大一无机化学课本答案docWord文档下载推荐.docx

1、It is. suppose that a mixture of IL H20 and 560L NH3 is soluble in water.Its mass is 1, 000 grams of water, so the volume of this ammonia solution is,9. Solution:In this case, the simplest form is:（2）,So,This is the molecular formula11. Answer: because CaC12 and NaCl electrolyte, CH3C00H for weak el

2、ectrolytes, C6H1206 as the electrolyte, the solution of the concentration of the solute （particle） size in the order: CaC12 NaCl CH3COOH C6H1206, so high and low order to freezing point C6H12O6 CaC12.Answer: so,Lets say I take this solution, 1,000 grams, and I have one 14.In the sum of the formula,N

3、o, because its too small to be accurateA: because of the Ag + overdose, the preparation of the Agl aerosol tape is positive, so it is the anion that makes up the glue. So the ability to get together is the order of magnitude K3 Fe （CN）.B: because of the I - excess, the prepared Agl sol tape is negat

4、ively charged. Therefore, it is the cation that makes the coagulation of the glue. So the ability to get together is in the order of A1C13 MgS04.Chapter ii problem solving2. Answer:（1）, so there is.（2）the initial pressure is,The volume of the intermediate state is zero（3）answer: because, therefore,（

5、constant pressure is） because the reaction is available（1）-215-296.8 + （-1） （-100）二-411.8Material - 412.-219-296.8 + （-1） （-100）二-416.8Material - 417.（2）（3） （4）. So you can look it up in the same way the reaction equation is:8.Answer: 1, 2, 3, 5, 4, 6 （because of the reduced gas component）（because o

6、f the increase in the gas component）（because the gas component increases）（because of the reduced gas component）（1）because the melting is an isothermal constant pressure reversible process, so there is,（2）because gasification is a reversible process such as isothermal pressure, so there is because me

7、lting and boiling are both isothermal and isostatic equilibrium processes, so there are because boiling is an isothermal constant pressure reversible process, so there is,The equilibrium state of freedom is equal to zeroIt is possible to make a mistake by mistake, mistake, mistake, mistake. the high

8、 temperature spontaneously.The lower temperature is spontaneous.At any temperature, its spontaneous.18. Answer:So, the reaction is never going to happen spontaneously. reaction20.Answer:So its higher than 111221.Answer:,so, not spontaneous at 25 C reaction.（2）, so, in 360 C can occur spontaneously.3

9、.Chapter iii problem solving（2）（3）A fG - 394.4-16.5-228.6-228. 6A rG = 1 x （394. 4） + （2） x （16. 5）, （228. 6） and （197. 15）= 1.65 KJ, mol, 1A rG = - RTln3,Equilibrium time n over moles 0.021 0.004 0. 0040. 025 * 0. 16 二 0. 164HBr （g） + 02 （g）= 2H20 （g） + 2 Br2 （g）（1）the rate equation for the reactio

10、n is zeroFrom the experimental data analysis:So the rate equation for this reaction is zero（1）the rate equation for the experimental data is: （3）Because the reaction is slow （v is small） and the time is short, it can be assumed that the reaction is approximately constant within the interval, so 12 i

11、s produced7.According toIn the case of milk, the rate at which the acid reacts is directly proportional to the rate constant of different temperatures, namelyReaction 2 so2 （g） + 02 （g） = 2 S03 （g） heat effectAccording to theThere areThe watch can calculate the reactionSo, balance moves to the left.

12、Or:18.（1）decrease; （2）; （3） increase; （4） : （5） : （6）:（7）; （8）; （9）; （10）; （12） remains the same.（1）PC15 （g）二 PC13 （g） + C12 （g）The initial mole number is 2, 0, 0Balance the number of moles 0. 5, 1. 5, 1. 5The concentration of each component in equilibrium is:（2）when set to balance, the decompositio

13、n PC15 is x moles.PC15 （g）二 PC13 （g） + C12 （g）The initial number of moles is 2, 0, 1. 0Balance the number of moles 2. 0 minus x, x, x, and x22.The initial Mohr is 0. 2 0. 2 0The initial Mohr is 0. 2 minus x 0. 2 minus x 2x12 pyrolysis rate for25. Solution:Chapter 5. Problem setsA. 2A. 3A. 4A. 5A. 6A

