ÌâÄ¿ÄÚÈÝ
15£®£¨1£©¸ÃŨÑÎËáÖÐHClµÄÎïÖʵÄÁ¿Å¨¶ÈΪ11.8mol•L-1£®
£¨2£©È¡ÓÃÈÎÒâÌå»ýµÄ¸ÃÑÎËáÈÜҺʱ£¬ÏÂÁÐÎïÀíÁ¿Öв»ËæËùÈ¡Ìå»ýµÄ¶àÉÙ¶ø±ä»¯µÄÊÇBD£®
A£®ÈÜÖʵÄCl${\;}^{_}$ÎïÖʵÄÁ¿
B£®ÈÜÒºµÄŨ¶È
C£®ÈÜÒºÖÐCl-µÄÊýÄ¿
D£®ÈÜÒºµÄÃܶÈ
£¨3£©Ä³Ñ§ÉúÓûÓÃÉÏÊöŨÑÎËáºÍÕôÁóË®ÅäÖÆ450mLÎïÖʵÄÁ¿Å¨¶ÈΪ0.4mol•L-1µÄÏ¡ÑÎËᣮ¸ÃѧÉúÐèÒªÁ¿È¡16.9 mLÉÏÊöŨÑÎËá½øÐÐÅäÖÆ£¨½á¹û±£ÁôСÊýµãºóһ룩£¬ÈôÓÃÁ¿Í²Á¿È¡Å¨ÑÎËáʱ¸©ÊÓ¹Û²ì°¼ÒºÃæ£¬ÔòËùÅäÖÆµÄÏ¡ÑÎËáÎïÖʵÄÁ¿Å¨¶È½«Æ«µÍ£¨Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©£¬ÏÂͬ£¬Èô¶¨ÈÝÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÓÖ²¹¼ÓË®ÖÁ¿Ì¶ÈÏߣ¬ÔòËùÅäÖÆµÄÏ¡ÑÎËáÎïÖʵÄÁ¿Å¨¶È½«Æ«µÍ£®
£¨4£©È¡100mL0.4mol•L-1µÄÑÎËáÓë100mL0.1mol•L-1µÄAgNO3ÈÜÒº»ìºÏ£¬»ìºÏºóµÄÌå»ý¿É½üËÆÎªÁ½ÈÜÒºµÄÌå»ýÖ®ºÍ£¬ÔòËùµÃÈÜÒºÖеÄCl-ÎïÖʵÄÁ¿Å¨¶ÈΪ0.15mol/L£®
·ÖÎö £¨1£©ÒÀ¾ÝC=$\frac{1000¦Ñ¦Ø}{M}$¼ÆËãŨÑÎËáµÄÎïÖʵÄÁ¿Å¨¶È£»
£¨2£©¸ù¾Ý¸ÃÎïÀíÁ¿ÊÇ·ñÓÐÈÜÒºµÄÌå»ýÓйØÅжϣ»
£¨3£©ÅäÖÆ450mLÎïÖʵÄÁ¿Å¨¶ÈΪ0.4mol•L-1µÄÏ¡ÑÎËᣬӦѡÔñ500mLÈÝÁ¿Æ¿£¬ÒÀ¾ÝÈÜҺϡÊ͹ý³ÌÖÐÈÜÖʵÄÎïÖʵÄÁ¿²»±ä¼ÆËãÐèҪŨÑÎËáµÄÌå»ý£»·ÖÎö²Ù×÷¶ÔÈÜÖʵÄÎïÖʵÄÁ¿ºÍÈÜÒºÌå»ýµÄÓ°Ï죬ÒÀ¾ÝC=$\frac{n}{V}$½øÐÐÎó²î·ÖÎö£»
£¨4£©100mL 0.4mol•L-1µÄÑÎËáÖÐHClµÄÎïÖʵÄÁ¿n=CV=0.4mol/L¡Á0.1L=0.04mol£¬100mL 0.1mol•L-1µÄAgNO3ÈÜÒºÖÐAgNO3µÄÎïÖʵÄÁ¿n=CV=0.1mol/L¡Á0.1L=0.01mol£¬Á½Õß»ìºÏºó·¢Éú·´Ó¦£ºAg++Cl-=AgCl¡ý£¬¸ù¾ÝÁ½ÕßµÄÁ¿À´·ÖÎöCl-µÄŨ¶È£®
½â´ð ½â£º£¨1£©ÖÊÁ¿·ÖÊý36.5%£¬ÃܶÈΪ1.18g/mLµÄÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈC=$\frac{1000¡Á1.18¡Á36.5%}{36.5}$=11.8mol/L£»
