Hitunglah konsentrasi akhir larutan HCI jika 50 mL (12M) ditambahkan 200 mL aquades
Answer (3)
1 ∫ ∞ x 2 ln x d x = t → ∞ lim 1 ∫ t x 2 ln x ∫ x 2 ln x d x = ( ∗ ) u = ln x , d u = x 1 d v = x 2 1 , v = − x 1 ( ∗ ) = ln x ⋅ ( − x 1 ) − ∫ ( x 1 ⋅ ( − x 1 )) d x = − x ln x + ∫ x 2 1 d x = − x ln x − x 1 + C t → ∞ lim 1 ∫ t x 2 ln x = t → ∞ lim [ − x ln x − x 1 ] 1 t = t → ∞ lim ( − t ln t − t 1 − ( − 1 ln 1 − 1 1 ) ) = t → ∞ lim ( − t ′ ( ln t ) ′ ) − 0 − ( − 1 ) = t → ∞ lim ( − 1 t 1 ) + 1 = t → ∞ lim ( − t 1 ) + 1 = 0 + 1 = 1
The integral 1 ∫ ∞ x 2 l n x d x evaluates to 1 after applying integration by parts and taking the limit as t → ∞ . The steps involve substitution, using the integration by parts formula, and simplifying the limit of the evaluated expression. Finally, both improper terms approach zero, leaving a value of 1 .
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Jawaban:Penjelasan:M1 = 12 M V1 = 50 ml. V2 = 50 + 200 = 250 ml. Dit: M2 M1 x V1 = M2 x V2 12 M x 50 ml = M2 x 250 ml. M2 = 600 / 250 = 2,4 M Jadi, konsentrasi ...
