ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÊµÑéÊÒÓÃNaOH¹ÌÌåÅäÖÆ240mL 1.2mol/LµÄNaOHÈÜÒº£¬Ìî¿ÕÇë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÅäÖÆ240mL 1.2mol/LµÄNaOHÈÜÒº

Ó¦³ÆÈ¡NaOHµÄÖÊÁ¿/g

Ñ¡ÓÃÈÝÁ¿Æ¿µÄ¹æ¸ñ/mL

³ýÈÝÁ¿Æ¿Í⻹ÐèÒªµÄÆäËü²£Á§ÒÇÆ÷

_________

________

_______

£¨2£©ÈÝÁ¿Æ¿ÉÏÐè±êÓÐÒÔÏÂÎåÏîÖеÄ_____________________£»

¢ÙŨ¶È ¢Ú ÎÂ¶È ¢ÛÈÝÁ¿ ¢Üѹǿ ¢Ý¿Ì¶ÈÏß

£¨3£©ÅäÖÆʱ£¬ÆäÕýÈ·µÄ²Ù×÷˳ÐòÊÇ£¨×Öĸ±íʾ£¬Ã¿¸ö×ÖĸֻÄÜÓÃÒ»´Î£©___£»

A£®½«ÒÑÀäÈ´µÄNaOHÈÜÒºÑز£Á§°ô×¢Èë250mLµÄÈÝÁ¿Æ¿ÖÐ

B£®ÓÃÌìƽ׼ȷ³ÆÈ¡ËùÐèµÄNaOHµÄÖÊÁ¿£¬¼ÓÈëÉÙÁ¿Ë®£¬Óò£Á§°ôÂýÂý½Á¶¯£¬Ê¹Æä³ä·ÖÈܽâ

C£®ÓÃ30mLˮϴµÓÉÕ±­2¡ª3´Î£¬Ï´µÓÒº¾ù×¢ÈëÈÝÁ¿Æ¿£¬Õñµ´

D£®½«ÈÝÁ¿Æ¿¸Ç½ô£¬µßµ¹Ò¡ÔÈ

E£®¸ÄÓýºÍ·µÎ¹Ü¼ÓË®£¬Ê¹ÈÜÒº°¼ÃæÇ¡ºÃÓë¿Ì¶ÈÏßÏàÇÐ

F£®¼ÌÐøÍùÈÝÁ¿Æ¿ÄÚСÐļÓË®£¬Ö±µ½ÒºÃæ½Ó½ü¿Ì¶È1¡ª2cm´¦

G ×°ÈëÌùºÃ±êÇ©µÄÊÔ¼ÁÆ¿´ýÓÃ

£¨4£©ÏÂÁÐÅäÖƵÄÈÜҺŨ¶ÈÆ«µÍµÄÊÇ__£»

A£®³ÆÁ¿NaOHʱ£¬ÔÚÂËÖ½ÉϳÆÁ¿

B£®ÅäÖÆÇ°£¬ÈÝÁ¿Æ¿ÖÐÓÐÉÙÁ¿ÕôÁóË®

C£®¼ÓÕôÁóˮʱ²»É÷³¬¹ýÁ˿̶ÈÏß

D£®ÏòÈÝÁ¿Æ¿ÖÐתÒÆÈÜҺʱ²»É÷ÓÐÒºµÎÈ÷ÔÚÈÝÁ¿Æ¿ÍâÃæ

E ÈÝÁ¿Æ¿ÔÚʹÓÃÇ°ºæ¸É

¡¾´ð°¸¡¿12 250ml£» ÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü ¢Ú¢Û¢Ý BACFEDG ACD

¡¾½âÎö¡¿

£¨1£©¸ù¾Ýn=cv¼ÆËãÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾Ým=nM¼ÆËãËùÐèÇâÑõ»¯ÄƵÄÖÊÁ¿£»ÇâÑõ»¯ÄƾßÓи¯Ê´ÐÔ¡¢Ò׳±½â£¬Ó¦·ÅÔÚÉÕ±­ÖÐѸËÙ³ÆÁ¿£»

