A cell contains 100 mM K+ and is surrounded by interstitial fluid containing 10 mM K+. The equilibrium potential for potassium is -58mV. What is the temperature of the cell, in degrees C? Enter your answer as a whole number. Do not use words or symbols.

determine the mol (!) amount of glucose present in the glycogen sample. Show all details of the working out of your calculation. Indicate all units and provide the answer with one decimal precision tube 1 :  26.73 µg/ml tube 2: 23.09 µg/ml tube 3 60.36 µg/ml tube 4 47.64 µg/ml

The following data give the constant dissociation, Ka (for acetic acid at various temperatures): T(°C) 0 Ка 1.657 x 10-5 10 1.729 x 10-5 15 1.745 x 10-5 25 1.753 × 10-5 kJ ΔΗ° = 1.53 mol Submit Previous Answers Part B Correct Plot Inka vs. 1/T and fit a line to the points. The slope will correspond to —▲H° / R. 1/T InKa 3.66 x 10-3-11.01 3.53 x 10-3-10.97 3.47 x 10-3-10.96 3.36 × 10-3-10.95 InKa = (−184 K) (±) — 10.3 = slope -184 K = AH° R AH° -184 K = 0.008314 kJ/(mol·K) AH 1.53 kJ/mol = Use these data, and the constant dissociation at 25 °C, to calculate AS° for acetic acid ionization. (InK Express your answer with the appropriate units. μÅ AS° = Value Units ? -AH° + ᎡᎢ. AS° R

How do I calculate the final concentration? I just need one solution to walk through the steps.    Prepare eight standards of glucose with your 1 mM glucose solution, each with a final volume of 10 mL. Complete the table on the next page, indicating what your final concentration of glucose will be for each standard, in mM. Volume of Potassium Ferrocyanide Solution Volume of Enzyme Solution Volume of 1 mM glucose solution Final volume of solution Final Concentration of glucose in final solution (mM) 2.50 mL 1.00 mL 0.00 mL 10.00 mL   2.50 mL 1.00 mL 0.50 mL 10.00 mL   2.50 mL 1.00 mL 1.00 mL 10.00 mL   2.50 mL 1.00 mL 2.00 mL 10.00 mL   2.50 mL 1.00 mL 3.00 mL 10.00 mL   2.50 mL 1.00 mL 4.00 mL 10.00 mL   2.50 mL 1.00 mL 5.00 mL 10.00 mL   2.50 mL 1.00 mL 6.00 mL 10.00 mL