MamyLaPoudre Posted December 21, 2018 Share Posted December 21, 2018 Bonjour, je voulais savoir si vous pourriez m'aider à comprendre la correction de l'item E du qcm 139 du livre de la prof voilà le qcm et sa correction Enfaite je ne comprend pas à partir de E2 = E 02 - 0.03 pH + ... Quote Link to comment Share on other sites More sharing options...
Solution Chat_du_Cheshire Posted December 21, 2018 Solution Share Posted December 21, 2018 Coucou, oui car en fait il faut savoir que E2 = E2°' + 0,03logX/Y, d'après le cours E2°' = E2° - 0,06xpH /n + 0,03logX/Y, car E2°' = E2° - 0,06xpH /n Quote Link to comment Share on other sites More sharing options...
MamyLaPoudre Posted December 21, 2018 Author Share Posted December 21, 2018 D’accord merci beaucoup !! @Chat_du_Cheshire Quote Link to comment Share on other sites More sharing options...
Chat_du_Cheshire Posted December 21, 2018 Share Posted December 21, 2018 il y a 25 minutes, millapauline a dit : D’accord merci beaucoup !! @Chat_du_Cheshire avec plaisir Pauline bon courage!! Quote Link to comment Share on other sites More sharing options...
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