DDWIN3 5 10Deriving a Solubility Curve Sandra Curry Temperature\u00B0C SensorMeter5Solubility(g) KNO3\u00A3Solubility(g) KNO3 100g waterhExponential fit of channel 5 against channel 1. y = 25.0e^0.0229x. Correlation coefficient is 0.992. Ch 1 + 273\u00B0C!Calculated as channel 1 plus 273.5Solubility(g) KNO3\u00A3Solubility(g) KNO3 100g wateriExponential fit of channel 2 against channel 3. y = 0.0481e^0.0229x. Correlation coefficient is 1.00.5Solubility(g) KNO3\u00A3Solubility(g) KNO3 100g water Keyboard data09:0716/03/01 0 4-1.2028067430947E-10 109.99999999988 1073714542 10737691020 1 .2UL39,17 3 20 150 939524096 12079595520 0 .1MA000 4 294.677429199219 347.064514160156 1073741824 10905190400 1 .2MA000 3 20 150 939524096 12079595520 0 .1MA000 3 20 150 939524096 12079595520 0 .1KY000-1 0 65472 0 654720 0 1XX-1 0 65472 0 654720 0 1XX-1 0 65472 0 654720 0 1XX00000000EX232-1 0 65472 0 654720 0 1XXTABLECHANS1,2,3,4,5,#,TABLEINC1,#,TIMESTYLE,#,X1is1R1TIM0P1OLARPlot0F1RGS0A1LGN0U1NSHARECALS0M1ARKERS0G1RGRID1G1RBARWIDTH0#,P1LOTCHANS1,2,5,#,E1VENTFREQUENCY0L1OGSCALE#,Z1ERO1,2,3,4,5,6,7,8,9,10,11,12,13,#,J1OIN4,5,#,S1HOW1,2,3,6,7,8,9,#,E1BAR#,B1ARSTYLE1,1,1,1,1,1,1,1,1,1,1,1,1,0,#,Xis3RTIM0POLARPlot0FRGS0ALGN0UNSHARECALS0MARKERS0GRGRID1GRBARWIDTH0#,PLOTCHANS2,3,4,#,EVENTFREQUENCY0LOGSCALE#,ZERO1,2,3,4,5,6,7,8,9,10,11,12,13,#,JOIN4,#,SHOW1,2,3,5,6,7,8,9,#,EBAR#,BARSTYLE1,1,1,1,1,1,1,1,1,1,1,1,1,0,#,n?.?n??$@H:#;Z<&`>@@$I@B@Aj:#;Z<]>p=:?;f