£¨12·Ö£©Na2S2O3£¨Ë׳Ʊ£ÏÕ·Û£©ÔÚÒ½Ò©¡¢Ó¡È¾ÖÐÓ¦Óù㷺£¬¿Éͨ¹ýÏÂÁз½·¨ÖƱ¸£ºÈ¡15.1 gNa2SO3ÈÜÓÚ80.0 mLË®¡£ÁíÈ¡5.0 gÁò·Û£¬ÓÃÉÙÐíÒÒ´¼Èóʪºó¼Óµ½ÉÏÊöÈÜÒºÖС£Ð¡»ð¼ÓÈÈÖÁ΢·Ð£¬·´Ó¦1Сʱºó¹ýÂË¡£ÂËÒºÔÚ100¡æ¾­Õô·¢¡¢Å¨Ëõ¡¢ÀäÈ´ÖÁ10¡æºóÎö³öNa2S2O3¡¤5H2O¡£
£¨1£©¼ÓÈëµÄÁò·ÛÓÃÒÒ´¼ÈóʪµÄÄ¿µÄÊÇ     ¡£
£¨2£©ÂËÒºÖгýNa2S2O3ºÍ¿ÉÄÜδ·´Ó¦ÍêÈ«µÄNa2SO3Í⣬×î¿ÉÄÜ´æÔÚµÄÎÞ»ú»¯ºÏÎïÔÓÖÊÊÇ     £»Æä¼ì²âµÄ·½·¨ÊÇ£ºÈ¡³öÉÙÐíÈÜÒº£¬¼ÓÑÎËáÖÁËáÐԺ󣬹ýÂ˳ýÈ¥S£¬ÔÙ¼ÓBaCl2ÈÜÒº¡£Ôò¼ÓÈëµÄÑÎËá·¢ÉúÁ½¸ö·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
Na2SO3+2HCl=SO2¡ü+H2O+2NaCl£»     ¡£
£¨3£©Ä³»·¾³¼à²âС×éÓú¬0.100 mol¡¤L-1Na2S2O3ÈÜÒº[º¬ÉÙÁ¿µÄNa2SO3£¬ÇÒn(Na2S2O3) ¡Ãn(Na2SO3)
= 5¡Ã1]²â¶¨Ä³¹¤³§·ÏË®ÖÐBa2+µÄŨ¶È¡£ËûÃÇÈ¡·ÏË®50.0 mL£¬¿ØÖÆÊʵ±µÄËá¶È¼ÓÈë×ãÁ¿µÄK2Cr2O7ÈÜÒº£¬µÃBaCrO4³Áµí£»³Áµí¾­Ï´µÓ¡¢¹ýÂ˺ó£¬ÓÃÊÊÁ¿µÄÏ¡ÁòËáÈܽ⣬´ËʱCrO42-È«²¿×ª»¯ÎªCr2O72-£»ÔÙ¼Ó¹ýÁ¿KIÈÜÒº£¬³ä·Ö·´Ó¦ºó£¬ÓÃÉÏÊöNa2S2O3ÈÜÒº½øÐе樣¬·´Ó¦Íêȫʱ£¬²âµÃÏûºÄNa2S2O3ÈÜÒºµÄÌå»ýΪ36.0 mL¡£
£¨ÒÑÖªÓйط´Ó¦µÄÀë×Ó·½³ÌʽΪ£º¢ÙCr2O72-+6I-+14H+   2Cr3++3I2+7H2O£»
¢ÚI2+2S2O32-   2I-+S4O62-£»¢ÛI2+SO32-+H2O   2I-+SO42-+2H+£©
ÔòµÎ¶¨¹ý³ÌÖпÉÓà    ×÷ָʾ¼Á¡£¼ÆËã¸Ã¹¤³§·ÏË®ÖÐBa2+µÄÎïÖʵÄÁ¿Å¨¶È¡£

