ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿I.ÔĶÁ¡¢·ÖÎöÏÂÁÐÁ½¸ö²ÄÁÏ£º

²ÄÁÏÒ»

²ÄÁ϶þ

ÎïÖÊ

ÈÛµã/¡æ

·Ðµã/¡æ

ÃܶÈ/ g/cm3

ÈܽâÐÔ

ÒÒ¶þ´¼(C2H6O2)

£­11.5

198

1.11

Ò×ÈÜÓÚË®ºÍÒÒ´¼

±ûÈý´¼(C3H8O3)

17.9

290

1.26

ÄܸúË®¡¢¾Æ¾«ÒÔÈÎÒâ±È»¥ÈÜ

»Ø´ðÏÂÁÐÎÊÌâ(Ìî×ÖĸÐòºÅ)£º

£¨1£©´Óº¬ÉÙÁ¿NaClµÄNa2CO3ÖÐÌá´¿Na2CO3µÄ²Ù×÷ΪÈܽ⡢_________¡¢________¡¢¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï

£¨2£©½«ÒÒ¶þ´¼ºÍ±ûÈý´¼Ï໥·ÖÀëµÄ×î¼Ñ·½·¨ÊÇ_________¡£

A£®ÕôÁ󷨡¡¡¡¡¡¡¡¡¡B£®ÝÍÈ¡·¨ C£®½á¾§·¨ D£®·ÖÒº·¨

II.ÇàÝïËØÊÇ×îºÃµÄµÖ¿¹Å±¼²µÄÒ©Î¿É´Ó»Æ»¨Ýï¾¥Ò¶ÖÐÌáÈ¡£¬ËüÊÇÎÞÉ«Õë×´¾§Ì壬¿ÉÈÜÓÚÒÒ´¼¡¢ÒÒÃѵÈÓлúÈܼÁ£¬ÄÑÈÜÓÚË®¡£³£¼ûµÄÌáÈ¡·½·¨ÈçÏÂ

²Ù×÷I¡¢IIÖУ¬²»»áÓõ½µÄ×°ÖÃÊÇ________£¨ÌîÐòºÅ£©¡£

¢ó.ʵÑéÊÒÓÃ98%¡¢ÃܶÈΪ1.84g/cm3µÄŨÁòËáÅäÖÆ0.2mol/LµÄH2SO4ÈÜÒº470mL.

£¨1£©ÐèҪʹÓõÄÖ÷ÒªÒÇÆ÷ÓÐÁ¿Í²¡¢ÉÕ±­¡¢²£Á§°ô¡¢___________¡¢__________¡£

£¨2£©Æä²Ù×÷²½Öè¿É·Ö½âΪÒÔϼ¸²½£º

A£®ÓÃÁ¿Í²Á¿È¡______mLŨÁòËᣬ»º»º×¢Èë×°ÓÐÔ¼50mLÕôÁóË®µÄÉÕ±­À²¢Óò£Á§°ô½Á°è¡£

B£®ÓÃÊÊÁ¿ÕôÁóË®·ÖÈý´ÎÏ´µÓÉÕ±­ºÍ²£Á§°ô£¬½«Ã¿´ÎµÄÏ´Òº¶¼ÒÆÈëÈÝÁ¿Æ¿Àï¡£

C£®½«Ï¡ÊͺóµÄÁòËáСÐĵØÓò£Á§°ôÒýÁ÷ÈÝÁ¿Æ¿Àï¡£

D£®¼ì²éÈÝÁ¿Æ¿ÊÇ·ñ©ˮ¡£

E£®½«ÕôÁóˮֱ½Ó¼ÓÈëÈÝÁ¿Æ¿£¬ÖÁÒºÃæ½Ó½ü¿Ì¶ÈÏß1-2cm´¦¡£

F£®¸Ç½ôÆ¿Èû£¬·´¸´µßµ¹Õñµ´£¬Ò¡ÔÈÈÜÒº¡£

G£®ÓýºÍ·µÎ¹ÜÏòÈÝÁ¿Æ¿ÀïÖðµÎ¼ÓÈëÕôÁóË®£¬µ½ÒºÃæ×îµÍµãÇ¡ºÃÓë¿ÌÏßÏàÇС£

Çë¾Ý´ËÌîд£º

¢ÙÍê³ÉÉÏÊö²½ÖèÖеĿհ״¦¡£

¢Ú²¹³äÍê³ÉÕýÈ·µÄ²Ù×÷˳Ðò£¨ÓÃ×ÖĸÌîд£©£º____________

£¨ D £©¡ú£¨ A £©¡ú ¡ú ¡ú ¡ú ¡ú£¨ F £©¡£

£¨3£©ÏÂÁвÙ×÷»áʹËùÅäÏ¡ÁòËáŨ¶ÈÆ«¸ßµÄÊÇ______________¡£

A¡¢È¡ÓÃŨÁòËáʱÑöÊӿ̶ÈÏß

B¡¢ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴµÓºóÓÃ1.2mol/LµÄÁòËáÈóÏ´

