Stability Essays

Security Essays Security Essay Security Essay Stable. Solid oxidizer contact with ignitable material may cause fire. Contrary with burnable materials, and solid lessening agents.ToxicologyHarmful whenever gulped. May cause conceptive disorders.AimThe point of this test is to decide the crystallization temperature of the arrangement potassium nitrate at various focuses and use data to discover the standard enthalpy of potassium nitrate.Equipment and reagents* Boiling tubes* Dark card* Bunsen burner* Thermometer* Weighing scale* Burette (50cm3)* Clamp* Stand* Potassium nitrate* Deionised water/refined waterSafety* Wear goggles for eye insurance at all times.* Laboratory coats must be worn at all times.* Wear gloves to stay away from substance contact to skin, potassium nitrate.* Long hair was tied back when Bunsen burners were used.Procedure1. 10g of potassium nitrate were weighed out and put into a bubbling cylinder, and afterward the specific mass was noted in an outcome table.2. Precisely 8.0cm3 of deionised water was added to the bubbling cylinder containing the potassium nitrate. This was finished by utilizing a burette.3. The cylinder was then warmed delicately until the precious stones were broken down and afterward the warmth source was removed.4. The cylinders were permitted to cool for security reasons. The temperature at which the gems were first showed up was noted. A dim card was utilized for this reason. The outcomes were recorded.5. A further 25cm3 of refined water was included and stages 3-4 were repeated.6. A further 25cm3 of refined water was included and stages 3-4 were rehashed again.7. A further 25cm3 of refined water was included and stages 3-4 were rehashed again.In my own sentiment I figure the above methodology will be an appropriate strategy as it won't require some investment to perform and furthermore it is very easy to complete. Be that as it may, it could be somewhat loose as we need to watch and decide when the gems of potassium nitrate structures and changes, at that point att empt to peruse the temperature of the thermometer. This trouble could bring about a slight incorrect estimations as significant seconds are squandered between observing the precious stones transforming and perusing the estimations on the thermometer, which implies that the temperature would have expanded further from when gems initially shaped as the Bunsen burner would at present be warming the bubbling tube.Alternative methods1. contrast my outcomes and different class understudies or with txt books results:AdvantagesdisadvantagesThis would be a suitable method to check if my outcomes are levelheaded. As though I had very various outcomes to various different understudies, at that point It would be improbable that every one of them would be wrong.It is difficult to be 100% exact in estimating temperature and reagents, in this way the outcomes will contrast marginally between students.2. Play out the test a few times for every volume of water and take an average.Advantagesdisadvant agesThis technique would be much more precise.It would take too long to even think about carrying out3. Conjecture the crystallization temperature.AdvantagesdisadvantagesFor this technique there wouldnt be any counts other than for the enthalpy, consequently this strategy would be straightforward.This would be very inaccurate.I think my technique didnt should be changed in any capacity to adjust to the test on the grounds that the technique I utilized went easily.ResultsVolume of watercnR.In cnCrystallisation temperatureReciprocal of outright temperature 1/T(cm3)(mol dm3)(R = 8.31 Jmol-1 K-1)( 0C) T(K)(K-1)8.012.3620.9062 3352.98 x 10-310.89.8919.0456 3293.04 x 10-312.08.2417.5347 3203.13 x 10-314.07.0616.2442 3153.17 x 10-3Table of qualities for the line of best fit.R.ln SReciprocal of total temperature 1/T(R = 8.31 J mol-1 K-1)(K-1)Y 1 and X121.350.00290Y 2 and X215.000.003255The estimation of S (the capture) was noted from these sets of values.If Y 1 = mX 1 + c and Y 2 = mX2 + cT hen (Y 1 c)/X1 =( Y 2 c )/X 2X2Y1 X2c = X1y2 X1cc = (X2Y1 X1Y2)/(X2 X1)The inclination, m = (Y1 Y2)/(X1 X2)The articulation angle = H was utilized to discover H, the enthalpy of arrangement of potassium nitrate.Intercept = S = 73.2 J mol-1 K-1Gradient = H = 17.9 K j mol-1By looking at the qualities I got from my outcomes to those of the line of best fit I had the option to appraise the SD of the qualities I had determined for S and H the accompanying table shows the results.R.in cn1/T esteems (a)1/T esteems determined from best fit line (b)Difference for every 1/T esteem, as a small amount of the best fit value.