ÌâÄ¿ÄÚÈÝ
19£®µâÊÇÈËÌå²»¿Éȱ·¦µÄÔªËØ£¬ÎªÁË·ÀÖ¹µâȱ·¦£¬ÏÖÔÚÊг¡ÉÏÁ÷ÐÐÒ»ÖÖ¼ÓµâÑΣ¬¾ÍÊÇÔÚ¾«ÑÎÖÐÌí¼ÓÒ»¶¨Á¿µÄKIO3½øÈ¥£®Ä³Ñо¿Ð¡×éΪÁ˼ì²âij¼ÓµâÑÎÖÐÊÇ·ñº¬Óе⣬²éÔÄÁËÓйصÄ×ÊÁÏ£¬·¢ÏÖÆä¼ì²âÔÀíÊÇ£ºKIO3+5KI+3H2SO4=3I2+3H2O+3K2SO4£¨1£©¸Ã·´Ó¦Ñõ»¯²úÎïÓ뻹ԲúÎïÖÊÁ¿±ÈÊÇ5£º1
£¨2£©ÏÈÈ¡ÉÙÁ¿µÄ¼ÓµâÑμÓÕôÁóË®Èܽ⣬Ȼºó¼ÓÈëÏ¡ÁòËáºÍKIÈÜÒº£¬×îºó¼ÓÈëÒ»¶¨Á¿µÄCCl4£¬Õñµ´£¬Õâʱºò£¬¹Û²ìµ½µÄÏÖÏóÊÇÈÜÒº·Ö³ÉÁ½²ã£¬ÉϲãÎÞÉ«£¬Ï²ã³Ê×Ï£¨ºì£©É«£»
£¨3£©Ä³Ñ§ÉúÔÚÒ»´Î·ÖÒº²Ù×÷Öз¢ÏÖÉÏϲãÈÜÒº¶¼ÊÇÎÞÉ«ÒºÌ壬ÎÞ·¨ÖªµÀ·ÖҺ©¶·ÖеÄÒºÌåÄÄÒ»²ãÊÇÓлú²ã£¬ÄÄÒ»²ãÊÇË®²ã£¬ÇëÄãÓüòµ¥µÄ·½·¨°ïËû¼ø±ð³öÀ´£¬Ð´³öÓйز½Öè¼°ÅжϷ½·¨´Ó·ÖҺ©¶·Ï¿ڷųöÉÙÁ¿Ï²ãÒºÌå¼ÓÈëСÊԹܣ¬ÔÙÍùÊÔ¹ÜÖмÓÈëÉÙÁ¿Ë®£¬ÈôÊÔ¹ÜÖÐÒºÌå·Ö²ã£¬ÔòÖ¤Ã÷ϲãÊÇÓлú²ã£»Èô²»·Ö²ã£¬ÔòÖ¤Ã÷ϲãÊÇË®²ã£»£®
£¨4£©ÔÚÈÝÁ¿Æ¿µÄʹÓ÷½·¨ÖУ¬ÏÂÁвÙ×÷ÕýÈ·µÄÊÇAD
A£®Ê¹ÓÃÈÝÁ¿Æ¿Ç°¼ìÑéÊÇ·ñ©ˮ
B£®ÈÝÁ¿Æ¿ÓÃˮϴ¾»ºó£¬ÔÙÓôýÅäÈÜҺϴµÓ
C£®ÅäÖÆÈÜҺʱ£¬ÈôÊÔÑùÊÇÒºÌ壬ÓÃÁ¿Í²È¡ÑùºóÓò£Á§°ôÒýÁ÷µ¹ÈëÈÝÁ¿Æ¿ÖУ¬»ºÂý¼ÓË®ÖÁ¿Ì¶ÈÏß1¡«2cm´¦£¬ÓýºÍ·µÎ¹Ü¼ÓÕôÁóË®ÖÁ¿Ì¶ÈÏߣ®
D£®¸ÇºÃÆ¿Èû£¬ÓÃʳָ¶¥×¡Æ¿Èû£¬ÁíÒ»Ö»ÊÖÍÐסƿµ×£¬°ÑÈÝÁ¿Æ¿·´¸´µ¹×ª¶à´Î£¬Ò¡ÔÈ£®
£¨5£©ÉÏÃæʵÑéÖÐÓõ½Ò»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÏ¡ÁòËᣬÈôÅäÖÆ0.5mol/LµÄÁòËáÈÜÒº450mL£¬ÐèÓÃÁ¿Í²Á¿È¡ÖÊÁ¿·ÖÊýΪ98%¡¢ÃܶÈΪ1.84g/cm3µÄŨÁòËáµÄÌå»ýΪ13.6 mL£¬Èç¹ûʵÑéÊÒÓÐ15mL¡¢20mL¡¢50mLÁ¿Í²£¬Ó¦Ñ¡ÓÃ15 mLÁ¿Í²×îºÃ£®
