¡¾ÌâÄ¿¡¿ÓÃËáʽµÎ¶¨¹Ü׼ȷÒÆÈ¡25.00mLijδ֪Ũ¶ÈµÄÑÎËáÈÜÓÚÒ»½à¾»µÄ׶ÐÎÆ¿ÖУ¬È»ºóÓÃ0.20mol¡¤L -1µÄÇâÑõ»¯ÄÆÈÜÒº(ָʾ¼ÁΪ·Ó̪).µÎ¶¨½á¹ûÈçÏ£º
NaOHÆðʼ¶ÁÊý | NaOHÖÕµã¶ÁÊý | |
µÚÒ»´Î | 0.10mL | 18.60mL |
µÚ¶þ´Î | 0.30mL | 18.00mL |
£¨1£©¸ù¾ÝÒÔÉÏÊý¾Ý¿ÉÒÔ¼ÆËã³öÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ mol¡¤L-1£®
£¨2£©´ïµ½µÎ¶¨ÖÕµãµÄ±êÖ¾ÊÇ
£¨3£©ÒÔϲÙ×÷Ôì³É²â¶¨½á¹ûÆ«¸ßµÄÔÒò¿ÉÄÜÊÇ ¡£
A£®ÅäÖƱê×¼ÈÜÒºµÄÇâÑõ»¯ÄÆÖлìÓÐNa2CO3ÔÓÖÊ
B£®µÎ¶¨ÖÕµã¶ÁÊýʱ£¬¸©Êӵζ¨¹ÜµÄ¿Ì¶È£¬ÆäËü²Ù×÷¾ùÕýÈ·
C£®Ê¢×°Î´ÖªÒºµÄ׶ÐÎÆ¿ÓÃÕôÁóˮϴ¹ý£¬Î´Óôý²âÒºÈóÏ´
D£®µÎ¶¨µ½ÖÕµã¶ÁÊýʱ·¢Ïֵζ¨¹Ü¼â×ì´¦Ðü¹ÒÒ»µÎÈÜÒº
E£®Î´Óñê×¼ÒºÈóÏ´¼îʽµÎ¶¨¹Ü
¡¾´ð°¸¡¿£¨1£©0.1448 mol/L £¨2£©ÎÞÉ«±ä·ÛºìÉ«ÇÒ30s²»±äÉ« £¨3£©ADE
¡¾½âÎö¡¿
ÊÔÌâ·ÖÎö£º£¨1£©¸ù¾Ý±í¸ñÖÐÊý¾Ý¿ÉÖª£¬µÚÒ»´ÎµÎ¶¨ÏûºÄÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ýΪ£º(18.60-0.10)mL=18.50mL£¬µÚ¶þ´ÎµÎ¶¨ÏûºÄÇâÑõ»¯ÄƵÄÌå»ýΪ£º(18.00-0.30)mL=17.70mL£¬ÏûºÄÇâÑõ»¯ÄÆÌå»ýµÄƽ¾ùֵΪmL=18.10 mL£¬
c(NaOH)=(0.20mol/L¡Á0.01810L)¡Â0.025L=0.1448 mol/L£»
£¨2£©µÎ¶¨½áÊø֮ǰ£¬ÈÜҺΪÎÞÉ«£¬µÎ¶¨½áÊøʱÈÜÒº±ä³ÉºìÉ«£¬ËùÒԵζ¨ÖÕµãÏÖÏóΪ£ºµÎ×îºóÒ»µÎÈÜÒºÓÉÎÞÉ«±ädzºìÉ«£¬°ë·ÖÖÓ²»ÍÊÉ«£»£¨3£©A£®ÅäÖƱê×¼ÈÜÒºµÄÇâÑõ»¯ÄÆÖлìÓÐNa2CO3ÔÓÖÊ£¬µ¼ÖÂV(NaOH)Æ«¸ß£¬¾Ýc(´ý²â)=·ÖÎö£¬c(´ý²â)Æ«¸ß£¬¹ÊAÕýÈ·£»B£®µÎ¶¨ÖÕµã¶ÁÊýʱ£¬¸©Êӵζ¨¹ÜµÄ¿Ì¶È£¬ÆäËü²Ù×÷¾ùÕýÈ·£¬µ¼ÖÂV(NaOH)Æ«µÍ£¬¾Ýc(´ý²â)=
·ÖÎö£¬c(´ý²â)Æ«µÍ£¬¹ÊB´íÎó£»C£®Ê¢×°Î´ÖªÒºµÄ׶ÐÎÆ¿ÓÃÕôÁóˮϴ¹ý£¬Î´Óôý²âÒºÈóÏ´£¬²Ù×÷ÕýÈ·£¬¶Ô½á¹ûÎÞÓ°Ï죬¹ÊC´íÎó£»D£®µÎ¶¨µ½ÖÕµã¶ÁÊýʱ·¢Ïֵζ¨¹Ü¼â×ì´¦Ðü¹ÒÒ»µÎÈÜÒº£¬µ¼ÖÂV(NaOH)Æ«¸ß£¬¾Ýc(´ý²â)=
