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Réponses aux questions des 2 exemples du cours2

Réponses aux questions des 2 exemples du cours2. Exemple 1 Je calcule le volume du bassin : V = 30x15x2 m 3 = 900 m 3 = 900000 dm 3 Nombre de molécules dans le bassin : N = 900000x3,35.10 25 molécules = 9. 10 5 x3,35. 10 25 = 9x3,35x10 5 10 25 = 30,15x10 30

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Réponses aux questions des 2 exemples du cours2

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  1. Réponses aux questions des 2 exemples du cours2 Exemple 1 Je calcule le volume du bassin : V = 30x15x2 m3 = 900 m3 = 900000 dm3 Nombre de molécules dans le bassin : N = 900000x3,35.1025 molécules = 9. 105 x3,35. 1025 = 9x3,35x1051025 = 30,15x1030 = 3,015x1031 Il y a 3,015x1031 molécules d’eau dans le bassin :

  2. le diamètre du noyau d’un atome d’hydrogène étant de 5 x 10-17 cm, donc le diamètre du noyau d’un atome 1000 fois plus petit serait : 5 x 10-17÷ 103 = 5 x 10-20 Le diamètre du noyau est 5 x 10-20 cm

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