¹Ê´ð°¸Îª£º11.8£»
£¨2£©A£®ÈÜÒºÖÐHClµÄÎïÖʵÄÁ¿=nV£¬ËùÒÔÓëÈÜÒºµÄÌå»ýÓйأ¬¹ÊA²»Ñ¡£»
B£®ÈÜÒº¾ßÓоùÒ»ÐÔ£¬ÈÜҺŨ¶ÈÓëÈÜÒºµÄÌå»ýÎ޹أ¬¹ÊBÑ¡£»
C£®ÈÜÒºÖÐCl-µÄÊýÄ¿=nNA=CVNA£¬ËùÒÔÓëÈÜÒºµÄÌå»ýÓйأ¬¹Êc²»Ñ¡£»
D£®ÈÜÒºµÄÃܶÈÓëÈÜÒºµÄÌå»ýÎ޹أ¬¹ÊDÑ¡£»
¹ÊÑ¡BD£»
£¨3£©ÅäÖÆ450mLÎïÖʵÄÁ¿Å¨¶ÈΪ0.4mol•L-1µÄÏ¡ÑÎËᣬӦѡÔñ500mLÈÝÁ¿Æ¿£¬ÉèÐèҪŨÑÎËáÌå»ýΪV£¬ÔòÒÀ¾ÝÈÜҺϡÊ͹æÂÉ£¬V¡Á11.8mol/L=0.4mol/L¡Á500mL£¬½âµÃV=16.9mL£»
ÈôÓÃÁ¿Í²Á¿È¡Å¨ÑÎËáʱ¸©ÊÓ¹Û²ì°¼ÒºÃæ£¬µ¼ÖÂÁ¿È¡µÄŨÑÎËáÌå»ýƫС£¬ÂÈ»¯ÇâµÄÎïÖʵÄÁ¿Æ«Ð¡£¬ÈÜҺŨ¶ÈÆ«µÍ£»Èô¶¨ÈÝÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÓÖ²¹¼ÓË®ÖÁ¿Ì¶ÈÏߣ¬µ¼ÖÂÈÜÒºÌå»ýÆ«´ó£¬ÈÜҺŨ¶ÈÆ«µÍ£¬
¹Ê´ð°¸Îª£º16.9£»Æ«µÍ£»Æ«µÍ£»
£¨4£©100mL 0.4mol•L-1µÄÑÎËáÖÐHClµÄÎïÖʵÄÁ¿n=CV=0.4mol/L¡Á0.1L=0.04mol£¬100mL 0.1mol•L-1µÄAgNO3ÈÜÒºÖÐAgNO3µÄÎïÖʵÄÁ¿n=CV=0.1mol/L¡Á0.1L=0.01mol£®
¸ù¾ÝÁ½Õß»ìºÏºó·¢Éú·´Ó¦£ºAg++Cl-=AgCl¡ý¿ÉÖª£¬HCl¹ýÁ¿£¬¹Ê·´Ó¦ºóÈÜÒºÖеÄn£¨Cl-£©=0.04mol-0.01mol=0.03mol£¬¹Ê·´Ó¦ºóµÄŨ¶Èc£¨Cl-£©=$\frac{0.03mol}{0.2L}$=0.15mol/L£¬
¹Ê´ð°¸Îª£º0.15mol/L£®
µãÆÀ ±¾Ì⿼²éÁËÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÅäÖÆ£¬Ã÷È·ÅäÖÆÔÀíºÍ²Ù×÷²½ÖèÊǽâÌâ¹Ø¼ü£¬×¢ÒâÎó²î·ÖÎöµÄ·½·¨ºÍ¼¼ÇÉ£¬ÌâÄ¿ÄѶȲ»´ó£®
¢ÙFeCl3ÈÜÒºÓëCu µÄ·´Ó¦¡¡¢Ú½«FeCl3ÈÜÒº¼ÓÈÈÕô¸É£¬²¢×ÆÉÕ×îÖյõ½Fe2O3
¢ÛFeCl3ÈÜÒºÓëKI µÄ·´Ó¦¡¡¢Ü±¥ºÍFeCl3ÈÜÒºµÎÈë·ÐË®ÖÐÖÆ±¸Fe£¨OH£©3½ºÌå
¢ÝFeCl3ÈÜÒºÓëH2S µÄ·´Ó¦¡¡¢ÞFeCl3ÈÜÒºÓëNaHCO3ÈÜÒºµÄ·´Ó¦
