Stability Essays

Soundness Essays Soundness Essay Soundness Essay Stable. Solid oxidizer contact with ignitable material may cause fire. Contradictory with flammable materials, and solid diminishing agents.ToxicologyHarmful whenever gulped. May cause conceptive disorders.AimThe point of this analysis 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 assurance at all times.* Laboratory coats must be worn at all times.* Wear gloves to maintain a strategic distance from concoction contact to skin, potassium nitrate.* Long hair was tied back when Bunsen burners were used.Procedure1. 10g of potassium nitrate were weighed out and set 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 tenderly 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 precious stones 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 supposition I figure the above methodology will be a reasonable technique as it won't require some investment to perform and furthermore it is very easy to do. In any case, it could be somewhat loose as we need to watch and decide when the precious stones of potassium nitrate structures and changes, at that point attempt to peruse the temperature of the thermometer. This trouble could bring about a slight erroneous estimations as important seconds are squandered between observing the gems improving and perusing the estimations on the thermometer, which implies that the temperature would have expanded further from when precious stones originally shaped as the Bunsen burner would even now 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 sane. As though I had very various outcomes to various different understudies, at that point It would be far-fetched that every one of them would be wrong.It is difficult to be 100% exact in estimating temperature and reagents, along these lines the outcomes will vary somewhat between students.2. Play out the examination a few times for every volume of water an d take an average.AdvantagesdisadvantagesThis strategy would be significantly more precise.It would take too long to even consider carrying out3. Supposition the crystallization temperature.AdvantagesdisadvantagesFor this strategy there wouldnt be any counts other than for the enthalpy, in this manner this technique would be straightforward.This would be very inaccurate.I think my strategy didnt should be changed in any capacity to adjust to the test in light of the fact that the strategy I utilized went easily.ResultsVolume of watercnR.In cnCrystallisation temperatureReciprocal of total 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 catch) was noted from these sets of values.If Y 1 = mX 1 + c and Y 2 = mX2 + cThen (Y 1 c)/X1 =( Y 2 c )/X 2X2Y1 X2c = X1y2 X1cc = (X2Y1 X1Y2)/(X2 X1)The slope, 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 assess 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-3Mol ar mass 101.11Concentration:Concentration = 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 : % blunder = 73.2 + 0.01% mistake = 0.01 x 100 = 0.01%H : % mistake = 17.9 + 0.01% blunder = 0.01 x100 = 0.06%17.9DiscussionOn the entire I accept that my analysis went very well since everything was directed by plan. I examined my trial and my outcomes with my talk and were supposed to be sensibly precise. I likewise accept that any incorrectness in my outcomes was because of freshness and human blunder, when estimations of weight, temperature and volume were made. My conclusive outcomes were acquired utilizing results from a line of best fit set on my diagram this could have likewise caused a slight of error in my outcomes. These outcomes could have been uncertain as incredibly little numbers were utilized which were hard to plot effectively on the diagram and I had evaluated where the focuses were to be put. Additionally I needed to appraise where to put my line of best fit. Also the mistake could have been because of the thermometer I utilized, as there was a little hole between observing the precious stone and recording the temperature. Therefore I couldnt be as exact as I would have been gotten a kick out of the chance to be as on the grounds that you must be fast in perusing the temperature which was continually evolving. Besides, it was very hard to recognize little particles that had shaped and air rises in the arrangement. In addition, it was difficult to really say what measure of particles shaped, was the perfect add up to take the arrangement off the warmth source and note the results.In my own assessment I imagine that my technique was an appropriate one since it gave me exact outcomes and didnt occupy an excess of time to proceed.ImprovementsThe enhancements that could have been made to make my test progressively precise are to be increasingly cautious in estimating reagents and furthermore to utilize a progressively precise thermometer. Likewise, a bigger scaled diagram could have been utilized as results would have been plotted all the more accurately.Confidently in future I would be progressively knowledgeable about completing the examination and would be increasingly exact at detecting the precious stones when they form.Conclusion and % errorThermometer 0.10620 : % mistake = 0.1 x 100 = 0.2%62560 : % blunder = 0.1 x 100 = 0.2%56470 : % blunder = 0.1 x 100 = 0.2%47420 : % mistake = 0.1 x 100 = 0.2%42Average % mistake = 0.2%Burette 0.1cm38cm3 : % blunder = 0.1 x 100 = 1.3%810cm3 : % mistake = 0.1 x 100 = 1%1012cm3 : % mistake = 0.1 x 100 = 0.8%1214cm3 : % blunder = 0.1 x 100 = 0.7%14Average % mistake = 1.3 + 1+ 0.8 +0.7 = 0.95%4Weighing scales + 0.01g% mistake = 0.01 x 100= 0.1% Show review as it were