Potassium chlorate is prepared by the electrolysis of KCl in basic solution 6OH⁻+Cl⁻→ClO₃⁻+3H₂O+6e⁻. If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 g of KClO₃ using a current of 2 A is [numerical value] (Given: F=96,500 Cmol⁻¹; molar mass of KClO₃=122 g mol⁻¹)
Potassium chlorate is prepared by the electrolysis of KCl in basic solution 6OH⁻+Cl⁻→ClO₃⁻+3H₂O+6e⁻. If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 g of KClO₃ using a current of 2 A is [numerical value] (Given: F=96,500 Cmol⁻¹; molar mass of KClO₃=122 g mol⁻¹)
n (KClO? )=10/122. No. of F = (10/122)*6. No. of F supplied= (100/122). t=11hrs.
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CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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