14、. 7EightA. 1 3.The change in the number of 1, 1, 2, 1, 3, 4, 1, 3, 4, 1, 3, 4, 1, 3, 4, 1, 3, 4（1）battery reaction:（2）battery reaction: AgCl （S） + e = + Cl -（3）electrode response:（4）negative poles:The positive:Electromotive force:Original battery reaction:8.1 negative:Electromotive forceCell respons

15、e:2 negative:Battery reaction: Ce4 + Fe2 + + + Ce3 +Negative 3:2H + + 2e 二 H2The positive pole: Cr2072 - + + 6e = 2Cr3 + + 7H20Electromotive force VThe battery reaction is: Cr2072 - + + + + 3112Question 15:（2）positive pole: negative pole:（3）battery reaction:（5）in a solution, it decreases, causing th

16、e electromotive force to rise. If you add ammonia to the solution, its going to go down,Its causing the electric force to go down.Problem 22: positive and negativeSo there areBecause it can be thought of as the anode and composition of the battery:Chapter 6 answers the atomic structure2. （1）By calcu

17、lation, the volatility of the macro object is so small that it can be ignored.6. （1） reasonable（2）unreasonable. Because L can，t be equal to n, L 二 o integer.（3）.（5）unreasonable. Because L = 0（6）its not reasonable because L can，t be greater than nThe seven electrons of the atom are arranged in one s2

18、2s22pxl2pyxl2pyxl2pzlTherefore, its quantum combination is 1S2 （1, 0, 0, 1/2） （1, 0, 0, -1/2）.2S2 （2, 0, 0, 1/2） （2, 0, 0, -1/2）2P3 （2, 0, 1/2） （2, 1, 1, 1/2） （2, 1, 1, 1/2） （1, 1, 1, 1/2） （1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,Or 2, 1, 0, minus 1/2, 2, 1, minus 1/2, 2, 1,

19、 minus 1, minus 1/2.9.Na: 1 s22s22p63slThe IS electron valid nuclear charge = 11 -= 11-0. 30 1 = 10. 702S electrons = 11 - = 11 -= 11-2 0. 85-7 0. 35 二 6. 85 （2S2P is the same group）3 3S electrons 二 11 - 二 11 -二 11-8 0. 85-2 1 二 2. 2 （2S2P is the same group）For multi-electron atoms, when the main qu

20、antum number n = 4, there are four energy levels, namely 4S, 4P, 4D and 4f. They correspond to the number of orbitals, 1, 3, 5, 7. The maximum number of electrons you can hold is 2n2 二 242 - 32 - 2 （1 + 3 + 5 + 7）For hydrogen, the E4s, , E3d because its a single electron system, energy levels are on

21、ly related to the number of quantum Numbers.In the case of K, because of the drilling effect and shielding effect, there is an energy level crossing, its E3d E4s2S Ar 3d54S2Xe 4fl45dlO6Sl, 4f145dl06S26P613. （1） errors Because of the Angle of the atomic orbital part of the graphical representation is

22、 the electron distribution outside the nucleus different spatial orientation of risk, not represent the actual movement of the electrons.（2）the error in the N shell （principle quantum number 二 4） there are 4 s, 4 p, 4 d, 4 f, a total of four energy level, the corresponding 1 hc-positie atomic orbita

23、l is a total of 16, of the principal quantum number N = 1, only Ils, can accommodate two opposite spin electronics.（3）correct because the hydrogen atom is a single electronic system, there is no interleaving phenomenon.（4）error because hydrogen is a single electronic system without the shielding of

24、electrons.Cu2 + Ar 3dl + Ar 3d5Pb2 + Xe 4fl45dlO6S2 - Ne 3S23P6（1） 24 （2） Ar 3d54sl （3） 3d54sl（4）the fourth period; VI. B; Cr03When n 4, its the Ens E （n - 1） dSo when you complete the ns and np orbitals, the rest of the electrons are going to have to fill in the E （n + 1） s orbital.So you can，t fil

25、l the outer layer with d electrons, so you can only have up to eight electrons in the outer layer of the atomic orbital. Similarly, due to the Ens E8s cannot be determined by n + 0. 7 L）The atomic electron layer of element 100 is Rn 7S25f 146dl07p2So this element is the fourth period of the seventh cycleSc Ca： The Sr AgFe2 + Fe3 +; Pb Pb2 +;