Answer Happy

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Here's the graph I did in excel, but it keeps telling me I'm
wrong.
The following data were obtained when a Ca2+ ISE was immersed in standard solutions whose ionic strength was constant at 2.0 M. Ca2+ (M) E (mv) 3.28x10-5 -75.1 3.28 x 10-4 -46.9 3.28 x 10-3 -18.8 3.28 x 10-2 +9.33 3.28 x 10-1 +37.5 (a) Prepare a calibration curve and find the least-squares slope and intercept and their standard deviations. slope 4.0 28.1 +49 28.1 intercept 51.12 х Sy 491 x 4951.12 (b) The Nernst Equation tells us the equation values based on theory, but the data we collect in the lab often doesn't exactly agree with theory. Calculate the value of the slope correction value ß in the equation below. Response of Ca2+ electrode: E = constant + 0.05916 V p(0.05 16 Vlog(Aca2+)(outside) 49951.419 X (C) For a measured potential, the calibration curve gives us log [Ca2+]. We can compute [Ca2+] = 10109 [Ca2+1 Using rules for propagation of uncertainty in the Summary of Rules for Propagation of Uncertainty table, calculate [Ca2+] and its associated uncertainty for a sample that gave a reading of -22.1 + 0.4 mv in four replicate measurements. 4.0 4,0 M

Chart Title 60 Ca2+ log Ca2+ E(mv) 3.28E-05 -4.48413 -75.1 3.28E-04 -3.48413 -46.9 3.28E-03 -2.48413 -18.8 3.28E-02 -1.48413 9.33 3.28E-01 -0.48413 37.5 40 y = 28.143x + 51.117 R2 = 1 20 -5 -4 -3 - 2 -1 0 -20 -40 -60 -80 -100 Series1 Series2 ......... Linear (Series1)