C¡¢×ªÒÆÈÜҺʱ£¬²»É÷ÓÐÉÙÁ¿ÈÜÒºÈ÷³ö

D¡¢¶¨ÈÝʱÑöÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß

E¡¢½«Ï¡ÊͺóµÄÏ¡ÁòËáÁ¢¼´×ªÈëÈÝÁ¿Æ¿ÇÒ½øÐкóÃæµÄʵÑé²Ù×÷

F¡¢¶¨Èݺ󣬰ÑÈÝÁ¿Æ¿µ¹ÖÃÒ¡ÔÈ£¬Õý·Åºó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÓÖ²¹³ä¼¸µÎ ÕôÁóË®ÖÁ¿Ì¶ÈÏß

¡¾´ð°¸¡¿Õô·¢Å¨Ëõ ÀäÈ´½á¾§£¨½µÎ½ᾧ£© A C 500mlÈÝÁ¿Æ¿ ½ºÍ·µÎ¹Ü 5.4ml CBEG ABE

¡¾½âÎö¡¿

¸ù¾Ý̼ËáÄƺÍÂÈ»¯ÄƵÄÈܽâ¶ÈËæζȵı仯²îÒì·ÖÎö£¬Ó¦²ÉÓýᾧ·¨·ÖÀë¡£»¥ÈܵÄÒºÌå»ìºÏÎïÀûÓ÷еã²îÒì²ÉÓÃÕôÁóµÄ·½·¨·ÖÀë¡£¸ù¾Ý×°ÖõÄÌصã·ÖÎö£¬AΪÕôÁó×°Öã¬BΪ¹ýÂË×°Öã¬CΪ¼ÓÈȹÌÌåµÄ×°Ö᣸ù¾ÝÅäÖÆÈÜÒºµÄ¹ý³ÌÖÐÈÜÖʵı仯ºÍÈÜÒºµÄÌå»ýµÄ±ä»¯½øÐÐÎó²î·ÖÎö¡£

I. (1)̼ËáÄƵÄÈܽâ¶ÈËæ×Åζȱ仯¼Ó´ó£¬ËùÒÔÓýᾧ·¨·ÖÀ룬ʵÑé²Ù×÷ΪÈܽ⣬Õô·¢Å¨Ëõ£¬ÀäÈ´½á¾§£¨½µÎ½ᾧ£©£¬¹ýÂË£¬Ï´µÓ£¬¸ÉÔï¡£

(2)ÒÒ¶þ´¼ºÍ±ûÈý´¼»¥ÈÜ£¬¶þÕ߷е㲻ͬ£¬ÀûÓÃÕôÁóµÄ·½·¨·ÖÀ룬¹ÊA£»

II.ÌáÈ¡¹ý³ÌÖÐÓйýÂ˺ÍÕôÁóµÈʵÑé¹ý³Ì£¬²»Éæ¼°¼ÓÈȹÌÌ壬ËùÒÔÑ¡C£»

¢ó. (1)¸ù¾ÝÈÜÒºµÄÌå»ýÑ¡Ôñ 500mlÈÝÁ¿Æ¿£¬ÁíÍ⻹ÐèҪʹÓýºÍ·µÎ¹Ü¡£

(2) ¢ÙÐèÒªµÄŨÁòËáµÄÌå»ýΪ= 5.4ml£»

¢Ú ʵÑéÇ°Ïȼì²éÈÝÁ¿Æ¿ÊÇ·ñ©ˮ£¬Á¿È¡Ò»¶¨Ìå»ýµÄŨÁòËá½øÐÐÏ¡ÊÍ£¬È»ºó½«Ï¡ÊͺóµÄÁòËáÒýÁ÷µ½ÈÝÁ¿Æ¿ÖУ¬²¢Ï´µÓÉÕ±­ºÍ²£Á§°ô£¬¼ÓË®£¬µ½½Ó½ü¿Ì¶ÈÏß1-2cm´¦£¬ÓýºÍ·µÎ¹ÜÏòÈÝÁ¿Æ¿ÀïÖðµÎ¼ÓÈëÕôÁóË®£¬µ½ÒºÃæ×îµÍµãÇ¡ºÃÓë¿ÌÏßÏàÇУ¬¸Ç½ôÆ¿Èû£¬·´¸´µßµ¹Õñµ´£¬Ò¡ÔÈÈÜÒº¹Ê²Ù×÷˳ÐòΪ£ºCBEG£»