£¨2£©ÈÝÁ¿Æ¿ÉϱêÓÐζȡ¢ÈÝÁ¿ºÍ¿Ì¶ÈÏߣ»

£¨3£©ÊµÑéÊÒÅäÖÆ240mL 1.2mol/L NaOHÈÜÒºµÄ²Ù×÷˳ÐòΪ¼ÆËã¡ú³ÆÁ¿¡úÈܽ⡢ÀäÈ´¡úÒÆÒº¡úÏ´µÓ¡ú¶¨ÈÝ¡úÒ¡ÔÈ¡ú×°Æ¿ÌùÇ©£»

£¨4£©ÒÀ¾Ýc=·ÖÎöÈÜÒºÌå»ýºÍÈÜÖʵÄÎïÖʵÄÁ¿ÊÇ·ñ·¢Éú±ä»¯·ÖÎö½â´ð¡£

(1)ʵÑéÊÒûÓÐ240mL¹æ¸ñµÄÈÝÁ¿Æ¿£¬ÅäÖÆ240mL 1.2mol/LµÄNaOHÈÜÒºÐèҪѡÔñ250mlÈÝÁ¿Æ¿£¬ÔòÓ¦³ÆÈ¡NaOHµÄÖÊÁ¿Îª£º0.25L¡Á1.2mol/L¡Á40g/mol=12g£¬³ý250mlÈÝÁ¿Æ¿Í⣬»¹ÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü£¬¹Ê´ð°¸Îª£º12£»250ml£»ÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü£»

£¨2£©ÈÝÁ¿Æ¿ÉϱêÓÐζȡ¢ÈÝÁ¿ºÍ¿Ì¶ÈÏߣ¬¹Ê´ð°¸Îª£º¢Ú¢Û¢Ý£»

£¨3£©ÊµÑéÊÒÅäÖÆ240mL 1.2mol/L NaOHÈÜÒºµÄ²Ù×÷˳ÐòΪ¼ÆËã¡ú³ÆÁ¿¡úÈܽ⡢ÀäÈ´¡úÒÆÒº¡úÏ´µÓ¡ú¶¨ÈÝ¡úÒ¡ÔÈ¡ú×°Æ¿ÌùÇ©£¬ËùÒÔ½«ÉÏÊöʵÑé²½ÖèAµ½G°´ÊµÑé¹ý³ÌÏȺó´ÎÐòÅÅÁÐΪBACFEDG£¬¹Ê´ð°¸Îª£ºBACFEDG£»

£¨4£©A¡¢³ÆÁ¿NaOHʱ£¬ÔÚÂËÖ½ÉϳÆÁ¿£¬ÇâÑõ»¯ÄÆ»áÎüÊÕ¿ÕÆøÖеĶþÑõ»¯Ì¼ºÍË®£¬»áµ¼ÖÂNaOHµÄÎïÖʵÄÁ¿¼õС£¬ËùÅäÈÜҺŨ¶È½«Æ«µÍ£¬¹ÊÕýÈ·£»

B¡¢ÓÉÏ¡ÊͶ¨ÂÉ¿ÉÖª£¬Ï¡ÊÍÇ°ºóNaOHµÄÎïÖʵÄÁ¿²»±ä£¬ÈôÅäÖÆÇ°£¬ÈÝÁ¿Æ¿ÖÐÓÐÉÙÁ¿ÕôÁóË®¶ÔËùÅäÈÜҺŨ¶ÈÎÞÓ°Ï죬¹Ê´íÎó£»

C¡¢¼ÓÕôÁóˮʱ²»É÷³¬¹ýÁ˿̶ÈÏߣ¬µ¼ÖÂÈÜÒºµÄÌå»ýÆ«´ó£¬ËùÅäÈÜҺŨ¶ÈÆ«µÍ£¬¹ÊÕýÈ·£»