·ÖÎö£¬c(´ý²â)Æ«¸ß£¬¹ÊDÕýÈ·£»E£®Î´Óñê×¼ÒºÈóÏ´¼îʽµÎ¶¨¹Ü£¬±ê×¼ÒºµÄŨ¶È½µµÍ£¬ÏûºÄ±ê×¼ÒºµÄÌå»ýÔö´ó£¬µ¼Ö´ý²âŨ¶ÈÆ«¸ß£¬¹ÊEÕýÈ·£»´ð°¸ÎªADE¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ºü¸Õ×ÓÔÚ¡°Á¶Ê¯µ¨È¡¾«»ª·¨¡±ÖÐ×÷ÈçÏÂÐðÊö£º¡°ÒÔÊ¿“ô(¼´×©Å÷)ÀÝ×÷Á½¸ö·½Í·Â¯£¬Ò»Â¯ÖÐÖøÍÅÌ£¬Ê¹¶¨£¬¼´ÃÜÄàÖ®£¬Ò»Â¯ÖÐÒÔÌ¿ÉÕʯµ¨(CuSO4¡¤5H2O)ʹ×÷ÑÌ£¬ÒÔÎïÉÈÖ®£¬Æ侫»ª¾¡ÈëÍÅÌ¡£Â¯ÖÐÈ´»ð´ýÀä¡¡¡±¡£ºü¸Õ×ÓËùÈ¡¡°¾«»ª¡±Îª
A. H2O B. SO2 C. Ï¡ÁòËá D. CuSO4ÈÜÒº
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿X¡¢Y¡¢Z¡¢P¡¢Q¡¢W¡¢R¡¢TÊÇÔªËØÖÜÆÚ±í¶ÌÆÚÖÐ×î³£¼ûµÄÖ÷×åÔªËØ£¬ÇÒÔ×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÏà¹ØÐÅÏ¢ÈçÏÂ±í£º
ÔªËØ | Ïà¹ØÐÅÏ¢ |
X | ÆäijһÖÖͬλËØ£¬±»ÓÃ×÷Ïà¶ÔÔ×ÓÖÊÁ¿µÄ±ê×¼ |
Y | Æä×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïÄÜÓëÆäÆø̬Ç⻯Îï·¢Éú»¯ºÏ·´Ó¦ |
P | ÊǶÌÖÜÆÚÖУ¨³ýÏ¡ÓÐÆøÌåÍ⣩Ô×Ӱ뾶×î´óµÄÔªËØ |
Q | ´æÔÚÖÊÁ¿ÊýΪ25¡¢ÖÐ×ÓÊýΪ13µÄºËËØ |
W | λÓÚÖÜÆÚ±íµÚ13ÁÐ |
R | ÓëZͬ×壬 ÇÒ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄŨÈÜÒº³£ÎÂÏÂÓëW µ¥ÖÊ»á³öÏÖ¶Û»¯ÏÖÏó |
(1)WÔªËØλÓÚÖÜÆÚ±íµÚ________ÖÜÆÚ______×壻TÀë×ӵĽṹʾÒâͼΪ____________¡£
(2)½«Q¡¢Z¡¢RµÄ¼òµ¥Àë×ӵİ뾶´Ó´óµ½Ð¡ÅÅÐò___________(ÓÃÀë×Ó·ûºÅ±íʾ£©£»R¡¢TÁ½ÖÖÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄËᣨ»ò¼î£©ÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòΪ__________(Óû¯Ñ§Ê½±íʾ)¡£