¹²12·Ö¡£
£¨1£©ÓÐÀûÓÚÁò·ÛºÍNa2SO3ÈÜÒº³ä·Ö½Ó´¥£¬¼Ó¿ì·´Ó¦ËÙÂÊ£¨2·Ö£©
£¨2£©Na2SO4£¨1·Ö£©¡¡    Na2S2O3+2HCl=S¡ý+SO2¡ü+H2O+2NaCl£¨2·Ö£©
£¨3£©µí·Û£¨1·Ö£©
n(Na2S2O3) =" 0.0360" L¡Á0.100 mol¡¤L-1 =" 0.0036" mol
Ôòn(Na2SO3) =" 0.00360" mol¡Â5 =" 0.00072" mol
¸ù¾ÝÌâÒ⣺2BaCrO4 ~ Cr2O72- ~ 3I2 ~ 6S2O32-
µÃn1(BaCrO4) = ==" 0.0012" mol
2BaCrO4 ~ Cr2O72- ~ 3I2 ~ 3SO32-
µÃn2(BaCrO4) = ==" 0.00048" mol
Ôòc(Ba2+) = = 3.36¡Á10-2 mol¡¤L-1£¨6·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÁòÒ×ÈÜÓÚÓлúÈܼÁ£¬ÒÒ´¼Ò×ÈÜÓÚË®£¬ËùÒÔÓÃÒÒ´¼ÈóʪÁò·Û£¬Ä¿µÄÊÇÔö´óÆäÓëË®µÄ½Ó´¥Ãæ»ý£¬ÓÐÀûÓÚÓëNa2SO3ÈÜÒº³ä·Ö½Ó´¥£¬¼Ó¿ì·´Ó¦ËÙÂÊ£»
£¨2£©ÑÇÁòËáÄÆÔÚ¼ÓÈȵĹý³ÌÖпÉÄܱ»Ñõ»¯¶øÉú³ÉÁòËáÄÆ£¬ËùÒÔ×î¿ÉÄÜ´æÔÚµÄÎÞ»ú»¯ºÏÎïÔÓÖÊÊÇNa2SO4£»ÁòËáÄÆ ÂÈ»¯±µÈÜÒº¼ìÑ飬ÑÇÁòËáÄÆÓëÔªËØ·´Ó¦Öª²úÉú¶þÑõ»¯ÁòÆøÌ壬¶øNa2S2O3ÓëÑÎËá·´Ó¦¼ÈÓжþÑõ»¯ÁòÉú³É£¬ÓÖÓÐÁòµ¥ÖÊÉú³É£¬ËùÒÔ¸ù¾ÝÏÖÏóµÄ²»Í¬¿ÉÒÔ¼ì²â£¬ÑÎËáµÄÁíÒ»×÷ÓõĻ¯Ñ§·½³ÌʽΪNa2S2O3+2HCl=S¡ý+SO2¡ü+H2O+2NaCl£»
£¨3£©Na2S2O3ÈÜÒº½øÐе樵ÄÊǵⵥÖÊ£¬ËùÒÔÑ¡Óõí·Û×÷ָʾ¼Á£¬µâÓëµí·ÛµÄ»ìºÏҺΪÀ¶É«£¬µÎ¶¨ÖÕµãʱÀ¶É«Ïûʧ£»¸ù¾ÝNa2S2O3ÈÜÒºµÄÌå»ý¿É¼ÆËãµâµ¥ÖʵÄÎïÖʵÄÁ¿£¬´Ó¶ø¿É¼ÆËãBaCrO4³ÁµíµÄÎïÖʵÄÁ¿£¬Ò²¾Í¿ÉÒÔ¼ÆËã±µÀë×ÓµÄŨ¶È£¬¸ù¾Ý·½³ÌʽµÃ³ö¹Øϵʽ£¬¾ßÌå²½Öè¼û´ð°¸¡£
¿¼µã£º¿¼²é¶ÔÎïÖÊÖƱ¸µÄ·ÖÎö£¬¹Øϵʽ·¨µÄÔËÓÃ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓÃÀ¨ºÅÄÚÊÔ¼Á¼°²Ù×÷·½·¨³ýÈ¥ÏÂÁи÷ÎïÖÊÖеÄÉÙÁ¿ÔÓÖÊ£¬²»ÕýÈ·µÄÊÇ

