Answer:
Option D = 0.2 Kj
Explanation:
Given data:
Mass of diethyl ether = 1.0 g
Hvap = 15.7 Kj / mol
Heat absorbed = ?
Solution:
Q = mass × Hvap / molar mass
Q = 1.0 g × 15.7 Kj / mol / 74.12 g/mol
Q = 15.7 Kj / 74.12
Q = 0.212 KJ
First of all the ethylamine is base so will be react with water to take the proton from the water (H⁺) , because water is amphoteric will react with bases as acid and acids as a base, so the water in this case will react as an acid and will gives the proton to the base.
In the picture you may see the chemical equation and the structure of the products.
0.300 M IKI represents the
concentration which is in molarity of a potassium iodide solution. This means
that for every liter of solution there are 0.300 moles of potassium iodide. Knowing
that molarity is a ratio of solute to solution.
By using a conversion factor:
100 ml x (1L / 1000 mL) x (0.300
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g
Therefore, in the first
conversion by simply converting the unit of volume to liter, Molarity is in L
where the volume is in liters. The next step is converted in moles from volume
by using molarity as a conversion factor which is similar to how density can be
used to convert between volume and mass. After converting to moles it is simply
used as molar mass of Kl which is obtained from periodic table to convert from
mole to grams.
In order to get the grams of IKI
to create a 100 mL solution of 0.600 M IKI, use the same formula as above:
100 ml x (1L / 1000 mL) x (0.600
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g
<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.
******************************
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).
*************************************
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>
the first law says that the change in internal energy of a system is given by:
δ<span>E = δq + δw</span>
where δ<span>E is the i change in internal energy, </span>
<span>δq is the amount of thermal energy added to the system from the surroundings </span>
<span>δw is the l work done *on* the system *by* the surroundings. </span>
<span>For a system only undergoing expansion work,
δw = -p</span>δ<span>V, so: </span>
δE = δq - p δ<span>V </span>
when δV = 0, then δe=δq