£¨6£©ÏÂÃæ²Ù×÷Ôì³ÉËùÅäÏ¡ÁòËáÈÜҺŨ¶ÈÆ«¸ßµÄÊÇACD
A£®ÈܽâµÄʱºòÈÜҺûÓÐÀäÈ´µ½ÊÒξÍתÒÆ
B£®×ªÒÆʱûÓÐÏ´µÓÉÕ±¡¢²£Á§°ô
C£®ÏòÈÝÁ¿Æ¿¼ÓË®¶¨ÈÝʱÑÛ¾¦¸©ÊÓÒºÃæ
D£®ÓÃÁ¿Í²Á¿È¡Å¨ÁòËáºóÏ´µÓÁ¿Í²²¢°ÑÏ´µÓҺתÒƵ½ÈÝÁ¿Æ¿
E£®Ò¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÓÖ¼ÓÕôÁóË®ÖÁ¿Ì¶ÈÏߣ®
·ÖÎö £¨1£©KIO3+5KI+3H2SO4¡ú3I2+3K2SO4+3H2OÖУ¬IÔªËصĻ¯ºÏ¼ÛÓÉ+5¼Û½µµÍΪ0£¬IÔªËصĻ¯ºÏ¼ÛÓÉ-1¼ÛÉý¸ßΪ0£¬»¯ºÏ¼ÛÉý¸ßÖµ=»¯ºÏ¼Û½µµÍÖµ=תÒƵç×ÓÊý£¬¸ù¾Ý»¯ºÏ¼Û±ä»¯È·¶¨µç×ÓתÒÆÇé¿ö£»
£¨2£©µâÒ×ÈÜÓÚÓлúÈܼÁ£¬ÔÚËÄÂÈ»¯Ì¼ÔÚµÄÈܽâ¶ÈÔ¶´óÓÚÔÚË®ÖУ¬¼ÓÈëËÄÂÈ»¯Ì¼ÝÍÈ¡Ë®Öеĵ⣬ËÄÂÈ»¯Ì¼ÓëË®²»»¥ÈÜ£¬ÈÜÒº·Ö³ÉÁ½²ã£¬ËÄÂÈ»¯Ì¼µÄÃܶȱÈË®£¬Óлú²ãÔÚϲ㣬µâÈÜÓÚËÄÂÈ»¯Ì¼³Ê×ϺìÉ«£¬Éϲ㼸ºõÎÞÉ«£»
£¨3£©ÀûÓÃÝÍÈ¡¼ÁÓëË®²»»¥ÈܽøÐÐÉè¼Æ£¬´Ó·ÖҺ©¶·Ï¿ڷųöÉÙÁ¿Ï²ãÒºÌå¼ÓÈëСÊÔ¹ÜÖУ¬ÔÙÍùÊÔ¹ÜÖмÓÈëÉÙÁ¿Ë®£¬ÈôÊÔ¹ÜÖÐÒºÌå·Ö²ã£¬ÔòÖ¤Ã÷ϲãÊÇÓлú²ã£»Èô²»·Ö²ã£¬ÔòÖ¤Ã÷ϲãÊÇË®²ã£»
£¨4£©¸ù¾ÝÈÝÁ¿Æ¿µÄʹÓ÷½·¨ºÍ×¢ÒâÊÂÏîÀ´»Ø´ðÎÊÌ⣻
£¨5£©¸ù¾ÝÏ¡ÊÍÇ°ºóÈÜÖʵÄÖÊÁ¿²»±äÁз½³ÌÀ´¼ÆË㣬¸ù¾ÝÁ¿Í²µÄÁ¿È¡¹æÔòÀ´Ñ¡Ôñ£»
£¨6£©·ÖÎö²Ù×÷¶ÔÈÜÖʵÄÎïÖʵÄÁ¿»ò¶ÔÈÜÒºµÄÌå»ýµÄÓ°Ï죬¸ù¾Ýc=$\frac{n}{V}$·ÖÎö¶ÔËùÅäÈÜҺŨ¶ÈµÄÓ°Ï죮
½â´ð ½â£º£¨1£©Ñõ»¯»¹Ô·´Ó¦KIO3+5KI+3H2SO4¨T3K2SO4+3I2+3H2OÖУ¬IÔªËصĻ¯ºÏ¼ÛÓÉ+5¼Û½µµÍΪ0£¬IÔªËصĻ¯ºÏ¼ÛÓÉ-1¼ÛÉý¸ßΪ0£¬»¯ºÏ¼ÛÉý¸ßÖµ=»¯ºÏ¼Û½µµÍÖµ=תÒƵç×ÓÊý=5£¬µç×ÓתÒÆÇé¿öÈçÏ£º£¬KIΪ»¹Ô¼Á£¬+5¼ÛµÄIÔªËر»»¹Ô£¬»¹Ô²úÎïÓëÑõ»¯²úÎï¾ùΪµâ£¬ÓÉÔ×ÓÊغ㼰·´Ó¦¿ÉÖª£¬ÎïÖʵÄÁ¿Ö®±ÈΪ1£º5£¬¹Ê´ð°¸Îª£º5£º1£»
