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听力原文: In the Middle Ages two was the official number of meals because of the clergymen's insistence on fasting. But this official pattern was frequently changed for some practical reasons. For example, for health reasons, laborers, the sick, the very young and the every old were permitted a third meal in the morning, hence the breaking of the fast or breakfast. By the seventh century three meals a day had become the norm at all levels of society. It remains the publicly adopted norm even today, and there may be two reasons for this general acceptance. According to the result of the computerized tests of the body's energy needs, physiologists conclude that the three-meal pattern is suitable nutritionally to the average adult. The second reason is related to the human association. The number 3 has had a mystical, comforting significance for Western minds and consequently we speak with great respect of the Trinity and we take the triangle as geometrically the perfect figure. In fact, considering the prevalence of snacking in America, it would be more correct to say that the norm is six or seven 'meals' a day. When did three meals a day become a widely-accepted norm at all levels of society?
A.
By the seventeenth century
B.
By the seventh century
C.
By the end of last century
D.
In the Middle Ages
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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 - )]
【简答题】肉桂酸甲酯(H)是常用于调制具有草莓、葡萄、樱桃、香子兰等香味的食用香精。它的分子式为C 10 H 10 O 2 ,且分子中只含有1个苯环,苯环上只有一个取代基。又知J为肉桂酸甲酯的一种同分异构体,其分子结构模型如下图所示(图中球与球之间连线表示单键或双键)。 试回答下列问题。 (1)①肉桂酸甲酯(H)的结构简式为_________________。②有关肉桂酸甲酯的叙述中,正确的是_______...
【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-)]
【多选题】有固定名称和固定位置的是
A.
经外奇穴
B.
阿是穴
C.
十二经穴
D.
任脉穴
E.
督脉穴
【多选题】有固定名称和固定位置的是
A.
经外奇穴
B.
募穴
C.
十四经穴
D.
交会穴
【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 - )
【判断题】直线在正投影图上的投影必定是直线。
A.
正确
B.
错误
【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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