Chart 1▾⠀HGfrAxis TitleFED1/T-1.20397 0.003356-0.59784 0.003145-0.16252 0.0030030.262364 0.002874LN KChart Title0.40.200.0034 0.0033 0.0032 0.0031 003 0.0029 0.00282-0.4-0.6-0.8-1-1.2-1.4TKAxis TitleCy=-3033.1x+8.9597R² = 0.999298318333348K-Series1Linear (Series1)COB0.30.550.851.3+2TA> FNM45125 2360 475 567891011121314151617181920Consider the following data to determine the molar entropy AS°:Temp (°C) Keq0.30.550.851.325456075a. 364.82 J/K.molb. 74.49 J/K.molc. 86.37 J/K.mold. 89.59 J/K.mole. 64.25 J/K.molClear my choice

The required thermodynamic equation to draw ln(Keq) versus 1/T plot is: ln(Keq) = -∆HoR*1T + ∆SoR…

Chart 1 > E H X V fx 1 Axis Title F E LN K 1/T -1.20397 0.003356 -0.59784 0.003145 -0.16252 0.003003 0.262364 0.002874 Chart Title 0.4 0.2 0 0.0034 0.0033 0.0032 0.0031 003 0.0029 0.0028 -0.4 -0.6 -0.8 -1 -1.2 -1.4 TK y=-3033.1x+8.9597 R² = 0.999 298 318 333 348 K --Series1 ......... Linear (Series1) O O B 0.3 0.55 0.85 1.3 + T A 1 25 2 45 3 60 4 75 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Consider the following data to determine the molar entropy AS: Temp (°C) 25 45 60 75 Keg 0.3 0.55 0.85 1.3 a. 364.82 J/K.mol b. 74.49 J/K.mol c. 86.37 J/K.mol d. 89.59 J/K.mol e. 64.25 J/K.mol Clear my choice

Chart 1 > E H X V fx 1 Axis Title F E LN K 1/T -1.20397 0.003356 -0.59784 0.003145 -0.16252 0.003003 0.262364 0.002874 Chart Title 0.4 0.2 0 0.0034 0.0033 0.0032 0.0031 003 0.0029 0.0028 -0.4 -0.6 -0.8 -1 -1.2 -1.4 TK y=-3033.1x+8.9597 R² = 0.999 298 318 333 348 K --Series1 ......... Linear (Series1) O O B 0.3 0.55 0.85 1.3 + T A 1 25 2 45 3 60 4 75 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Consider the following data to determine the molar entropy AS: Temp (°C) 25 45 60 75 Keg 0.3 0.55 0.85 1.3 a. 364.82 J/K.mol b. 74.49 J/K.mol c. 86.37 J/K.mol d. 89.59 J/K.mol e. 64.25 J/K.mol Clear my choice

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Convert kPa/sec to uL/min for the following values. -0.0019 kPa/sec -0.0029 kPa/sec -0.0028 kPa/sec -0.0021 kPa/sec -0.0026 kPa/sec -0.0034 kPa/sec
4. Ksp=4.1x10-36 for Pb3(AsO4)2Calculate Eofor Pb3(AsO4)2(s) + 6e-  3Pb(s) + 2AsO43-Pb2+ +2e-  Pb(s) Eo= -0.126 V
image https://cxp.cengage.com/contentservice/assets/owms01h/references/chemtables/gen_chem/eo.html ^ table
The gravimetric factor used to express CoCBr6·H20 in a sample that is finally weighed as PbClBr is choose below: FW PbClBr / 6 x FW CoCBr6·H20 FW CoCBr6·H20 / FW PbClBr FW CoCBr6·H20 / 6 x FW PbClBr 6 x FW PbClBr / FW CoCBr6·H20
NH4+ {aq) + NO2(aq) -> N2(g) +2H2O{l} Data Initial [NH4+] Initial [NO2-] rate 1 0.0100 0.200 5.4 x10-7 2 0.0200 0.200 10.8x10-7 3 0.0400 0.200 21.5x10-7 4 0.200 0.0202 10.8x10-7 5 0.200 0.0404 21.6x10-7 6 0.200 0.0808 43.3x10-7 Find x,y,k
Calculate the Constant Weight (in grams) of the Empty Crucibles. Show your solution. Crucible No. 1 2 3 Weighing 1 22.6035 22.0223 24.1535 Weighing 2 22.6017 22.0204 24.1533 Weighing 3 22.5994 22.0199 - Weighing 4 22.5992 - - Constant Weight, g ??? ??? ???
Calculate ΔrG∘ for the following reaction:4CO(g)+2NO2(g)→4CO2(g)+N2(g).Use the following reactions and given ΔrG∘values:2NO(g)+O2(g)→2NO2(g), ΔrG∘= - 72.6 kJmol−12CO(g)+O2(g)→2CO2(g), ΔrG∘= - 514.4 kJmol−11/2 O2(g)+1/2 N2(g)→NO(g), ΔrG∘= 87.6 kJmol−1 Express your answer using one decimal place.
When two predictors are highly correlated, which of the following statements is true? -Ridge regression will usually have non-zero coefficients for both predictors. -Ridge Regression will drive coefficients to zero for both features. -Lasso regression will usually have non-zero coefficients for both predictors. -Elastic net will usually have non-zero coefficients for both predictors select the correct option
Sample IdentificationCodeConcentration of M%TA=2-log(%T)Q50004.00 x 10-417.90.75Q50013.20 x 10-425.00.6Q50022.40 x 10-435.70.46Q50031.60 x 10-450.20.3Q50048.000 x 10-570.80.15SampleIdentificationCode%TA=2-log(%T)AMQ0210150143.70.359518560.360.000192Q0210150244.10.355561410.360.00018Q0210150343.80.358525890.360.00017Q0210150444.10.355561410.360.00018Q0210150543.80.358525890.360.00017What was their percent error?43%Does Batch 021015 meet legal requirements?No, because it is not between 2.85 * 10(4) and 3.15 * 10(4)Well #DropsBluedye1234567891012345678910Drops 9Distilled water876543210Concne 0.26tration0.52Test Tube #0.781.041.3Solutions3Concentration (M)2.082.32.6Concentration (ppm)1:1 dilution11.82Starting Dilution21.562:1 dilution0A.Zero standard0Was your calibration curve as linear as you expected?B.Did you experience any â€œdriftâ€ of the resistance readings?C.What is the equation of your best-fit line?D.What commercial drink did you analyze?E.Assuming…
From this given value of Kc: 2H2S (g) <-------> 2H2 + S2 Kc=2.3 X 10-4 at 1404K What is Kc for: H2 + 1/2S2 <--------> H2S
How to convert 4.76pm to um? Showing the work. 4.76pm x___________m x_____________ um =________________um How to convert 25.0mL to cL? Showing the work. 25.0mL x______________ x ___________ = ______________cL
1. What is the mEq of silver nitrate? 170 0.0170 0.170 17.0 2. How much (mL) silver nitrate is required for the standardization? 4.995 49.950 49.590 45.990 * MW of silver nitrate = 170 g/mol * Please take note that Im looking for the silver nitrate