£¨2£©µâÒ×ÈÜÓÚÓлúÈܼÁ£¬ÔÚËÄÂÈ»¯Ì¼ÔÚµÄÈܽâ¶ÈÔ¶´óÓÚÔÚË®ÖУ¬¼ÓÈëËÄÂÈ»¯Ì¼ÝÍÈ¡Ë®Öеĵ⣬ËÄÂÈ»¯Ì¼ÓëË®²»»¥ÈÜ£¬ÈÜÒº·Ö³ÉÁ½²ã£¬ËÄÂÈ»¯Ì¼µÄÃܶȱÈË®£¬Óлú²ãÔÚϲ㣬µâÈÜÓÚËÄÂÈ»¯Ì¼³Ê×ϺìÉ«£¬Éϲ㼸ºõÎÞÉ«£¬¹Ê´ð°¸Îª£ºÈÜÒº·Ö³ÉÁ½²ã£¬ÉϲãÎÞÉ«£¬Ï²ã³Ê×Ï£¨ºì£©É«£»
£¨3£©´Ó·ÖҺ©¶·Ï¿ڷųöÉÙÁ¿Ï²ãÒºÌå¼ÓÈëСÊÔ¹ÜÖУ¬¼ÓÈëÉÙÁ¿Ë®£¬ÈôÊÔ¹ÜÖÐÒºÌå·Ö²ã£¬ÔòÖ¤Ã÷ϲãÊÇÓлú²ã£»Èô²»·Ö²ã£¬ÔòÖ¤Ã÷ϲãÊÇË®²ã£¬
¹Ê´ð°¸Îª£º´Ó·ÖҺ©¶·Ï¿ڷųöÉÙÁ¿Ï²ãÒºÌå¼ÓÈëСÊԹܣ¬ÔÙÍùÊÔ¹ÜÖмÓÈëÉÙÁ¿Ë®£¬ÈôÊÔ¹ÜÖÐÒºÌå·Ö²ã£¬ÔòÖ¤Ã÷ϲãÊÇÓлú²ã£»Èô²»·Ö²ã£¬ÔòÖ¤Ã÷ϲãÊÇË®²ã£»
£¨4£©A£®Ê¹ÓÃÈÝÁ¿Æ¿Ç°¼ì²éÆäÊÇ·ñ©ˮ£¬·ñÔòÅäÖÆÈÜÒºµÄŨ¶ÈÓÐÎó²î£¬¹ÊAÕýÈ·£»
B£®ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴ¾»ºó£¬ÔÙÓôýÅäÈÜҺϴµÓ£¬µ¼ÖÂÈÜÖʵÄÎïÖʵÄÁ¿Æ«´ó£¬ÅäÖÆÈÜÒºµÄŨ¶ÈÆ«¸ß£¬¹ÊB´íÎó£»
C£®ÈÝÁ¿Æ¿Ö»ÄÜÓÃÓÚÅäÖÆÒ»¶¨Å¨¶ÈµÄÈÜÒº£¬²»ÄÜÔÚÈÝÁ¿Æ¿ÖÐÈܽâÈÜÖÊ£¬¹ÊC´íÎó£»
D£®¸ÇºÃÆ¿¸Ç£¬ÓÃʳָ¶¥×¡Æ¿Èû£¬ÓÃÁíÒ»Ö»ÊÖµÄÊÖÖ¸ÍÐסƿµ×£¬°ÑÈÝÁ¿Æ¿µ¹×ªºÍÒ¡¶¯¶à´Î£¬Ä¿µÄÊÇÒ¡ÔÈÈÜÒº£¬¹ÊDÕýÈ·£»
¹Ê´ð°¸Îª£ºAD£»
£¨5£©ÉèŨÁòËáµÄÌå»ýΪVmL£¬Ï¡ÊÍÇ°ºóÈÜÖʵÄÖÊÁ¿²»±ä£¬Ôò£º98%¡Á1.84g/cm3V=0.5mol/L¡Á0.50L¡Á98g/mol£¬½âµÃV=13.6mL£¬
Ϊ¼õСÎó²î£¬Ó¦Ñ¡ÓÃ15mLµÄÁ¿Í²£¬
¹Ê´ð°¸Îª£º13.6£»15£»
£¨6£©A£®Î´ÀäÈ´µ½ÊÒξͽ«ÈÜҺתÒƵ½ÈÝÁ¿Æ¿²¢¶¨ÈÝ£¬ÈÜÒºÀäÈ´ºóÌå»ýƫС£¬ÅäÖƵÄÈÜҺŨ¶ÈÆ«¸ß£¬¹ÊAÕýÈ·£»