(3) X¡¢Y¡¢ZÈýÖÖÔªËصļòµ¥Ç⻯ÎïÖУ¬Îȶ¨ÐÔ×îÇ¿µÄ»¯ºÏÎïµÄµç×ÓʽΪ_________£¬»¯ºÏÎïP2RÈÜÒºÔÚ¿ÕÆøÖг¤ÆÚ·ÅÖ㬻áºÍÑõÆø·´Ó¦Éú³ÉÓë¹ýÑõ»¯ÄƵĽṹÏàËƵÄÎïÖÊP2R2£¬ÆäÈÜÒºÏÔ»ÆÉ«£¬ÔòP2R2µÄµç×ÓʽΪ______________¡£
(4)¸ù¾ÝϱíÖеÄÐÅÏ¢£¬ÅжÏÏÂÁл¯ºÏÎï¹Ì̬ʱµÄ¾§ÌåÀàÐÍ£¨Ìî¡°Àë×Ó¾§Ì塱¡¡µÈ£©¡£
»¯ºÏÎï | x3Y4 | R2T2 |
Ö÷ÒªÎïÀíÐÔÖÊ | Ó²¶È¿ÉÒԺͽð¸ÕʯÏàæÇÃÀ£¬ÈÛµã>3600¡æ | ³£ÎÂÏÂÊÇÒ»ÖֳȻÆÉ«ÒºÌ壬ÈÛµãΪ-76¡ãC |
¾§ÌåÀàÐÍ | ________ | _________ |
(5)QµÄµ¥ÖÊÄÜÓëX¡¢ZÐγɵÄijÖÖ»¯ºÏÎï·´Ó¦£¬Ð´³öÆ仯ѧ·½³Ìʽ£º______________¡£
(6)XÓëQÐγɵÄlmol»¯ºÏÎïQ2X3ÓëË®·´Ó¦£¬Éú³É2molMg(OH)2ºÍ1molijÆø̬Ìþ£¬¸ÃÌþ·Ö×ÓÖÐ̼ÇâÖÊÁ¿±ÈΪ9: 1£¬Ð´³ö¸ÃÆø̬ÌþµÄ½á¹¹¼òʽ_______£¬²¢Ð´³öQ2X3ÓëË®·´Ó¦µÄ»¯Ñ§·½³Ìʽ________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A. ÆûÓÍ¡¢²ñÓͺÍÖ²ÎïÓͶ¼ÊÇ̼Ç⻯ºÏÎȼÉÕÖ»Éú³ÉCO2ºÍH2O
B. ÓÍÖ¬Ç⻯¿ÉÓÃÓÚÖÆ×÷ÈËÔì»ÆÓÍ
C. ±ûÈý´¼³£ÓÃÓÚÖÆ×÷Æû³µ·À¶³Òº
D. µ°°×ÖÊÊǽöÓÉ̼¡¢Çâ¡¢ÑõÔªËØ×é³ÉµÄÎïÖÊ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÊÂʵÓ뽺ÌåÐÔÖÊÎ޹صÄÊÇ
A£®Óü¤¹â±ÊÕÕÉäÖó·Ð¹ýµÄFeCl3ÈÜÒº¿É¹Û²ìµ½Ã÷ÁÁ¹â·
B£®´óÎíÌìÆøÓÃÆû³µÔ¶¹âµÆÕÕÉäÇ°·½³£³öÏÖÃ÷ÁÁ¹â·
C£®Ã÷·¯¿ÉÓÃ×÷¾»Ë®¼Á
D£®³ôÑõ¡¢Cl2¾ù¿ÉÓÃÓÚË®µÄÏû¶¾¾»»¯
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨ £©
A.1 mol H2OµÄÖÊÁ¿Îª18g/mol
B.CH4µÄĦ¶ûÖÊÁ¿Îª16g
C.6.02¡Á1023¸öSO2·Ö×ÓµÄÖÊÁ¿Îª64g
D.±ê×¼×´¿öÏ£¬1 molÈκÎÎïÖÊÌå»ý¾ùΪ22.4L
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÎïÖÊÖÐËùº¬µÄÔÓÖÊ£¨À¨ºÅÖÐÊÇÔÓÖÊ£©£¬Ê¹ÓÃNaOHÈÜÒº²»ÄܳýÈ¥µÄÊÇ