A£®±½Öеı½·Ó£¨Br2Ë®¡¢¹ýÂË£©
B£®ÒÒËáÒÒõ¥ÖеÄÒÒËᣨ±¥ºÍ̼ËáÄÆÈÜÒº¡¢·ÖÒº£©
C£®äåÒÒÍéÖеÄÒÒ´¼£¨Ë®¡¢·ÖÒº£©
D£®äå±½ÖеÄä壨NaOHÈÜÒº¡¢·ÖÒº£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

½ñÓÐÏÂÁÐÆøÌ壺H2¡¢Cl2¡¢HCl¡¢NH3¡¢NO¡¢H2S¡¢SO2£¬ÓÃÈçͼËùʾµÄ×°ÖýøÐÐʵÑ飬Ìî¿ÕÏÂÁпհףº

£¨1£©ÉÕÆ¿¸ÉÔïʱ£¬´ÓA¿Ú½øÆø¿ÉÊÕ¼¯µÄÆøÌåÊÇ________£¬´ÓB¿Ú½øÆø¿ÉÊÕ¼¯µÄÆøÌåÊÇ______________¡£
£¨2£©ÉÕÆ¿ÖгäÂúˮʱ£¬¿ÉÓÃÀ´²âÁ¿___________µÈÆøÌåµÄÌå»ý¡£
£¨3£©µ±ÉÕÆ¿ÖÐ×°ÈëÏ´Òº£¬ÓÃÓÚÏ´Æøʱ£¬ÆøÌåÓ¦´Ó________¿Ú½øÈëÉÕÆ¿¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

£¨12·Ö£©Ä³»¯Ñ§ÐËȤС×éÓÃÈçÏÂͼ1ËùʾװÖýøÐÐ̽¾¿ÊµÑ飬ÒÔÑéÖ¤²úÎïÖÐÓÐÒÒÏ©Éú³ÉÇÒÒÒÏ©¾ßÓв»±¥ºÍÐÔ¡£µ±Î¶ÈѸËÙÉÏÉýºó£¬¿É¹Û²ìµ½ÊÔ¹ÜÖÐäåË®ÍÊÉ«£¬ÉÕÆ¿ÖÐŨH2SO4ÓëÒÒ´¼µÄ»ìºÏÒºÌå±äΪ×غÚÉ«¡£