(3) A¡¢È¡ÓÃŨÁòËáʱÑöÊӿ̶ÈÏߣ¬ÔòŨÁòËáµÄÌå»ýÆ«´ó£¬Å¨¶ÈÆ«´ó£¬¹ÊÕýÈ·£»B¡¢ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴµÓºóÓÃ1.2mol/LµÄÁòËáÈóÏ´£¬Ôì³ÉÁòËáÔö¶à£¬Å¨¶ÈÆ«´ó£»C¡¢×ªÒÆÈÜҺʱ£¬²»É÷ÓÐÉÙÁ¿ÈÜÒºÈ÷³ö£¬ÈÜÖÊÓÐËðʧ£¬Å¨¶ÈƫС£»D¡¢¶¨ÈÝʱÑöÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏߣ¬ÈÜÒºµÄÌå»ýÆ«´ó£¬Å¨¶ÈƫС£»E¡¢½«Ï¡ÊͺóµÄÏ¡ÁòËáÁ¢¼´×ªÈëÈÝÁ¿Æ¿ÇÒ½øÐкóÃæµÄʵÑé²Ù×÷£¬Ã»ÓÐÀäÈ´£¬µ±ÀäÈ´ºóÈÜÒºµÄÌå»ýƫС£¬Å¨¶ÈÆ«´ó£»F¡¢¶¨Èݺ󣬰ÑÈÝÁ¿Æ¿µ¹ÖÃÒ¡ÔÈ£¬Õý·Åºó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÓÖ²¹³ä¼¸µÎÕôÁóË®ÖÁ¿Ì¶ÈÏߣ¬ÈÜÒºÌå»ýÆ«´ó£¬Å¨¶ÈƫС¡£¹ÊÑ¡ABE¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÖйúÕþ¸®³Ðŵ£¬µ½2020Ä꣬µ¥Î»GDP¶þÑõ»¯Ì¼ÅŷűÈ2005ÄêϽµ40%¡«50%¡£CO2¿Éת»¯³ÉÓлúÎïʵÏÖ̼ѭ»·£¬ÓÐЧ½µµÍ̼ÅÅ·Å¡£

£¨1£©ÔÚÌå»ýΪ1 LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬³äÈë1 mol CO2ºÍ3 mol H2£¬Ò»¶¨Ìõ¼þÏ·´Ó¦£ºCO2(g)£«3H2(g)CH3OH(g)£«H2O(g)¡¡¦¤H£½£­49.0 kJmol£­1£¬²âµÃCO2ºÍCH3OH(g)Ũ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ¡£

¢Ù¼ÆË㣺´Ó0minµ½3min£¬H2µÄƽ¾ù·´Ó¦ËÙÂÊv(H2)£½_______mol¡¤L£­1¡¤min£­1£¬·´Ó¦ÖÁƽºâʱ£¬·Å³öµÄÈÈÁ¿ÊÇ_______kJ¡£

¢ÚÄÜ˵Ã÷ÉÏÊö·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ________(Ìî±àºÅ)¡£

A£®c(CO2) ¡Ãc(CH3OH)£½1¡Ã1

B£®»ìºÏÆøÌåµÄÃܶȲ»ÔÙ·¢Éú±ä»¯

C£®µ¥Î»Ê±¼äÄÚÏûºÄ3mol H2£¬Í¬Ê±Éú³É1mol H2O

D£®CO2µÄÌå»ý·ÖÊýÔÚ»ìºÏÆøÌåÖб£³Ö²»±ä

¢Û·´Ó¦´ïµ½Æ½ºâºó£¬±£³ÖÆäËûÌõ¼þ²»±ä£¬Äܼӿ췴ӦËÙÂÊÇÒʹÌåϵÖÐÆøÌåµÄÎïÖʵÄÁ¿¼õÉÙ£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ________(Ìî±àºÅ)¡£