A. MgO¹ÌÌ壨Al2O3£© B. Na2CO3¹ÌÌ壨NaHCO3£©
C. FeCl2ÈÜÒº£¨CuCl2£© D. Fe2O3¹ÌÌ壨SiO2£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÎïÖÊÊôÓÚ´¿¾»ÎïµÄÊÇ£¨ £©
A.ÂÈË®
B.Ư°×·Û
C.ÒºÂÈ
D.СËÕ´òÈÜÒº
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿
£¨1£©ÅäÖÆ100 mL 0.10 mol¡¤L£1NaOH±ê×¼ÈÜÒº¡£
¢ÙÖ÷Òª²Ù×÷²½Ö裺¼ÆËã¡ú³ÆÁ¿¡úÈܽâ¡ú___ ___(ÀäÈ´ºó)¡úÏ´µÓ(²¢½«Ï´µÓÒºÒÆÈëÈÝÁ¿Æ¿)¡ú___ ___¡ú___ ___¡ú½«ÅäÖƺõÄÈÜÒºµ¹ÈëÊÔ¼ÁÆ¿ÖУ¬ÌùÉϱêÇ©¡£
¢Ú³ÆÁ¿________gÇâÑõ»¯ÄƹÌÌåËùÐèÒÇÆ÷ÓУºÍÐÅÌÌìƽ(´øíÀÂë¡¢Ä÷×Ó)¡¢________¡¢________¡£
£¨2£©È¡20.00 mL´ý²âÑÎËá·ÅÈë׶ÐÎÆ¿ÖУ¬²¢µÎ¼Ó2¡«3µÎ·Ó̪×÷ָʾ¼Á£¬ÓÃÒÑÅäÖƵÄNaOH±ê×¼ÈÜÒº½øÐеζ¨¡£Öظ´ÉÏÊöµÎ¶¨²Ù×÷2¡«3´Î£¬¼Ç¼Êý¾ÝÈçÏ¡£
ʵÑé | NaOHÈÜÒºµÄŨ¶È(mol¡¤L£1) | µÎ¶¨Íê³Éʱ£¬NaOHÈÜÒºµÎÈëµÄÌå»ý(mL) | ´ý²âÑÎËáµÄÌå»ý(mL) |
1 | 0.10 | 22.62 | 20.00 |
2 | 0.10 | 22.72 | 20.00 |
3 | 0.10 | 22.80 | 20.00 |
¢ÙµÎ¶¨´ïµ½ÖÕµãµÄ±êÖ¾ÊÇ___________________________________________________________________¡£
¢Ú¸ù¾ÝÉÏÊöÊý¾Ý£¬¿É¼ÆËã³ö¸ÃÑÎËáµÄŨ¶ÈԼΪ_____________(±£ÁôÁ½Î»ÓÐЧÊý×Ö)¡£
¢ÛÅÅÈ¥¼îʽµÎ¶¨¹ÜÖÐÆøÅݵķ½·¨Ó¦²ÉÓÃÈçͼËùʾ²Ù×÷ÖеÄ________£¬È»ºóÇáÇἷѹ²£Á§Çòʹ¼â×첿·Ö³äÂú¼îÒº¡£
¢ÜÔÚÉÏÊöʵÑéÖУ¬ÏÂÁвÙ×÷(ÆäËû²Ù×÷ÕýÈ·)»áÔì³É²â¶¨½á¹ûÆ«¸ßµÄÓÐ________(Ìî×ÖĸÐòºÅ)¡£
A£®µÎ¶¨ÖÕµã¶ÁÊýʱ¸©ÊÓ
B£®ËáʽµÎ¶¨¹ÜʹÓÃÇ°£¬Ë®Ï´ºóδÓôý²âÑÎËáÈóÏ´
C£®×¶ÐÎƿˮϴºóδ¸ÉÔï
D£®³ÆÁ¿NaOH¹ÌÌåÖлìÓÐNa2CO3¹ÌÌå
E£®¼îʽµÎ¶¨¹Ü¼â×첿·ÖÓÐÆøÅÝ£¬µÎ¶¨ºóÏûʧ
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com