ͼ1                  Í¼2
£¨1£©Ð´³ö¸ÃʵÑéÖÐÉú³ÉÒÒÏ©µÄ»¯Ñ§·½³Ìʽ                      ¡£
£¨2£©¼×ͬѧÈÏΪ£º¿¼Âǵ½¸Ã»ìºÏÒºÌå·´Ó¦µÄ¸´ÔÓÐÔ£¬äåË®ÍÊÉ«µÄÏÖÏó²»ÄÜÖ¤Ã÷·´Ó¦ÖÐÓÐÒÒÏ©Éú³ÉÇÒÒÒÏ©¾ßÓв»±¥ºÍÐÔ£¬ÆäÀíÓÉÕýÈ·µÄÊÇ            ¡£
a£®ÒÒÏ©ÓëäåË®Ò×·¢ÉúÈ¡´ú·´Ó¦
b£®Ê¹äåË®ÍÊÉ«µÄ·´Ó¦£¬Î´±ØÊǼӳɷ´Ó¦
c£®Ê¹äåË®ÍÊÉ«µÄÎïÖÊ£¬Î´±ØÊÇÒÒÏ©
£¨3£©ÒÒͬѧ¾­¹ýϸÖ¹۲ìºóÊÔ¹ÜÖÐÁíÒ»ÏÖÏóºó²¢½èÓÃpHÊÔÖ½²â¶¨£¬Ö¤Ã÷·´Ó¦ÖÐÓÐÒÒÏ©Éú³É£¬Çë¼òÊö
                      ¡£
£¨4£©±ûͬѧ¶ÔÉÏÊöʵÑé×°ÖýøÐÐÁ˸Ľø£¬ÔÚ¢ñºÍ¢òÖ®¼äÔö¼ÓÉÏͼ2×°ÖÃÒÔ³ýÈ¥ÒÒ´¼ÕôÆøºÍSO2£¬ÔòAÖеÄÊÔ¼ÁÊÇ          £¬BÖеÄÊÔ¼ÁΪ             ¡£
£¨5£©´¦ÀíÉÏÊöʵÑéºóÉÕÆ¿ÖзÏÒºµÄÕýÈ··½·¨ÊÇ              ¡£
a£®ÀäÈ´ºóµ¹ÈëÏÂË®µÀÖÐ  b£®ÀäÈ´ºóµ¹Èë¿Õ·ÏÒº¸×ÖÐ c£®ÀäÈ´ºó¼ÓˮϡÊÍ£¬µ¹Èë·ÏÒº¸×ÖУ¬¼Ó·Ï¼îÖкÍ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ÒÑÖªÒÒ´¼¿ÉÒÔºÍÂÈ»¯¸Æ·´Ó¦Éú³É΢ÈÜÓÚË®µÄCaCl2¡¤6C2H5OH¡£ÓйصÄÓлúÊÔ¼ÁµÄ·ÐµãÈçÏ£ºCH3COOC2H5Ϊ77.1¡æ£»C2H5OHΪ78.3¡æ£»C2H5OC2H5£¨ÒÒÃÑ£©Îª34.5¡æ£»CH3COOHΪ118¡æ¡£ÊµÑéÊҺϳÉÒÒËáÒÒõ¥´Ö²úÆ·µÄ²½ÖèÈçÏ£ºÔÚÕôÁóÉÕÆ¿ÄÚ½«¹ýÁ¿µÄÒÒ´¼ÓëÉÙÁ¿Å¨ÁòËá»ìºÏ£¬È»ºó¾­·ÖҺ©¶·±ßµÎ¼Ó´×Ëᣬ±ß¼ÓÈÈÕôÁó¡£ÓÉÉÏÃæµÄʵÑé¿ÉµÃµ½º¬ÓÐÒÒ´¼¡¢ÒÒÃÑ¡¢´×ËáºÍË®µÄÒÒËáÒÒõ¥´Ö²úÆ·¡£
£¨1£©·´Ó¦ÖмÓÈëµÄÒÒ´¼ÊǹýÁ¿µÄ£¬ÆäÄ¿µÄÊÇ                                    ¡£
£¨2£©±ßµÎ¼Ó´×Ëᣬ±ß¼ÓÈÈÕôÁóµÄÄ¿µÄÊÇ                      ¡£
½«´Ö²úÆ·ÔÙ¾­ÏÂÁв½Ö辫ÖÆ£º
£¨3£©Îª³ýÈ¥ÆäÖеĴ×Ëᣬ¿ÉÏò²úÆ·ÖмÓÈë         £¨Ìî×Öĸ£©¡£
A.ÎÞË®ÒÒ´¼        B.̼ËáÄÆ·ÛÄ©        C.ÎÞË®´×ËáÄÆ
£¨4£©ÔÙÏòÆäÖмÓÈë±¥ºÍÂÈ»¯¸ÆÈÜÒº£¬Õñµ´£¬·ÖÀ룬ÆäÄ¿µÄÊÇ                   ¡£
£¨5£©È»ºóÔÙÏòÆäÖмÓÈëÎÞË®ÁòËáÍ­£¬Õñµ´£¬ÆäÄ¿µÄÊÇ                  ¡£×îºó£¬½«¾­¹ýÉÏÊö´¦ÀíºóµÄÒºÌå¼ÓÈëÁíÒ»¸ÉÔïµÄÕôÁóÆ¿ÄÚ£¬ÔÙÕôÁó£¬ÆúÈ¥µÍ·ÐµãÁó·Ö£¬ÊÕ¼¯·ÐµãÔÚ76¡æ~78¡æÖ®¼äµÄÁó·Ö¼´µÃ´¿¾»µÄÒÒËáÒÒõ¥¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

(14·Ö)²ÝËáÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬¹ã·ºÓÃÓÚÒ©ÎïÉú²ú¡¢¸ß·Ö×Ӻϳɵȹ¤Òµ£¬²ÝËᾧÌåÊÜÈȵ½100¡æʱʧȥ½á¾§Ë®£¬³ÉΪÎÞË®²ÝËᡣijѧϰС×éµÄͬѧÄâÒÔ¸ÊÕáÔüΪԭÁÏÓÃË®½â¡ªÑõ»¯¡ªË®½âÑ­»·½øÐÐÖÆÈ¡²ÝËá¡£

aaaaaa

 

