Net Voltage (V) 0.03 0.02 y = 7.7821x + 0.0005 R² = 0.9939 0.01 0.00 0 0.001 0.002 Sulfate Concentration (M) 0.003

Standard Reduction (Electrode) Potentials at 25 Half-Cell Reaction * (volts) F2(8) + 2 e2F (aq) 287 Ce* (aq) +e Ce (aq) 1.61 Mn0, (aq) + BH'(aq) +5e Mn"( "(aq) + 4 H20(1) 1.51 Cla(g) +2e -2 C (aq) 1.36 Cr20, (aq) + 14 H'(aq) - 6e2 Cr*"(aq) + 7 H20() 1.33 Oz() + 4 H'(aq) + 4e2 H20(1) 1.229 Br20) + 2 e2 Br (aq) 1.08 NO3 (aq) + 4 H'(aq) + 3eNO(g) + 2 H20() Hg"(aq) 0.96 2 Hg (aq) + 2e 0.920 Hg"(aq) + 2 e- Hg(1) 0.855 Ag (aq) +e Ag(s) 0.799 Hg2" (aq) - 2e-2 Hg(1) 0.789 Fe" (aq) +e Fe"(ag) 0.771 12(s) + 2e 21 (aq) 0.535 Fe(CN)"(aq) + e Fe(CN), (aq) 0.48 Cu (aq) + 2eCu(s) 0.337 Cu"(aq) +e- Cu (aq) 0.153 S(s) + 2 H'(aq) + 2 e H2S(aq) 0.14 2H'(aq) + 2e-H2(8) 0.0000 Pb (aq) + 2e- Ph(s) -0126 Sn (aq) + 2 e Sn(s) -014 Ni (aq) + 2 e- Ni(s) -0.25 Co (aq) + 2 e Co(s) -0.28 "(aq) + 2e Cd(s) -0.403 Cr"(aq) +e "(aq) -041 Fe"(aq) + 2 e Fe(s) -044 Cr"(aq) +3e-Cr(s) -0.74 Zn"(aq) +2 e- Zn(s) -0.763 2 H20() +2e Hz(z) + 2 OH (aq) -0.83 Mn"(aq) + 2e - Mn(s) -1.18 A" (aq) + 3e- Al(s) -1.66 Mg (aq) + 2 e- Mg(s)…

A circuit is constructed with a 6.00 V battery connected across a 2900.0 Ω resistor.(i) Determine the moles of electrons that flow through the circuit per second for 26.98 min.(ii) Determine the number of joules produced by the circuit for the 26.98 min.

Can you write the answer in a way that fills the empty cells, please? Thank you.

Write the net cell equation for this electrochemical cell. Phases are optional. Do not include the concentrations. Cu(s)|Cu²+(aq, 0.0155 M)||Ag*(aq, 3.50 M)|Ag(s) ,2+ net cell equation: Calculate Ee, AGxn, AGrxn, and Ecell at 25.0 °C, using standard potentials as needed. 'cell V 'cell kJ AGixn = mol

Consider the reaction Cr₂O72-(aq) + I¯(aq) → Cr³+ (aq) + 1₂(s). (a) Balance this equation assuming the reaction occurs in acidic solution. (Include states-of-matter under the given conditions in your answer.) chemPad Help XX ->> Cr₂O2(aq)+14H+ (aq) +61- Cr_20_7^2-(aq)+14H^+(aq)+61^--->2Cr^3+(aq)+31_2(s)+7H_20(1) Your answer appears to be missing one or more states-of-matter. (b) Using standard reduction potentials, calculate the Ecell, AGO, and K for the reaction at 298.15 K. Ecell = AG° = K = Greek 123 Tutorial 2Cr3+ (aq)+31₂(s)+7H₂O(1) V J (Use 298.15 as the temperature and round answer to two significant figures) (c) Calculate the cell potential if [I] = 3.59x10-3 M, [Cr₂O72- ] = 2.37x10-2 M, [H+] = 2.29x10-3 M, and [Cr³+] = 4.38 x 10-2 M. V

Balance the following equation by selecting the coefficients in the dropdown boxes.The coefficient for C5H12 isa. 1b. 2c. 3d. 4The coefficient for O2 isa. 1b. 2c. 3d. 4e. 5f. 6g. 7h. 8i. 9j. 10The coefficient for CO2 is a. 1b. 2c. 3d. 4e. 5f. 6g. 7h. 8i. 9j. 10The coefficient for H2O is a. 1b. 2c. 3d. 4e. 5f. 6g. 7h. 8i. 9j. 10

R = 8.314 mol·K F = 96,485 mot AG = AG° + RT · In(Q) ΔΕΔΕ-).In(O) Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e¯ → 2 H20 (1) 1.229 2 z" (aq) Z2 (s) + 2 e 3+ (aq) + 3 е 0.426 A (s) 0.292 2 H20 (1) + 2 e G2+ (aq) + 2 e H2 (g) + 2 OH (aq) - 0.828 G (s) - 1.245 > M2+ (aq) + 2 e - 1.893 → M (s) A student constructs a voltaic electrochemical cell with two metal electrodes [metal G and metal A] in their respective aqueous nitrate solutions [G(NO3)2 and A(NO3)3]. Use this information, as well as the reduction potentials in the table above to complete each statement below. Consider the same cell from the above prompt. Calculate AG° in kJ/mol for this galvanic cell. Report your answer with 4 significant figures. You do not need to report units with your answer. If your value is negative, make sure to include a "-" symbol.

Given the line notation:       Pt(s) | X+(aq), X2+(aq) || Y3+(aq) | Y(s)                         where: X2+ + e-  X+              Eo = 0.93 V Y3+ + 3 e-  Y             Eo = -2.48 V Pt4+ + 4 e-  Pt           Eo = 1.20 V                       Calculate the EMF.

The resistance of a 0.008 mol dm-³ solution of NaCl (aq) at 300 K was found to be 2.75 x 10³ . In a separate experiment, using the same conductivity cell, the resistance for a 0.1 mol dm-³ KCI aqueous solution was determined as 125.6 Q. Given that the conductivity of 0.1 mol dm-³ KCl aqueous solution at 300 K is 0.01302 S cm-1, calculate the molar conductivity of the NaCl aqueous solution.