B£®×ªÒÆʱûÓÐÏ´µÓÉÕ±¡¢²£Á§°ô£¬µ¼ÖÂÅäÖƵÄÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿Æ«Ð¡£¬ÅäÖƵÄÈÜҺŨ¶ÈÆ«µÍ£¬¹ÊB´íÎó£»
C£®¶¨ÈÝʱ¸©Êӿ̶ÈÏߣ¬µ¼Ö¼ÓÈëµÄÕôÁóË®Ìå»ýƫС£¬ÅäÖƵÄÈÜÒºÌå»ýƫС£¬ÈÜҺŨ¶ÈÆ«¸ß£¬¹ÊCÕýÈ·£»
D£®Ï´µÓÁ¿Í²£¬ÈÜÖʵÄÎïÖʵÄÁ¿Æ«´ó£¬ÅäÖƵÄÈÜҺŨ¶ÈÆ«¸ß£¬¹ÊDÕýÈ·£»
E£®ÓÖ¼ÓÕôÁóË®ÖÁ¿Ì¶ÈÏߣ¬ÅäÖƵÄÈÜÒºÌå»ýÆ«´ó£¬ÈÜҺŨ¶ÈƫС£¬¹ÊE´íÎó£®
¹Ê´ð°¸Îª£ºACD£®
µãÆÀ ±¾Ì⿼²éÝÍÈ¡¡¢Ñõ»¯»¹Ô·´Ó¦¡¢Ò»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÅäÖÆÓëÓйؼÆËãµÈ£¬ÄѶÈÖеȣ¬ÊǶÔËùѧ֪ʶµÄ×ÛºÏÔËÓã¬ÐèҪѧÉú¾ß±¸ÔúʵµÄ»ù´¡ÖªÊ¶Óë×ÛºÏÔËÓÃ֪ʶ·ÖÎöÎÊÌâ¡¢½â¾öÎÊÌâµÄÄÜÁ¦£¬Ñ§Ï°ÖÐÈ«Ãæ°ÑÎÕ»ù´¡ÖªÊ¶£®
A£® | Èõµç½âÖÊ | B£® | Ç¿µç½âÖÊ | C£® | ·Çµç½âÖÊ | D£® | »ìºÏÎï |
A£® | ̼̼˫¼ü²»¿ÉÐýתËùÒÔÏ©ÌþÒ»¶¨ÓÐ˳·´Òì¹¹ | |
B£® | CH2Cl2ÎÞͬ·ÖÒì¹¹Ìå¿ÉÒÔÖ¤Ã÷CH4ÊÇÕýËÄÃæÌåµÄ½á¹¹ | |
C£® | ÁÚ¶þ¼×±½Ö»ÓÐÒ»Öֽṹ¿ÉÒÔÖ¤Ã÷±½·Ö×ÓÖÐÎÞµ¥Ë«½¡µÄ½»Ìæ½á¹¹ | |
D£® | ÒÒÏ©¡¢ÒÒȲÊÇƽÃæÐÍ·Ö×Ó¿ÉÍÆÖªCH3-CH=C£¨CH3£©-C¡ÔC-CH3·Ö×ÓÖÐËùÓеÄ̼Ô×Ó¹²Ãæ |
A£® | ¸ÃÔ×ÓµÄĦ¶ûÖÊÁ¿ÊÇaNAg/mol | |
B£® | Wg¸ÃÔ×ÓµÄÎïÖʵÄÁ¿Ò»¶¨ÊÇ$\frac{W}{{a{N_A}}}$mol | |
C£® | Wg¸ÃÔ×ÓÖк¬ÓÐ$\frac{w}{a}$¸ö¸ÃÔ×Ó | |
D£® | ÓÉÒÑÖªÐÅÏ¢¿ÉµÃ£ºNA=$\frac{b}{12}$ |
A£® | 25% | B£® | 35% | C£® | 75% | D£® | 50% |
A£® | MgSO4¨TMg2++SO42- | B£® | Fe£¨OH£©3¨TFe3++3OH- | ||
C£® | NaHCO3¨TNa++HCO3- | D£® | KAl£¨SO4£©2¨TK++Al3++2SO42- |