D¡¢ÏòÈÝÁ¿Æ¿ÖÐתÒÆÈÜҺʱ²»É÷ÓÐÒºµÎÈ÷ÔÚÈÝÁ¿Æ¿ÍâÃ棬»áµ¼ÖÂNaOHµÄÎïÖʵÄÁ¿¼õС£¬ËùÅäÈÜҺŨ¶È½«Æ«µÍ£¬¹ÊÕýÈ·£»

E¡¢ÈÝÁ¿Æ¿ÔÚʹÓÃÇ°ºæ¸É¶ÔNaOHµÄÎïÖʵÄÁ¿ºÍÈÜÒºµÄÌå»ýÎÞÓ°Ï죬¶ÔËùÅäÈÜҺŨ¶ÈÎÞÓ°Ï죬¹Ê´íÎó£»

ACDÕýÈ·£¬¹Ê´ð°¸Îª£ºACD¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿»¯Ñ§ÊÇÒ»ÃÅÒÔʵÑéΪ»ù´¡µÄѧ¿Æ£¬»¯Ñ§ËùÈ¡µÃµÄ·á˶³É¹ûÓëʵÑéµÄÖØÒª×÷Ó÷ֲ»¿ª¡£½áºÏÏÂÁÐʵÑé×°ÖÃͼ»Ø´ðÎÊÌ⣺

£¨1£©Ð´³öÉÏÊöͼÖÐÒÇÆ÷µÄÃû³Æ£º¢Ù________£»¢Ú________£»¢Ü________¡£

£¨2£©ÈôÀûÓÃ×°Öâñ·ÖÀëÒÒËá(·Ðµã118 ¡æ)ºÍÒÒËáÒÒõ¥(·Ðµã77.1 ¡æ)µÄ»ìºÏÎ»¹È±ÉÙµÄÒÇÆ÷ÓÐ________£¬½«ÒÇÆ÷²¹³äÍêÕûºó½øÐеÄʵÑé²Ù×÷µÄÃû³ÆΪ________£»ÊµÑéʱÒÇÆ÷¢ÚÖÐÀäÈ´Ë®µÄ½ø¿ÚΪ________(Ìî¡°f¡±»ò¡°g¡±)¡£

£¨3£©ÏÖÐèÅäÖÆ250 mL 0.2 mol¡¤L£­1 NaClÈÜÒº£¬×°ÖâòÊÇijͬѧתÒÆÈÜÒºµÄʾÒâͼ£¬Í¼ÖÐÓÐÁ½´¦´íÎó·Ö±ðÊÇ________________£¬________________¡£