(a-b)/bDifference2(R = 8.21 Jmol-1 K-1)K-1K-120.900.002980.0029250.01883.5344 x10-419.040.003040.003033.300 x10-31.089 x10-517.530.003130.0031135.461 x10-32.9822521 x10-516.240.003190.00319-6.2696 x10-33.93 x10-5CalculationsMolar mass (KNO) =K= 39N= 14O= 3 X16= 39 + 14 + 48= 101.11gNumber of moles :Amount = mass = 10.00 = 0.0989 mol dm-3Molar mass 101.11Concentration:Concentr ation = amountVolume(a) 8 cm3 : C = 0.0989 = 12.36 mol dm-38/1000(b) 10 cm3 : C = 0.0989 = 9.89 mol dm-310/1000(c) 12 cm3 : C = 0.0989 = 8.24 mol dm-312/1000(d) 14 cm3 : C = 0.0989 = 7.06 mol dm-314/1000R.In cn:(a) 12.36 mol dm-3 : ln (12.36) = 2.514= R X ln (concentration)=8.31 x 2.51=20.90(b) 9.89 mol dm-3 : ln (9.89) = 2.2915= R X ln (concentration)=8.31 x 2.29=19.04(c) 8.24 mol dm-3 : ln (8.24) = 2.1090= R X ln (concentration)=8.31 x 2.11=17.53(d) 7.06 mol dm-3 : ln (7.06) = 1.954= R X ln (concentration)=8.31 x 1.95=16.24Crystallisation temperature in KelvinK = (0C ) + 273(a) 620C : 62+ 273 = 335K(b) 560C : 56 + 273 = 329K(c) 470C : 47+ 273 = 320K(d) 420C : 42+ 273 = 315KReciprocal 1/T(a) 20.90 : 1/335 = 2.98 X 10-3(b) 19.04 : 1/329 = 3.04 X 10-3(c) 17.53 : 1/320 = 313 X 10-3(d) 16.24 : 1/315 = 317 X 10-3GradientY = mx + cm = (y2 y1)(x2 x1)m = 15.00 21.350.003255 0.00290m = - 6.35 .3.55 x 10-4m = - 17,887.32394 Kj mol= - 17.89 Kj molH = 17.89 Kj molS = (0.003255 x 21.35 0.00290 x 15.00)(0.003255 0.00290)S = 0.06949425 0.04353.55 x 10-4S = 73.223= 73.2Differences for every 1/T valueD = a-b/b(1) D = 0.00298 0.002925 = 0.01880.02925(2) D = 0.00304 0.00303 = 3.300 x 10-30.00303(3) D = 0.00313 0.003113 = 5.461 x 10-30.003113(4) D = 0.00317 0.00319 = - 6.2696 x 10-30.00319Differences2(1) (0.0188)2 = 3.5344 x 10-4(2) (3.300 x 10-3)2 = 1.089 x 10-5(3) (5.461 x 10-3)2 = 2.9822521 x 10-5(4) (- 6.2696 x 10-3)2 = 3.93 x 10-5? (contrast) 2 = 4.33 x 10-4Standard deviationSD = V? (- X) 2 = V 4.33 X 10-4 = 0.01n 4S : % mistake = 73.2 + 0.01% blunder = 0.01 x 100 = 0.01%H : % mistake = 17.9 + 0.01% mistake = 0.01 x100 = 0.06%17.9DiscussionOn the entire I accept that my investigation went very well since everything was directed by plan. I talked about my test and my outcomes with my talk and were supposed to be sensibly precise. I additionally accept that any mistake in my outcomes was because of freshness and human blunder, when estimations of weight, temperature and volum e were made. My conclusive outcomes were acquired utilizing results from a line of best fit put on my diagram this could have likewise caused a slight of incorrectness in my outcomes. These outcomes could have been loose as incredibly little numbers were utilized which were hard to plot accurately on the diagram and I had assessed where the focuses were to be set. Likewise I needed to evaluate where to put my line of best fit. Moreover the error could have been because of the thermometer I utilized, as there was a little hole between observing the precious stone and recording the temperature. Hence I couldnt be as exact as I would have been gotten a kick out of the chance to be as in light of the fact that you must be snappy in perusing the temperature which was continually evolving. Besides, it was very hard to recognize little particles that had framed and air rises in the arrangement. Furthermore, it was difficult to really say what measure of particles framed, was the perfect ad d up to take the arrangement off the warmth source and note the results.In my own conclusion I feel that my strategy was a reasonable one since it gave me exact outcomes and didnt occupy an excessive amount of time to proceed.ImprovementsThe upgrades that could have been made to make my investigation progressively precise are to be increasingly cautious in estimating reagents and furthermore to utilize a progressively exact thermometer. What's more, a bigger scaled diagram could have been utilized as results would have been plotted all the more accurately.Confidently in future I would be increasingly knowledgeable about doing the test and would be progressively exact at detecting the gems when they form.Conclusion and % errorThermometer 0.10620 : % blunder = 0.1 x 100 = 0.2%62560 : % mistake = 0.1 x 100 = 0.2%56470 : % mistake = 0.1 x 100 = 0.2%47420 : % mistake = 0.1 x 100 = 0.2%42Average % mistake = 0.2%Burette 0.1cm38cm3 : % mistake = 0.1 x 100 = 1.3%810cm3 : % mistake = 0.1 x 10 0 = 1%1012cm3 : % blunder = 0.1 x 100 = 0.8%1214cm3 : % mistake = 0.1 x 100 = 0.7%14Average % mistake = 1.3 + 1+ 0.8 +0.7 = 0.95%4Weighing scales + 0.01g% blunder = 0.01 x 100= 0.1% Show see as it were