ÒÑ֪ˮÔÚ25¡æºÍ95¡æʱ£¬ÆäµçÀëƽºâÇúÏßÈçÓÒͼËùʾ£º

(1)Ôò25¡æʱˮµÄµçÀëƽºâÇúÏßӦΪ________(Ìî¡°A¡±»ò¡°B¡±)£¬Çë˵Ã÷ÀíÓÉ________________________________________________

_____________________________________________________¡£

(2)25¡æʱ£¬½«pH£½9µÄNaOHÈÜÒºÓëpH£½4µÄH2SO4ÈÜÒº»ìºÏ£¬ÈôËùµÃ»ìºÏÈÜÒºµÄpH£½7£¬ÔòNaOHÈÜÒºÓëpH£½4µÄH2SO4ÈÜÒºµÄÌå»ý±ÈΪ__________¡£

(3)95¡æʱ£¬Èô100Ìå»ýpH1£½aµÄijǿËáÈÜÒºÓë1Ìå»ýpH2£½bµÄijǿ¼îÈÜÒº»ìºÏºóÈÜÒº³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°£¬¸ÃÇ¿ËáµÄpH1ÓëÇ¿¼îµÄpH2Ö®¼äÓ¦Âú×ãµÄ¹ØϵÊÇ____________¡£

(4)ÇúÏßB¶ÔӦζÈÏ£¬pH£½2µÄijHAÈÜÒººÍpH£½10µÄNaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬»ìºÏÈÜÒºµÄpH£½5¡£Çë·ÖÎöÆäÔ­Òò£º______________________________________

________________________________________________________________________¡£


½âÎö£º±¾ÌâµÄ¹Ø¼üÊǸãÇå³þζȶÔË®µÄµçÀëƽºâ¡¢Ë®µÄÀë×Ó»ýºÍÈÜÒºpHµÄÓ°Ïì¡£

(1)µ±Î¶ÈÉý¸ßʱ£¬´Ù½øË®µÄµçÀ룬ˮµÄÀë×Ó»ýÒ²Ôö´ó£¬Ë®ÖÐÇâÀë×ÓŨ¶È¡¢ÇâÑõ¸ùÀë×ÓŨ¶È¶¼Ôö´ó£¬Ë®µÄpH¼õС£¬µ«ÈÜÒºÈÔÈ»³ÊÖÐÐÔ¡£Òò´Ë½áºÏͼÏñÖÐA¡¢BÇúÏ߱仯Çé¿ö¼°ÇâÀë×ÓŨ¶È¡¢ÇâÑõ¸ùÀë×ÓŨ¶È¿ÉÒÔÅжϣ¬25¡æʱˮµÄµçÀëƽºâÇúÏßӦΪA£¬ÀíÓÉΪˮµÄµçÀëÊÇÎüÈȹý³Ì£¬Éý¸ßζȣ¬Ë®µÄµçÀë³Ì¶ÈÔö´ó¡£

(2)25¡æʱËùµÃ»ìºÏÈÜÒºµÄpH£½7£¬ÈÜÒº³ÊÖÐÐÔ¼´Ëá¼îÇ¡ºÃÖкͣ¬¼´n(OH£­)£½n(H£«)£¬ÔòV(NaOH)¡¤10£­5 mol/L£½V(H2SO4)¡¤10£­4 mol/L£¬µÃV(NaOH)¡ÃV(H2SO4)£½10¡Ã1¡£

(3)ҪעÒâÊÇ95¡æʱ£¬Ë®µÄÀë×Ó»ýΪ1¡Á10£­12£¬¼´c(H£«)¡¤c(OH£­)£½1¡Á10£­12£¬ÔòµÈÌå»ýÇ¿Ëᡢǿ¼î·´Ó¦ÖÁÖÐÐÔʱpH(Ëá)£«pH(¼î)£½12¡£¸ù¾Ý95¡æʱ»ìºÏºóÈÜÒº³ÊÖÐÐÔ£¬pH2£½bµÄijǿ¼îÈÜÒºÖÐc(OH£­)£½10b£­12£»ÓÐ100¡Á10£­a£½1¡Á10b£­12£¬¿ÉµÃ10£­a£«2£½10b£­12£¬ËùÒÔ£¬ÓÐÒÔϹØϵ£ºa£«b£½14»òpH1£«pH2£½14¡£

(4)ÔÚÇúÏßB¶ÔӦζÈÏ£¬ÒòpH(Ëá)£«pH(¼î)£½12£¬¿ÉµÃËá¼îÁ½ÈÜÒºÖÐc(H£«)£½c(OH£­)£¬ÈçÊÇÇ¿Ëᡢǿ¼î£¬Á½ÈÜÒºµÈÌå»ý»ìºÏºóÈÜÒºÓ¦³ÊÖÐÐÔ£»ÏÖ»ìºÏÈÜÒºµÄpH£½5£¬¼´µÈÌå»ý»ìºÏºóÈÜÒºÏÔËáÐÔ£¬ËµÃ÷H£«ÓëOH£­ÍêÈ«·´Ó¦ºóÓÖÓÐеÄH£«²úÉú£¬¼´Ëá¹ýÁ¿£¬ËùÒÔ˵ËáHAÊÇÈõËá¡£

´ð°¸£º(1)A¡¡Ë®µÄµçÀëÊÇÎüÈȹý³Ì£¬Î¶ȵÍʱ£¬µçÀë³Ì¶ÈС£¬c(H£«)¡¢c(OH£­)С

(2)10¡Ã1¡¡(3)a£«b£½14»òpH1£«pH2£½14

(4)ÇúÏßB¶ÔÓ¦95¡æ£¬´ËʱˮµÄÀë×Ó»ýΪ1¡Á10£­12£¬HAΪÈõËᣬHAÖкÍNaOHºó£¬»ìºÏÈÜÒºÖл¹Ê£Óà½Ï¶àµÄHA·Ö×Ó£¬¿É¼ÌÐøµçÀë³öH£«£¬Ê¹ÈÜÒºpH£½5


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÓйØÐðÊö£ºÄÜ˵Ã÷M±ÈNµÄ·Ç½ðÊôÐÔÇ¿µÄÐðÊöÊÇ                            

¢Ù·Ç½ðÊôµ¥ÖÊMÄÜ´ÓNµÄ»¯ºÏÎïÖÐÖû»³ö·Ç½ðÊôµ¥ÖÊN¡£ 

¢ÚMÔ­×Ó±ÈNÔ­×ÓÈÝÒ׵õ½µç×Ó¡£

¢Ûµ¥ÖÊM¸úH2·´Ó¦±ÈN¸úH2·´Ó¦ÈÝÒ׵öࡣ  

¢ÜÆø̬Ç⻯ÎïË®ÈÜÒºµÄËáÐÔHmM£¾HnN¡£

¢Ý×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄËáÐÔHmMOx£¾HnNOy¡£

¢ÞÈÛµãM£¾N¡£

A.¢Ù¢Ú¢Û¢Ý                           B.¢Ú¢Ý     

C.¢Ù¢Ú¢Û¢Ü¢Ý                         D.È«²¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÔÏ·´Ó¦×î·ûºÏÂÌÉ«»¯Ñ§Ô­×Ó¾­¼ÃÐÔÒªÇóµÄÊÇ(¡¡¡¡)

A£®ÒÒÏ©¾ÛºÏΪ¾ÛÒÒÏ©¸ß·Ö×Ó²ÄÁÏ

B£®¼×ÍéÓëÂÈÆøÖƱ¸Ò»Âȼ×Íé

C£®ÒÔÍ­ºÍŨÏõËáΪԭÁÏÉú²úÏõËáÍ­

D£®ÓÃSiO2ÖƱ¸¸ß´¿¹è

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ijѧÉúÓûÓÃÒÑÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÑÎËáÀ´²â¶¨Î´ÖªÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒº£¬Ñ¡Ôñ¼×»ù³È×÷ָʾ¼Á¡£ÇëÌîдÏÂÁпհףº

(1)Óñê×¼µÄÑÎËáµÎ¶¨´ý²âµÄNaOHÈÜҺʱ£¬×óÊÖÎÕËáʽµÎ¶¨¹ÜµÄ»îÈû£¬ÓÒÊÖÒ¡¶¯×¶ÐÎÆ¿£¬ÑÛ¾¦×¢ÊÓ___________________________________________________¡£

Ö±µ½Òò¼ÓÈëÒ»µÎÑÎËáºó£¬ÈÜÒºÓÉ»ÆÉ«±äΪ³ÈÉ«£¬²¢________________Ϊֹ¡£

(2)ÏÂÁвÙ×÷ÖпÉÄÜʹËù²âNaOHÈÜÒºµÄŨ¶ÈÊýֵƫµÍµÄÊÇ________¡£

A£®ËáʽµÎ¶¨¹ÜδÓñê×¼ÑÎËáÈóÏ´¾ÍÖ±½Ó×¢Èë±ê×¼ÑÎËá

B£®µÎ¶¨Ç°Ê¢·ÅNaOHÈÜÒºµÄ׶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»ºóûÓиÉÔï

C£®ËáʽµÎ¶¨¹ÜÔڵζ¨Ç°ÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ

D£®¶ÁÈ¡ÑÎËáÌå»ýʱ£¬¿ªÊ¼ÑöÊÓ¶ÁÊý£¬µÎ¶¨½áÊøʱ¸©ÊÓ¶ÁÊý

(3)ÈôµÎ¶¨¿ªÊ¼ºÍ½áÊøʱ£¬ËáʽµÎ¶¨¹ÜÖеÄÒºÃæÈçͼËùʾ£¬ÔòÆðʼ¶ÁÊýΪ________mL£¬ÖÕµã¶ÁÊýΪ__________mL£»

ËùÓÃÑÎËáÈÜÒºµÄÌå»ýΪ______mL¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ(¡¡¡¡)

A£®ÎÞÂÛÊÇ´¿Ë®£¬»¹ÊÇËáÐÔ¡¢¼îÐÔ»òÖÐÐÔÏ¡ÈÜÒº£¬ÔÚ³£ÎÂÏ£¬Æäc(H£«)¡¤c(OH£­)£½1¡Á10£­14

B£®c(H£«)µÈÓÚ1¡Á10£­7 mol/LµÄÈÜÒº²»Ò»¶¨ÊÇÖÐÐÔÈÜÒº

C£®0.2 mol/L CH3COOHÈÜÒºÖеÄc(H£«)ÊÇ0.1 mol/L CH3COOHÈÜÒºÖеÄc(H£«)µÄ2±¶

D£®ÈκÎŨ¶ÈµÄÈÜÒº¶¼¿ÉÒÔÓÃpHÀ´±íʾÆäËáÐÔµÄÇ¿Èõ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÑÖª1 gÇâÆøÍêȫȼÉÕÉú³ÉË®ÕôÆøʱ·Å³öÈÈÁ¿121 kJ£¬ÇÒÑõÆøÖÐ1 mol O=O¼üÍêÈ«¶Ï  ÁÑʱÎüÊÕÈÈÁ¿496 kJ£¬ÇâÆøÖÐ1 mol H¨DH¼ü¶ÏÁÑʱÎüÊÕÈÈÁ¿Îª436 kJ£¬ÇóË®ÕôÆøÖÐ1 mol  H¨DO¼üÐγÉʱ·Å³öÈÈÁ¿    (       )

                                                                    

A£® 463kJ      B£® 557 kJ     C£®  486kJ       D£®188 kJ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Èȼ¤»îµç³Ø¿ÉÓÃ×÷»ð¼ý¡¢µ¼µ¯µÄ¹¤×÷µçÔ´¡£Ò»ÖÖÈȼ¤»îµç³ØµÄ»ù±¾

½á¹¹ÈçͼËùʾ£¬ÆäÖÐ×÷Ϊµç½âÖʵÄÎÞË®LiCl-KCl»ìºÏÎïÊÜÈÈÈÛÈں󣬵ç³Ø¼´¿É˲¼äÊä³öµçÄÜ¡£¸Ãµç³Ø×Ü·´Ó¦Îª£ºPbSO4+2LiCl+Ca £½CaCl2+Li2SO4+Pb¡£ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ:  (ÒÑÖª£ºPbµÄÔ­×ÓÁ¿ 207 )       (   )

A£®Õý¼«·´Ó¦Ê½£ºCa+2Cl- - 2e-  £½CaCl2

B£®·Åµç¹ý³ÌÖУ¬Li£« Ïò¸º¼«Òƶ¯

C£®Ã¿×ªÒÆ0.1molµç×Ó£¬ÀíÂÛÉÏÉú³É20.7gPb

D£®³£ÎÂʱ£¬ÔÚÕý¸º¼«¼ä½ÓÉϵçÁ÷±í»ò¼ìÁ÷¼Æ£¬Ö¸Õ벻ƫת

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¸ù¾ÝijÖÖ¹²ÐÔ£¬¿É½«SO3¡¢CO2¹éΪͬÀàÑõ»¯Îï¡£ÏÂÁÐÎïÖÊÖÐÒ²ÊôÓÚÕâÀàÑõ»¯ÎïµÄÊÇ£¨    £©

A£®CaCO3          B£®SO     C£®KMnO4         D£®Na2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐ×ö·¨ÕýÈ·µÄÊÇ

A£®½«Å¨ÏõËá±£´æÔÚÎÞÉ«²£Á§Æ¿ÖÐ

B£®½ðÊôÄƺͼر£´æÔÚúÓÍÖÐ

C£®Na2CO3¿ÉÒÔ±£´æÔÚ²£Á§ÈûµÄ²£Á§Æ¿ÖÐ

D£®NaOH¹ÌÌå·ÅÔÚÂËÖ½ÉϳÆÁ¿

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