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【单选题】
As a new diplomat, he often thinks of ______ he can react more appropriately on such occasions.A.
what
B.
which
C.
that
D.
how
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【单选题】CaCl 2 的无限稀释摩尔电导率与其离子的摩尔电导率的关系是:
A.
Λ ∞ (CaCl 2 ) = λ m (Ca 2+ ) + λ m (Cl - )
B.
Λ ∞ (CaCl 2 ) = 1/2 λ m (Ca 2+ ) + λ m (Cl - )
C.
Λ ∞ (CaCl 2 ) = λ m (Ca 2+ ) + 2 λ m (Cl - )
D.
Λ ∞ (CaCl 2 ) = 2 [ λ m (Ca 2+ ) + λ m (Cl - )]
【单选题】无限稀释的CaCl2摩尔电导率与其离子的摩尔电导率的关系是:
A.
Λm(CaCl2 ) = λm(Ca2+) + λm(Cl-) ;
B.
Λm(½CaCl2) = λm(½Ca2+) + 2λm(Cl-)
C.
Λm(CaCl2) = λm(Ca2+) + 2λm(Cl-) ;
D.
Λm(CaCl2) = 2 [λm(Ca2+) + λm(Cl-)]
【单选题】CaCl 2 无限稀释摩尔电导率与其离子的无限稀释摩尔电导率的关系是 ( )。
A.
L ¥ (CaCl 2 ) = l m (Ca 2+ ) + l m (Cl - )
B.
L ¥ (CaCl 2 ) = ½ l m (Ca 2+ ) + l m (Cl - )
C.
L ¥ (CaCl 2 ) = l m (Ca 2+ ) + 2 l m (Cl - )
D.
L ¥ (CaCl 2 ) = 2 [ l m (Ca 2+ ) + l m (Cl - )]
【单选题】CaCl 2 无限稀释摩尔电导率与其离子的无限稀释摩尔电导率的关系是 ( )。
A.
Λ ∞ (CaCl 2 ) =λ m (Ca 2+ ) + λ m (Cl - )
B.
Λ ∞ (CaCl 2 ) = ½ λ m (Ca 2+ ) + λ m (Cl - )
C.
Λ ∞ (CaCl 2 ) = λ m (Ca 2+ ) + 2 λ m (Cl - )
D.
Λ ∞ (CaCl 2 ) = 2 [ λ m (Ca 2+ ) + λ m (Cl - )]
【多选题】无限稀释的CaCl 2 摩尔电导率与其离子的摩尔电导率的关系是:
A.
Λ m (CaCl 2 ) = λ m (Ca 2+ ) + λ m (Cl - )
B.
Λ m (½CaCl 2 ) = λ m (½Ca 2+ ) + 2λ m (Cl - )
C.
Λ m (CaCl 2 ) = λ m (Ca 2+ ) + 2λ m (Cl - )
D.
Λ m (CaCl 2 ) = 2 [λ m (Ca 2+ ) + λ m (Cl - )]
E.
Λ m (½CaCl 2 ) = λ m (½Ca 2+ ) + λ m (Cl - )
【单选题】CaCl2 无限稀释摩尔电导率与其离子的无限稀释摩尔电导率的关系是 ( )。
A.
L¥(CaCl2) = lm(Ca2+) + lm(Cl-)
B.
L¥(CaCl2) = ½ lm(Ca2+) + lm(Cl-)
C.
L¥(CaCl2) = lm(Ca2+) + 2lm(Cl-)
D.
Pt|Sn4+,Sn2+
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