£¨4£©ÏÂÁйØÓÚÒÇÆ÷¢ÜµÄʹÓ÷½·¨ÖУ¬ÕýÈ·µÄÊÇ________(ÌîÏÂÁÐÑ¡ÏîµÄ±àºÅ×Öĸ)¡£

A£®Ê¹ÓÃÇ°Ó¦¼ì²éÊÇ·ñ©Һ

B£®Ê¹ÓÃÇ°±ØÐëºæ¸É

C£®²»ÄÜÓÃ×÷ÎïÖÊ·´Ó¦»òÈܽâµÄÈÝÆ÷

D£®ÈÈÈÜÒº¿ÉÖ±½ÓתÒƵ½ÆäÖÐ

£¨5£©ÏÂÁвÙ×÷»áʹÅäÖƵÄÈÜҺŨ¶ÈÆ«¸ßµÄÊÇ________(ÌîÏÂÁÐÑ¡ÏîµÄ±àºÅ×Öĸ)¡£

A£®Ã»Óн«Ï´µÓҺתÒƵ½ÈÝÁ¿Æ¿

B£®×ªÒƹý³ÌÖÐÓÐÉÙÁ¿ÈÜÒº½¦³ö

C£®Ò¡ÔȺó£¬ÒºÃæϽµ£¬²¹³äË®

D£®¶¨ÈÝʱ¸©Êӿ̶ÈÏß

¡¾ÌâÄ¿¡¿¼×ÍéÔÚ¼ÓÈÈÌõ¼þÏ¿ɻ¹Ô­Ñõ»¯Í­£¬ÆøÌå²úÎï³ýË®ÕôÆøÍ⣬»¹ÓÐ̼µÄÑõ»¯Îij»¯Ñ§Ð¡×éÀûÓÃÈçͼװÖÃ̽¾¿Æä·´Ó¦²úÎï¡£

[²éÔÄ×ÊÁÏ]¢ÙCOÄÜÓëÒø°±ÈÜÒº·´Ó¦£ºCO£«2[Ag(NH3)2]£«£«2OH£­===2Ag¡ý£«2NH4+£«CO32£­£«2NH3¡£

¢ÚCu2OΪºìÉ«£¬²»ÓëAg+·´Ó¦£¬ÄÜ·¢Éú·´Ó¦£ºCu2O£«2H£«===Cu2+£«Cu£«H2O¡£

£¨1£©×°ÖÃAÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________________________________¡£

£¨2£©°´ÆøÁ÷·½Ïò¸÷×°ÖôÓ×óµ½ÓÒµÄÁ¬½Ó˳ÐòΪA¡ú__________________¡£(Ìî×Öĸ±àºÅ)

£¨3£©ÊµÑéÖеμÓÏ¡ÑÎËáµÄ²Ù×÷Ϊ______________________________________________¡£

£¨4£©ÒÑÖªÆøÌå²úÎïÖк¬ÓÐCO£¬Ôò×°ÖÃCÖпɹ۲쵽µÄÏÖÏóÊÇ________________£»×°ÖÃFµÄ×÷ÓÃΪ_________________________________________¡£

£¨5£©µ±·´Ó¦½áÊøºó£¬×°ÖÃD´¦ÊÔ¹ÜÖйÌÌåÈ«²¿±äΪºìÉ«¡£

¢ÙÉè¼ÆʵÑéÖ¤Ã÷ºìÉ«¹ÌÌåÖк¬ÓÐCu2O£º______________________________________________¡£

¢ÚÓûÖ¤Ã÷ºìÉ«¹ÌÌåÖÐÊÇ·ñº¬ÓÐCu£¬¼×ͬѧÉè¼ÆÈçÏÂʵÑ飺ÏòÉÙÁ¿ºìÉ«¹ÌÌåÖмÓÈëÊÊÁ¿0.1mol¡¤L1AgNO3ÈÜÒº£¬·¢ÏÖÈÜÒº±äÀ¶£¬¾Ý´ËÅжϺìÉ«¹ÌÌåÖк¬ÓÐCu¡£ÒÒͬѧÈÏΪ¸Ã·½°¸²»ºÏÀí£¬ÓûÖ¤Ã÷¼×ͬѧµÄ½áÂÛ£¬»¹ÐèÔö¼ÓÈç϶ԱÈʵÑ飬Íê³É±íÖÐÄÚÈÝ¡£

ʵÑé²½Öè(²»ÒªÇóд³ö¾ßÌå²Ù×÷¹ý³Ì)

Ô¤ÆÚÏÖÏóºÍ½áÂÛ

__________________

Èô¹Û²ìµ½ÈÜÒº²»±äÀ¶£¬ÔòÖ¤Ã÷ºìÉ«¹ÌÌåÖк¬ÓÐCu£»Èô¹Û²ìµ½ÈÜÒº±äÀ¶£¬Ôò²»ÄÜÖ¤Ã÷ºìÉ«¹ÌÌåÖк¬ÓÐCu

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

¾«Ó¢¼Ò½ÌÍø