5c
For Trial 3
M = 409g, error = 0.5g
m = 463-409, error = sqrt(0.5^2+0.5^2) = 0.5*sqrt(2)
n =(516-463)-(448-409)=14, error = 0.5*sqrt(4) = 1.0g
K = (mc+MC)=383, error = sqrt[2*(0.5*4.18)^2+(0.5*0.385)^2] = 2.962
L = K*T/n
% errors are
K: 3/383 x 100% = 0.77
T: 0.5/20 x 100% = 2.5
n: 1.0/14 x 100% = 7.14
% errors in K and T are << error in n, so we can ignore them.
% error in L = same as in n = 7% x 547.4 = 40 (always round final error to 1 sig fig).
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The result is (i) L= 549 +/- 3 J/g or (ii) L = 550 +/- 40 J/g.
Both are very far above accepted figure of 334 J/g, so there is at least
one systematic error in the experiment or the calculations.
eg calorimeter may not be copper, so C is not 0.385 J/gK. (If it was
polystyrene, which absorbs/ transmits little heat, the effective value
of C would be very low, reducing L.)
Using +/- 40 is probably best (more cautious).
However, the spread in the actual results is much smaller; try to explain this discrepancy - eg
* measurements were "fiddled" to get better results; other Trials were made but only best 3 were chosen.
* measurements were more accurate than I assumed (eg masses to nearest 0.1g but rounded to 1g when written down).
Other sources of error:
L=(mc+MC)T/n is too high, so n (ice melted) may be too small, or T (temp fall) too high - why?
* it is suspicious that all final temperatures were 0c - was this
actually measured or just guessed? a higher final temp would reduce L.
* we have assumed initial and final temperature of ice was 0c, it may
actually have been colder, so less ice would melt - this could explain
small values of n
* some water might have been left in container when unmelted ice was
weighed (eg clinging to ice) - again this could explain small n;
* poor insulation - heat gained from surroundings, melting more ice,
increasing n - but this would reduce measured L below 334 J/g not
increase it.
* calorimeter still cold from last trial when next one started, not
given time to reach same temperature as water - this would reduce n.
Hope This Helps :)
</span>
Answer:
107.55 J, 88.67 cal
Explanation:
1 Joule = 0.239 Calories
450 cal x 0.239 = 107.55 J
3.5 x 106 = 371
371 J = 371 x 0.239 cal
371 J = 88.67 cal
<span>Methyl groups are electron-donating, so the more CH3 groups attached to the +ve carbon, the more stable the carbocation becomes.</span>
