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Calculate the resistance of an electrochemical cell constructed by using a copper electrode in a 1M copper (II) sulfate solution and a zinc electrode in a 1M zinc sulfate solution. The measured potential difference of the cell is 1.45 volts. The area of each electrode is 10cm² and the length of w

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To calculate the resistance of the electrochemical cell, we need to consider the resistance of the solutions and the resistance of the wire connecting the electrodes.First, let's convert the given units to meters and square meters for easier calculations:Area of each electrode = 10 cm = 0.001 mLength of wire = 30 cm = 0.3 mCross-sectional area of wire = 1 mm = 0.000001 mNow, let's calculate the resistivity    of each solution using the given conductivity percentages:_CuSO4 = 1 /  0.983 * _CuSO4  = 1 /  0.983 * 5.95 S/m  = 1 / 5.84635 S/m  0.1711 m_ZnSO4 = 1 /  0.899 * _ZnSO4  = 1 /  0.899 * 4.29 S/m  = 1 / 3.8581 S/m  0.2591 mNow, we can calculate the resistance of each solution using the formula R =  * L / A:R_CuSO4 = 0.1711 m * 0.3 m / 0.001 m  51.33 R_ZnSO4 = 0.2591 m * 0.3 m / 0.001 m  77.73 Next, let's calculate the resistance of the wire. The resistivity of copper is approximately 1.68  10^-8  m. Using the formula R =  * L / A:R_wire = 1.68  10^-8  m * 0.3 m / 0.000001 m  5.04 Finally, we can calculate the total resistance of the electrochemical cell by adding the resistances of the solutions and the wire:R_total = R_CuSO4 + R_ZnSO4 + R_wire  51.33  + 77.73  + 5.04   134.1 Therefore, the resistance of the electrochemical cell is approximately 134.1 ohms.

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