大学无机化学总复习题英语

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1、精选优质文档-倾情为你奉上. Monochoice questions 1.consider the following reaction: H2(g) + 2ICl(g) =2HCl(g)+I2(g)The rate law for this reaction is first order in both H2 and ICl: Rate=kH2ICl. Which of the following mechanisms are consistent with the observed rate law ? ( )（A）H2(g) + ICl(g) =HI(g)+ HCl (g)(slow

2、) ; HI (g) + ICl(g) = HCl (g)+I2(g) (fast );（B）H2(g) + ICl(g) =HI(g)+ HCl (g) (fast ) ; HI (g) + ICl(g) = HCl (g)+I2(g) (slow );（C）H2(g) + 2ICl(g) =2HCl(g)+I2(g) （D）H2(g) + ICl(g) = HICl (g) + H (g) (slow ) ;H (g) + ICl(g) =HCl(g)+I (g) (fast ); HICl (g)= HCl(g)+I (g) (fast ); I (g)+I (g)= I2(g) (fa

3、st )2. Which of the following is not 4 for the number of significant figures ( )?.（A）3.200 109 （B）pH=2.365 （C）0. （D）26.78% 3. Which substance can use as indicator to show the end point for the titrations: 0.1000 mol/L CH3COOH versus 0.1000 mol/L NaOH.? ( )（A）methyl orange .（B）methyl red.（C）phenolpht

4、halein.（D）chrome black T.4. the pH at the equivalence point for the titrations: 0.1000 mol/L HCl versus 0.1000 mol/L NH3 ( ).pKb = 4.75（A）9.72 （B）5.28 （C）7 （D）4.755. Which of the following statements is true?( ).（A）.knowledge of a rate law is useful in understanding how a reaction takes place at the

5、 molecular level.（B）a plot of concentration versus time for a first-order reaction is linear.（C）the overall order of a reaction equals the largest exponent in the rate law expression.（D）a catalyst increases the reaction rate by increasing the activation energy of the reaction.6.Which of the followin

6、g statements concerning the second main energy level is incorrect? ( )(A) its principal quantum number is 2.(B) it can contain an electron having the quantum number n=2,l=1,m=1,ms=1/2. (C) it has a bilobed (dumbbell) shape. (D) it cannot contain any f orbitals.7. Which of the following sets of quant

7、um numbers is permissible for an orbital ( ). （A）4,-2,0,-1/2 （B）4,4,0,-1/2 （C）3,0,-1,-1/2 （D）1,0,0, 1/2 8. Silicon has the following number of valence electrons ( ).（A）0 （B） 2 （C） 4 （D）69. Look at the following orbital diagrams and electron configuration. Which one is possible according to the Pauli

8、 exclusion principle.( )（A）()( )( )()() （B）()( )( )( )( ) 1s 2s 2p 1s 2s 2p（C）1s22s22p63s23p83d10 （D）1s22s42p6 10. which one is correct for the following in order of increasing atomic radius? ( )（A）Na Be Mg （B）Mg Be Na （C）Be Na Mg （D）BeMgI2/ I -Br2/Br- ; (B) I2/ I -Fe3+/Fe2+ Br2/Br-;(C) Br2/Br- I2/

9、I -Fe3+/Fe2+； (D) Br2/Br- Fe3+/Fe2+ I2/ I -(D)67Which molecular or ion according to the electron configuration KK(2S)2 (*2S)2 (2py)2(2pz)2 (2px)2(A)N2 (B)N2 (C)NO (D)N2+(B)68. Of the following sets of four quantum numbers(n,l,ml,ms),identify the ones that cannot exist for an electron in an atom:(A)

10、4,2,-1,+1/2 (B) 5,0,-1,+1/2 (C) 4,3,-1,+1/2 (D) 3,1,0,-1/2(B)69. Which of the following species has the greatest number of parallel spin electrons in the ground state?(A)Cr (B)Mn (C)Fe3+ (D) Co2+.(A)70. Which of the following species contains an element in an oxidation state that is not a fraction?(

11、A)VO43 (B) Mn2O3 (C) S4O62 (D)Cr2O72.(C)71. Which of the following molecules has the smallest bond angle between its atoms? (A) H2O (B)NH3 (C)XeF4 (D)CH4(A)72. A solution of iodine was prepared by dissolving 12.70g of I2 and 20g of KI in water, and making the volume up to 1 L. A 10.00 mL aliquot of

12、this solution was titrated with standard 0.0500 molL-1 sodium thiosulphate solution, according to the following equation;I2 + 2Na2S2O3 = Na2S4O6 + 2NaIThe volume of sodium thiosulphate used was 18.34mL. The molarity of the I2 solution in molL-1 was therefore: (A). 0.04585 (B). 0.05004 (C). 0.05453 (

13、D). 0.1001 (E). 0.1205(A)73. Consider the reaction 2A 4BC. What is the order of this reaction from the following straight line plot ?lgAtime(A) half-order reaction .(B) second-order reaction ; (C) zero-order reaction ;(D) first-order reaction ; (D)74. For a certain first order reaction, the time req

14、uired for half of an initial amount to decompose is 2 minutes. If the initial concentration of A is 0.8 molL-1, the time required to reduce the concentration of A to 0.2molL-1 is:(A)4.0min ; (B)1.0 min ; (C)8.0 min ; (D)2.5 min . (A)75. A double bond between two carbon atoms is formed when:(A) one e

15、lectron is shared (B) two electrons are shared(C) four electrons are shared (D)two electrons are transferred (C)76. The Lambert-Beers Law can NOT be represented by(B)(A) -lgIt/I0 = Kbc (B) lgIt/I0 = bc(C) lg1/T = bc (D) A = Kbc77. A 0.1 molL-1 solution of potassium acetate, KC2H3O2, has a lower pH t

16、han a 0.1 molL-1 solution of potassium cyanide, KCN. From this, you can correctly conclude that (A) hydrocyanic acid, HCN, is a weaker acid than acetic acid, HC2H3O2.(B) hydrocyanic acid, HCN, is less soluble in water than acetic acid, HC2H3O2.(C) the cyanide ion, CN, is a weaker base than the acetate ion, C2H3O2.(D) cyanides are less soluble in water than acetates.(E) acetate ion, C2H3O2, partially dissociates to form hydronium ion, H3O+.(A)78. A metal M displaces copper from an aqueous solution of copper (II) sulfate but does not react with an aqueous solution of zinc nitra