¢ßÅäÖÆFeCl3ÈÜÒºÐè¼ÓÈëÒ»¶¨Á¿µÄÑÎËᣨ¡¡¡¡£©
| A£® | ¢Ù¢Ü¢Þ | B£® | ¢Ú¢Û¢Ý¢Þ | C£® | ¢Ú¢Ü¢Þ¢ß | D£® | ¢Ù¢Ú¢Û¢Ü¢Ý¢Þ¢ß |
| A£® | ½« CO2 ͨÈë´ÎÂÈËá¸ÆÈÜÒº¿ÉÉú³É´ÎÂÈËá | |
| B£® | ÄÆÔÚÑõÆøÖÐȼÉÕÖ÷ÒªÉú³ÉNa2O | |
| C£® | Na2O¡¢Na2O2 ×é³ÉÔªËØÏàͬ£¬Óë CO2 ·´Ó¦²úÎïÒ²Ïàͬ | |
| D£® | ÐÂÖÆÂÈË®ÏÔËáÐÔ£¬ÏòÆäÖеμÓÉÙÁ¿×ÏɫʯÈïÊÔÒº£¬³ä·ÖÕñµ´ºóÈÜÒº³ÊºìÉ« |
| A£® | Àë×ӵϹÔÐÔÇ¿Èõ£ºFe2+£¾Br-£¾Cl- | |
| B£® | µ±ÂÈÆøÉÙÁ¿Ê±£¬·¢ÉúµÄÀë×Ó·´Ó¦£º2Fe2++Cl2¨T2Fe3++2Cl- | |
| C£® | µ±a=bʱ£¬·´Ó¦ºóÈÜÒºµÄÀë×ÓŨ¶È£ºc£¨Fe3+£©£ºc£¨Br-£©£ºc£¨Cl-£©=1£º2£º2 | |
| D£® | µ±ÂÈÆø¹ýÁ¿Ê±£¬·¢ÉúÀë×Ó·´Ó¦£º2Fe2++4Br-+3Cl2¨T2Fe3++2Br2+6Cl- |
| A£® | ÒÒϩͨÈëËáÐÔ¸ßÃÌËá¼ØÈÜÒºÖÐ | |
| B£® | ¹âÕÕÉä¼×ÍéÓëÂÈÆøµÄ»ìºÏÆøÌå | |
| C£® | ÔÚÄø×÷´ß»¯¼ÁµÄÌõ¼þÏ£¬±½ÓëÇâÆø·´Ó¦ | |
| D£® | ±½ÓëÒºäå»ìºÏºóÈöÈëÌú·Û |
¢Ù¼ÓÈÈÊÔ¹Üʱ£¬ÏȾùÔȼÓÈÈ£¬ºó¾Ö²¿¼ÓÈÈ
¢Ú×öH2»¹ÔCuOʵÑéʱ£¬ÏÈͨH2£¬ºó¼ÓÈÈCuO£¬·´Ó¦Íê±Ïºó£¬Ïȳ·¾Æ¾«µÆ´ýÊÔ¹ÜÀäÈ´£¬ºóֹͣͨH2
¢ÛÖÆÈ¡ÆøÌåʱ£¬Ïȼì²é×°ÖÃÆøÃÜÐÔ£¬ºó×°Ò©Æ·
¢Üµãȼ¿ÉȼÐÔÆøÌåÈçH2¡¢COµÈʱ£¬ÏȼìÑ鯸Ìå´¿¶È£¬ºóµãȼ£®
| A£® | ¢Ù¢Ú | B£® | ¢Ù¢Ú¢Û | C£® | ¢Ú¢Û¢Ü | D£® | È«²¿ |
¢Ù
| A£® | 0ÖÖ | B£® | 1ÖÖ | C£® | 2ÖÖ | D£® | 3ÖÖ |
HIn £¨ÈÜÒº£©?H+£¨ÈÜÒº£©+In-£¨ÈÜÒº£©
ºìÉ« »ÆÉ«
Ũ¶ÈΪ0.02mol/LµÄÏÂÁÐÈÜÒº £¨1£©ÑÎËá £¨2£©NaOHÈÜÒº £¨3£©NaHSO4ÈÜÒº £¨4£©NaHCO3ÈÜÒº£¨5£©°±Ë®£¬ÆäÖÐÄÜʹָʾ¼ÁÏÔºìÉ«µÄÊÇ£¨¡¡¡¡£©
| A£® | £¨1£©£¨4£©£¨5£© | B£® | £¨2£©£¨5£© | C£® | £¨1£©£¨3£© | D£® | £¨2£©£¨3£©£¨5£© |