A£®Éý¸ßÎÂ¶È B£®ËõСÈÝÆ÷Ìå»ý

C£®ÔÙ³äÈëCO2ÆøÌå D£®Ê¹ÓúÏÊʵĴ߻¯¼Á

£¨2£©ÒÑÖªA(g)£«B(g) C(g)£«D(g) ¦¤H£¬·´Ó¦µÄƽºâ³£ÊýºÍζȵĹØϵÈçÏ£º

ζÈ/ ¡æ

700

800

830

1000

1200

ƽºâ³£Êý

1.7

1.1

x

0.6

0.4

»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ù¸Ã·´Ó¦ÊÇ_____·´Ó¦£¨Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±)¡£

¢Ú830¡æʱ£¬ÏòÒ»¸ö2LµÄÃܱÕÈÝÆ÷ÖгäÈë0.20molµÄAºÍ0.80molµÄB£¬·´Ó¦ÖÁ10sʱ´ïµ½Æ½ºâ£¬²âµÃAµÄת»¯ÂÊΪ80%£¬¼ÆËã830¡æʱ£¬·´Ó¦µÄƽºâ³£Êýx£½____¡£

¡¾ÌâÄ¿¡¿ÊµÑéÊÒÓûÓÃNaOH¹ÌÌåÅäÖÆ1.0 mol/LµÄNaOHÈÜÒº240 mL£»

(1)ÅäÖÆÈÜҺʱ£¬Ò»°ã¿ÉÒÔ·ÖΪÒÔϼ¸¸ö²½Ö裺

¢Ù³ÆÁ¿¡¡¢Ú¼ÆËã¡¡¢ÛÈܽ⡡¢ÜÒ¡ÔÈ¡¡¢ÝתÒÆ¡¡¢ÞÏ´µÓ¡¡¢ß¶¨ÈÝ¡¡¢àÀäÈ´¡¡

ÆäÕýÈ·µÄ²Ù×÷˳ÐòΪ__________________________________£¬

±¾ÊµÑé±ØÐëÓõ½µÄÒÇÆ÷ÓÐÌìƽ¡¢Ò©³×¡¢²£Á§°ô¡¢ÉÕ±­¡¢____________________¡£

(2)ijͬѧÓû³ÆÁ¿NaOHµÄÖÊÁ¿£¬ËûÏÈÓÃÍÐÅÌÌìƽ³ÆÁ¿ÉÕ±­µÄÖÊÁ¿£¬ÌìƽƽºâºóµÄ״̬ÈçÏÂͼËùʾ¡£ÉÕ±­µÄʵ¼ÊÖÊÁ¿Îª________g£¬ÒªÍê³É±¾ÊµÑé¸ÃͬѧӦ³Æ³ö________g NaOH¡£

(3)ÏÂÁжÔÈÝÁ¿Æ¿¼°ÆäʹÓ÷½·¨µÄÃèÊöÖÐÕýÈ·µÄÊÇ£¨____£©

A£®ÈÝÁ¿Æ¿ÉϱêÓÐÈÝ»ý¡¢Î¶ȺÍŨ¶È

B£®ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴ¾»ºó£¬±ØÐëºæ¸É

C£®ÅäÖÆÈÜҺʱ£¬°ÑÁ¿ºÃµÄŨÁòËáСÐĵ¹ÈëÈÝÁ¿Æ¿ÖУ¬¼ÓÈëÕôÁóË®µ½½Ó½ü¿Ì¶ÈÏß1¡«2 cm´¦£¬¸ÄÓýºÍ·µÎ¹ÜµÎ¼ÓÕôÁóË®ÖÁ¿Ì¶ÈÏß

D£®Ê¹ÓÃÇ°Òª¼ì²éÈÝÁ¿Æ¿ÊÇ·ñ©ˮ

(4)ÔÚÅäÖƹý³ÌÖУ¬ÆäËû²Ù×÷¶¼ÊÇÕýÈ·µÄ£¬ÏÂÁвÙ×÷»áÒýÆðÎó²îÆ«¸ßµÄÊÇ_______________¡£

¢ÙתÒÆÈÜҺʱ²»É÷ÓÐÉÙÁ¿È÷µ½ÈÝÁ¿Æ¿ÍâÃæ

¢Ú¶¨ÈÝʱ¸©Êӿ̶ÈÏß

¢ÛδÀäÈ´µ½ÊÒξͽ«ÈÜҺתÒƵ½ÈÝÁ¿Æ¿²¢¶¨ÈÝ

¢Ü¶¨ÈݺóÈûÉÏÆ¿Èû·´¸´Ò¡ÔÈ£¬¾²Öúó£¬ÒºÃæµÍÓڿ̶ÈÏߣ¬ÔÙ¼ÓË®ÖÁ¿Ì¶ÈÏß

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø