¡¾ÌâÄ¿¡¿ÑÎËá¹ã·ºÓ¦ÓÃÔÚÏ¡ÓнðÊôµÄʪ·¨Ò±½ð¡¢Æ¯È¾¹¤Òµ¡¢½ðÊô¼Ó¹¤¡¢ÎÞ»úÒ©Æ·¼°ÓлúÒ©ÎïµÄÉú²úµÈÁìÓòÖС£HCl ¼«Ò×ÈÜÓÚË®£¬¹¤ÒµÉÏÓà HCl ÆøÌåÈÜÓÚË®µÄ·½·¨ÖÆÈ¡ÑÎËá¡£

£¨1£©Óà 12.0mol/L ŨÑÎËáÅäÖà 230mL 0.3mol/L µÄÏ¡ÑÎËᣬÐèÒªÁ¿È¡Å¨ÑÎËáµÄÌå»ýΪ___mL£»

£¨2£©ÈÜҺϡÊ͹ý³ÌÖÐÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢²£Á§°ô¡¢Á¿Í²¡¢____________¡¢___________£»

£¨3£©ÈÜҺϡÊ͹ý³ÌÖÐÓÐÒÔϲÙ×÷£º

a£®Á¿È¡Å¨ÑÎËáºÍÒ»¶¨Ìå»ýµÄË®£¬ÔÚÉÕ±­ÖÐÏ¡ÊÍ

b£®¼ÆËãËùÐèŨÑÎËáµÄÌå»ý

c£®ÉÏϵߵ¹Ò¡ÔÈ

d£®¼ÓÕôÁóË®ÖÁ¿Ì¶ÈÏß 1-2cm µØ·½£¬¸ÄÓýºÍ·µÎ¹Ü¼ÓÕôÁóË®ÖÁ°¼ÒºÃæÓë¿Ì¶ÈÏßÏàÇÐ

e£®½«Ï¡ÊÍҺתÒÆÈëÈÝÁ¿Æ¿£¬Ï´µÓÉÕ±­ºÍ²£Á§°ô£¬²¢½«Ï´µÓҺתÒÆÈëÈÝÁ¿Æ¿£¬Õñµ´

ÒÔÉÏÕýÈ·µÄ²Ù×÷˳ÐòΪ____________________________________________(ÌîÐòºÅ)£»

£¨4£©ÊµÑé¹ý³ÌÖеÄÒÔϲÙ×÷»áµ¼ÖÂ×îÖÕËùÅäÈÜҺŨ¶È£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±£©

a£®Á¿È¡Å¨ÑÎËáʱ¸©ÊÓ£º______________________£»

b£®Á¿È¡Å¨ÑÎËáºó£¬ÇåÏ´ÁËÁ¿Í²²¢½«Ï´µÓҺתÒÆÈëÈÝÁ¿Æ¿£º______________________£»

c£®ÊµÑéÇ°£¬ÈÝÁ¿Æ¿ÖÐÓÐÉÙÁ¿²ÐÁôÕôÁóË®£º______________________£»

£¨5£©±ê×¼×´¿ö£¬1L Ë®ÖÐͨÈë aL HCl ÆøÌ壬ºöÂÔÑÎËáÈÜÒºÖÐ HCl µÄ»Ó·¢£¬µÃµ½µÄÑÎËáÈÜÒºÃܶÈΪ b g/mL£¬ÎïÖʵÄÁ¿Å¨¶ÈΪ ______________________mol/L¡£

¡¾´ð°¸¡¿6.3 250mLÈÝÁ¿Æ¿ ½ºÍ·µÎ¹Ü b¡¢a¡¢e¡¢d¡¢c ƫС Æ«´ó ²»±ä

¡¾½âÎö¡¿

(1)ÒÀ¾ÝÅäÖÆÈÜÒºÌå»ýÑ¡ÔñÈÝÁ¿Æ¿µÄ¹æ¸ñ£¬ÔÙÒÀ¾ÝÈÜҺϡÊ͹ý³ÌÖÐÈÜÖʵÄÎïÖʵÄÁ¿²»±ä¼ÆËãÐèҪŨÑÎËáµÄÌå»ý£»

(2)ÒÀ¾ÝÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÒ»°ã²½ÖèÑ¡ÔñÐèÒªÒÇÆ÷£»

(3)ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÒ»°ã²½Ö裺¼ÆËã¡¢Á¿È¡¡¢Ï¡ÊÍ¡¢ÀäÈ´¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔÈ¡¢×°Æ¿Ìù±êÇ©£¬¾Ý´ËÅÅÐò£»

(4)·ÖÎö²Ù×÷¶ÔÈÜÖʵÄÎïÖʵÄÁ¿ºÍÈÜÒºÌå»ýµÄÓ°Ï죬ÒÀ¾Ýc=½øÐÐÎó²î·ÖÎö£»

(5)Ïȸù¾Ýn=¼ÆËã³ö±ê×¼×´¿öÏÂaLÂÈ»¯ÇâµÄÎïÖʵÄÁ¿£¬È»ºó¸ù¾Ým=nM¼ÆËã³öÂÈ»¯ÇâµÄÖÊÁ¿£¬1LË®µÄÖÊÁ¿Ô¼Îª1000g£¬´Ó¶ø¿ÉÖªÈÜÒºÖÊÁ¿£¬ÔÙ¸ù¾ÝV=¼ÆËã³öÈÜÒºµÄÌå»ý£¬×îºó¸ù¾Ýc=¼ÆËã³ö¸ÃÑÎËáµÄÎïÖʵÄÁ¿Å¨¶È¡£

(1)ÓÃ12.0mol/LŨÑÎËáÅäÖÃ230mL 0.3mol/LµÄÏ¡ÑÎËᣬӦѡÔñ250mLÈÝÁ¿Æ¿£¬ÉèÐèҪŨÑÎËáÌå»ýV£¬ÔòÒÀ¾ÝÈÜҺϡÊ͹ý³ÌÖÐÈÜÖʵÄÎïÖʵÄÁ¿²»±ä¼ÆËãµÃ£º12.0mol/L¡ÁV=0.250L¡Á0.3mol/L£¬½âµÃV=0.0063L=6.3mL£¬¹Ê´ð°¸Îª£º6.3£»

(2)ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÒ»°ã²½Ö裺¼ÆËã¡¢Á¿È¡¡¢Ï¡ÊÍ¡¢ÀäÈ´¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔÈ£¬Óõ½µÄÒÇÆ÷£ºÁ¿Í²¡¢½ºÍ·µÎ¹Ü¡¢ÉÕ±­¡¢²£Á§°ô¡¢ÈÝÁ¿Æ¿£¬ÅäÖÆ230mL 0.3mol/LµÄÏ¡ÑÎËᣬӦѡÔñ250mLÌå»ýÈÝÁ¿Æ¿£¬»¹È±ÉÙµÄÒÇÆ÷£º250mLÈÝÁ¿Æ¿¡¢½ºÍ·µÎ¹Ü£¬¹Ê´ð°¸Îª£º250mLÈÝÁ¿Æ¿£»½ºÍ·µÎ¹Ü£»

(3)ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÒ»°ã²½Ö裺¼ÆËã¡¢Á¿È¡¡¢Èܽâ»òÏ¡ÊÍ¡¢ÀäÈ´¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔÈ¡¢×°Æ¿Ìù±êÇ©£¬ËùÒÔÕýÈ·µÄ˳ÐòΪ£ºbaedc£¬¹Ê´ð°¸Îª£ºbaedc£»

(4)a£®Á¿È¡Å¨ÑÎËáʱ¸©ÊÓ£¬µ¼ÖÂÁ¿È¡µÄŨÑÎËáÌå»ýƫС£¬ÈÜÖʵÄÎïÖʵÄÁ¿Æ«Ð¡£¬ÈÜҺŨ¶ÈƫС£¬¹Ê´ð°¸Îª£ºÆ«Ð¡£»

b£®Á¿È¡Å¨ÑÎËáºó£¬ÇåÏ´ÁËÁ¿Í²²¢½«Ï´µÓҺתÒÆÈëÈÝÁ¿Æ¿£¬µ¼ÖÂÁ¿È¡µÄŨÑÎËáÌå»ýÆ«´ó£¬ÈÜÖʵÄÎïÖʵÄÁ¿Æ«´ó£¬ÈÜҺŨ¶ÈÆ«´ó£»¹Ê´ð°¸Îª£ºÆ«´ó£»

c£®ÊµÑéÇ°£¬ÈÝÁ¿Æ¿ÖÐÓÐÉÙÁ¿²ÐÁôÕôÁóË®£¬¶ÔÈÜÖʵÄÎïÖʵÄÁ¿ºÍÈÜÒºÌå»ý¶¼²»²úÉúÓ°Ï죬ÈÜҺŨ¶È²»±ä£»¹Ê´ð°¸Îª£º²»±ä£»

(5)±ê×¼×´¿öϵÄaLÂÈ»¯ÇâÆøÌåµÄÎïÖʵÄÁ¿Îª£ºn(HCl)= =mol£¬¸ÃHClµÄÖÊÁ¿Îª£º36.5g/mol¡Ámol=g£¬1LË®µÄÖÊÁ¿Ô¼Îª1000g£¬Ôò¸ÃÑÎËáÖÊÁ¿Îª£º1000g+g£¬¸ÃÑÎËáµÄÌå»ýΪ£º=mL=L£¬ËùÒÔ¸ÃÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ£ºc(HCl)= =mol/L£¬¹Ê´ð°¸Îª£º ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿C1O2×÷ΪһÖÖÇ¿Ñõ»¯¼Á£¬Êǹú¼ÊÉϹ«ÈϵĸßЧÏû¶¾Ãð¾ú¼Á£¬µ«ÒòÆäÒ×±¬Óж¾£¬³£ÓÃNaClO2Ìæ´ú¡£³£ÎÂÏ£¬½«NaOH¹ÌÌå¼ÓÈëµ½ÓÉ0.1molC1O2ÈÜÓÚË®Åä³ÉµÄ1LÈÜÒºÖС£ÈÜÒºpH¼°²¿·Ö×é·Öº¬Á¿±ä»¯ÇúÏßÈçͼ£¬ÏÂÁÐÐðÊö²»ÕýÈ·µÄÊÇ

(ÒÑÖª£º2C1O2+H2OHC1O2+H++C1O3-)

A. Ka(HClO2)¡Ö10-4.5

B. ²»ÄÜÓÃpHÊÔÖ½²â¸ÃÈÜÒºµÄpH

C. ËáÐÔ£ºHClO2<HClO3£¬Ñõ»¯ÐÔ£ºHClO2>HClO3

D. ¸ÃͼÏñÉÏÈκÎÒ»µã£¬¶¼ÓÐc£¨C1O2-£©+c£¨HC1O2£©+c£¨C1O3-£©=0.1mol¡¤L-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁз´Ó¦ÐèÒª¼ÓÈ뻹ԭ¼Á²ÅÄÜ·¢ÉúµÄÊÇ£¨ £©

A.Cl¡úCl2B.Fe3+¡úFe2+C.SO32¡úSO42-D.CO32¡úCO2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿³ýÈ¥»ìÔÚ̼ËáÄÆ·ÛÄ©ÖеÄÉÙÁ¿Ì¼ËáÇâÄÆ£¬×îºÏÀíµÄ·½·¨ÊÇ£¨¡¡¡¡£©
A.¼ÓÈÈ
B.¼ÓÈëÇâÑõ»¯ÄÆÈÜÒº
C.¼ÓÈëÑÎËá
D.¼ÓÈëCaCl2ÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ò»ÖÖ´Óº¬ÂÁï®îÜ·ÏÁÏ[ÂÁ²­¡¢CoOºÍCo2O3(Ö»ÈÜÓÚËᣬ²»ÈÜÓÚ¼î)¼°LiCoO2]ÖлØÊÕÑõ»¯îܵŤÒÕÁ÷³ÌÈçÏÂ:

»Ø´ðÏÂÁÐÎÊÌâ:

(1)²½ÖèI¡°¼îÈÜ¡±Ê±£¬ÎªÌá¸ßÂÁµÄÈ¥³ýÂÊ£¬³ýÉý¸ßζȺͲ»¶Ï½Á°èÍâ,»¹¿É²ÉÈ¡µÄ´ëÊ©ÊÇ____(ÁоÙ1µã)£¬¡°¼îÈÜ¡±Ê±·¢ÉúÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______£»²½ÖèII¡°¹ýÂË¡¢Ï´µÓ¡±¹ý³ÌÖÐÂËÔü»¹»áÓÐÉÙÁ¿µÄAl(OH)3£¬ÊÇÒòΪ________(ÌîÀë×Ó·½³Ìʽ)¡£

(2)²½ÖèIII¡°ËáÈÜ¡±Ê±,Co2O3 ת»¯ÎªCoSO4 µÄÀë×Ó·½³ÌʽΪ_________¡£

(3)²½ÖèV¡°³ýÂÁºÍﮡ±Ê±,µ÷½ÚpHµÄÊÊÓ÷¶Î§ÊÇ_______(ÒÑÖª¸ÃÌõ¼þÏÂ,Al3+¿ªÊ¼³ÁµíʱµÄpHΪ4.1,³ÁµíÍêȫʱµÄpHΪ4.7.Co2+¿ªÊ¼³ÁµíʱµÄpHΪ6.9¡£³ÁµíÍêȫʱµÄpHΪ9.4)£»²½ÖèVIËùµÃÂËÔüµÄ³É·ÖΪ__________¡£

(4)ìÑÉÕCoC2O4ʱ·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬ÔòCoC2O4·Ö½âµÄ»¯Ñ§·½³ÌʽÊÇ___£»Èô½ö´Ó³Áµíת»¯½Ç¶È¿¼ÂÇ£¬ÄÜ·ñÀûÓ÷´Ó¦CoCO3+ C2O42-=== CoC2O4 + CO32- ½«CoCO3 ת»¯ÎªCoC2O4?___ (Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±),˵Ã÷ÀíÓÉ:________[ÒÑÖªKsp(CoCO3) =1.4¡Á10-13,Ksp(CoC2O4)=6.3¡Á10-8]

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¶þïÌú[(C5H5)2Fe](³È»ÆÉ«·ÛÄ©£¬²»ÈÜÓÚË®£¬Ò×ÈÜÓÚÒÒÃѵÈÓлúÈܼÁ)¼°ÆäÑÜÉúÎïÔÚ¹¤Òµ¡¢Å©Òµ¡¢Ò½Ò©¡¢º½Ìì¡¢½ÚÄÜ¡¢»·±£µÈÐÐÒµ¾ßÓй㷺µØÓ¦Óá£Ä³Ð£¿ÎÍâС×éÒÀ¾Ý·´Ó¦2KOH+2C5H6+FeCl2=(C5H5)2Fe+2KCl+2H2OÔÚ¾ø¶ÔÎÞË®¡¢ÎÞÑõµÄÌõ¼þÏÂÖƱ¸¶þïÌú¡£

»Ø´ðÏÂÁÐÎÊÌâ:

(1)¼××éͬѧÄâÖƱ¸ÎÞË®FeCl2,Ö÷ҪʵÑéÁ÷³ÌΪ:

²½ÖèIÓñ¥ºÍNa2CO3ÈÜÒº½þÅݵÄÄ¿µÄÊÇ_______£»²½ÖèII ÖÐÌúмÊǹýÁ¿µÄ,ÆäÄ¿µÄÊÇ_____£»²½ÖèIVÍÑË®µÄ·½·¨ÊÇ____________¡£

(2)ÒÒ×éÓû·Îì¶þÏ©¶þ¾ÛÌå(·ÐµãΪ174¡æ)ÖƱ¸»·Îì¶þÏ©(·ÐµãΪ419¡æ),ÒÑÖª:(C5H6)2(»·Îì¶þÏ©¶þ¾ÛÌå)2C5H6(»·Îì¶þÏ©)¡£·ÖÀëµÃµ½»·Îì¶þÏ©µÄ²Ù×÷·½·¨Îª________¡£

(3)±û×éͬѧÖƱ¸¶þïÌúµÄ×°ÖÃÈçͼËùʾ(¼Ð³Ö×°ÖÃÒÑÂÔÈ¥)¡£

ÒÑÖª:¶þ¼×»ùÑÇí¿µÄ½á¹¹Ê½Îª£¬ÈÈÎȶ¨ÐԺã¬ÄÜÈܽâ´ó¶àÊýÓлúÎï¡£

¢ÙͼÖÐÀäÈ´Ë®´Ó½Ó¿Ú______½øÈë(Ìî¡°a¡±»ò¡°b¡±)¡£

¢Ú×°Ò©Æ·Ç°¼°Õû¸ö¹ý³ÌÐèͨÈë¸ÉÔïµÄA£¬×°Ò©Æ·Ç°Í¨ÈëN2µÄÄ¿µÄÊÇ_________¡£

¢Û·´Ó¦ºó·ÖÀë³öÉϲã³È»ÆÉ«ÒÒÃÑÇåÒº£¬ÏÈÓÃÑÎËáÏ´µÓ£¬ÆäÄ¿µÄÊÇ_______£»ÔÙÓÃˮϴ£¬Ë®Ï´Ê±ÄÜ˵Ã÷ÒÑÏ´µÓ¸É¾»µÄÒÀ¾ÝÊÇ_________£»Ï´µÓºóµÃµ½µÄ¶þïÌúÒÒÃÑÈÜÒº»ñµÃ¶þïÌú¹ÌÌå¿É²ÉÓõIJÙ×÷·½·¨ÊÇ_____________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚAl2(SO4)3ºÍMgSO4µÄ»ìºÏÈÜÒºÖУ¬µÎ¼ÓNaOHÈÜÒº£¬Éú³É³ÁµíµÄÁ¿ÓëµÎÈëNaOHÈÜÒºµÄÌå»ý¹ØϵÈçÓÒͼËùʾ£¬ÔòÔ­»ìºÏÒºÖÐAl2(SO4)3ÓëMgSO4µÄÎïÖʵÄÁ¿Å¨¶ÈÖ®±ÈΪ( )

A. 6£º1 B. 3£º1 C. 2£º1 D. 1£º2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿C¡¢TiµÄµ¥Öʼ°Æ仯ºÏÎïÔÚÏÖ´úÉç»áÓй㷺ÓÃ;¡£

£¨1£©»ù̬îÑÔ­×ӵĵç×ÓÅŲ¼Ê½Îª________¡£

£¨2£©CS2·Ö×ÓÖк¬ÓЦҼüºÍ¦Ð¼üÖ®±ÈΪ______£»NO2+ÓëCO2Êǵȵç×ÓÌ壬NO2+µÄµç×ÓʽΪ___£¬¼ü½ÇΪ_______¡£

£¨3£©CH3CHO·ÐµãµÍÓÚCH3CH2OHµÄÔ­ÒòÊÇ_____£»CH3CHO·Ö×ÓÖÐ̼ԭ×ÓÔÓ»¯ÀàÐÍΪ_____¡£

£¨4£©îÑËá±µ(BaTiO3)¾§ÌåµÄijÖÖ¾§°ûÈçͼËùʾ¡£NAΪ°¢·ü¼ÓµÂÂÞ³£ÊýÖµ£¬Ba2+¡¢O2-¡¢Ti4+µÄ°ë¾¶·Ö±ðΪapm¡¢bpm¡¢cpm¡£

¢ÙÓë±µÀë×ӵȾàÀëÇÒ×î½üµÄÑõÀë×ÓÓÐ______¸ö£»

¢Ú¼ÙÉ辧ÌåÖеÄTi4+¡¢Ba2+·Ö±ðÓëO2-»¥Ïà½Ó´¥£¬Ôò¸Ã¾§ÌåµÄÃܶȱí´ïʽΪ______g.cm-3¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿´¿¼îÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬ÔÚÒ½Ò©¡¢Ò±½ð¡¢»¯¹¤¡¢Ê³Æ·µÈÁìÓò±»¹ã·ºÊ¹Óá£

I. Óô¿¾»µÄ̼ËáÄƹÌÌåÅäÖÆ500mL 0.40mol/L Na2CO3ÈÜÒº¡£

(1)³ÆÈ¡Na2CO3¹ÌÌåµÄÖÊÁ¿ÊÇ______________________g¡£

(2)ÅäÖÆÈÜҺʱ£¬½øÐÐÈçϲÙ×÷£¬°´ÕÕ²Ù×÷˳Ðò£¬µÚ4²½ÊÇ_________(Ìî×Öĸ)¡£

a. ¶¨ÈÝ b. ¼ÆËã c. Èܽâ d. Ò¡ÔÈ e. תÒÆ f. Ï´µÓ g. ³ÆÁ¿

(3)ÏÂÁÐ˵·¨ÖУ¬ÕýÈ·µÄÊÇ_____________________(Ìî×Öĸ)¡£

a. ¶¨ÈÝʱ£¬ÑöÊӿ̶ÈÏߣ¬»áµ¼ÖÂÅäÖƵÄÈÜҺŨ¶ÈƫС

b. ¶¨ÈÝʱ£¬Èç¹û¼ÓË®³¬¹ý¿Ì¶ÈÏߣ¬ÒªÓõιÜÎü³ö

c. תÒÆʱ£¬ÈÜÒºµ¹³öÈÝÁ¿Æ¿Í⣬ҪÖØÐÂÅäÖÆÈÜÒº

d. Ò¡ÔȺó£¬ÒºÃæµÍÓڿ̶ÈÏߣ¬ÒªÔÙ¼ÓË®ÖÁ¿Ì¶ÈÏß

II. ijʵÑéС×éµÄͬѧģÄâºîµÂ°ñÖƼÖÆÈ¡´¿¼î£¬Á÷³ÌÈçÏ£º

(1)¹¤ÒµÉú²ú´¿¼îµÄµÚÒ»²½ÊdzýÈ¥±¥ºÍʳÑÎË®µÄÖÐSO42¨D¡¢Ca2+Àë×Ó£¬ÒÀ´Î¼ÓÈëµÄÊÔ¼Á¼°ÆäÓÃÁ¿ÊÇ ______________¡¢ _______________¡¢ (¹ýÂË)¡¢ _______________¡£

(2)ÒÑÖª£º¼¸ÖÖÑεÄÈܽâ¶È

NaCl

NH4HCO3

NaHCO3

NH4Cl

Èܽâ¶È(20¡ãC£¬100gH2Oʱ)

36.0

21.7

9.6

37.2

¢Ùд³ö×°ÖÃIÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ________________________________________¡£

¢Úд³ö×°ÖÃIIÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ________________________________¡£

(3)¸ÃÁ÷³ÌÖпÉÑ­»·ÀûÓõÄÎïÖÊÊÇ__________________¡£

(4)ÖƳöµÄ´¿¼îÖÐÖ»º¬ÓÐÔÓÖÊNaCl¡£

¢Ù¼ìÑéÓøô¿¼îÅäÖƵÄÈÜÒºÖк¬ÓÐCl¨DµÄ·½·¨ÊÇ_________________________¡£

¢Ú²â¶¨¸Ã´¿¼îµÄ´¿¶È£¬ÏÂÁз½°¸ÖпÉÐеÄÊÇ__________(Ìî×Öĸ)¡£

a. Ïòm¿Ë´¿¼îÑùÆ·ÖмÓÈë×ãÁ¿CaCl2ÈÜÒº£¬³Áµí¾­¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ³ÆÆäÖÊÁ¿Îªb g

b. Ïòm¿Ë´¿¼îÑùÆ·ÖмÓÈë×ãÁ¿Ï¡ÑÎËᣬÓüîʯ»Ò(Ö÷Òª³É·ÖÊÇCaOºÍNaOH)ÎüÊÕ²úÉúµÄÆøÌ壬¼îʯ»ÒÔöÖØb g

c. Ïòm¿Ë´¿¼îÑùÆ·ÖмÓÈë×ãÁ¿AgNO3ÈÜÒº£¬²úÉúµÄ³Áµí¾­¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ³ÆÆäÖÊÁ¿Îªb g

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