Çë¸ú¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Í¼Ê¾¢Ù¢ÚµÄÑõ»¯¡ªË®½â¹ý³ÌÊÇÔÚÉÏͼ1µÄ×°ÖÃÖнøÐеģ¬Ö¸³ö×°ÖÃAµÄÃû³Æ                  ¡£
£¨2£©Í¼Ê¾¢Ù¢ÚµÄÑõ»¯¡ªË®½â¹ý³ÌÖУ¬ÔÚÏõËáÓÃÁ¿¡¢·´Ó¦µÄʱ¼äµÈÌõ¼þ¾ùÏàͬµÄÇé¿öÏ£¬¸Ä±ä·´Ó¦Î¶ÈÒÔ¿¼²ì·´Ó¦Î¶ȶԲÝËáÊÕÂʵÄÓ°Ï죬½á¹ûÈçÉÏͼ2Ëùʾ£¬ÇëÑ¡Ôñ×î¼ÑµÄ·´Ó¦Î¶ÈΪ          £¬ÎªÁ˴ﵽͼ2ËùʾµÄζȣ¬Ñ¡Ôñͼ1µÄˮԡ¼ÓÈÈ£¬ÆäÓŵãÊÇ                           ¡£
£¨3£©ÔÚͼʾ¢Û¢ÜÖеIJÙ×÷Éæ¼°µ½³éÂË£¬Ï´µÓ¡¢¸ÉÔÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ         ¡£
A.ÔÚʵÑé¹ý³ÌÖУ¬Í¨¹ý¿ìËÙÀäÈ´²ÝËáÈÜÒº£¬¿ÉÒԵõ½½Ï´óµÄ¾§Ìå¿ÅÁ££¬±ãÓÚ³éÂË
B.ÔÚÏ´µÓ³Áµíʱ£¬Ó¦¹ØСˮÁúÍ·£¬Ê¹Ï´µÓ¼Á»º»ºÍ¨¹ý³ÁµíÎï
C.ΪÁ˼ìÑéÏ´µÓÊÇ·ñÍêÈ«£¬Ó¦°ÎÏÂÎüÂËÆ¿Ó밲ȫƿ֮¼äÏðƤ¹Ü£¬´ÓÎüÂËÆ¿ÉÏ¿Úµ¹³öÉÙÁ¿ÂËÒºÓÚÊÔ¹ÜÖнøÐÐÏà¹ØʵÑé¡£
D.ΪÁ˵õ½¸ÉÔïµÄ¾§Ì壬¿ÉÒÔÑ¡ÔñÔÚÛáÛöÖÐÖ±½Ó¼ÓÈÈ£¬²¢ÔÚ¸ÉÔïÆ÷ÖÐÀäÈ´¡£
£¨4£©Òª²â¶¨²ÝËᾧÌ壨H2C2O4¡¤2H2O£©µÄ´¿¶È£¬³ÆÈ¡7.200gÖƱ¸µÄ²ÝËᾧÌåÈÜÓÚÊÊÁ¿Ë®Åä³É250mLÈÜÒº£¬È¡25.00mL²ÝËáÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬ÓÃ0.1000mol/LËáÐÔ¸ßÃÌËá¼ØÈÜÒºµÎ¶¨
£¨5H2C2O4+2MnO4£­+6H+£½2Mn2++10CO2¡ü+8H2O£©£¬
¢ÙÈ¡25.00mL²ÝËáÈÜÒºµÄÒÇÆ÷ÊÇ                        £¬
¢ÚÔÚ²ÝËá´¿¶È²â¶¨µÄʵÑé¹ý³ÌÖУ¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£º             ¡£
A.ÈóÏ´µÎ¶¨¹Üʱ£¬Ó¦´ÓµÎ¶¨¹ÜÉÏ¿Ú¼ÓÂúËùÐèµÄËá»ò¼îÒº£¬Ê¹µÎ¶¨¹ÜÄÚ±Ú³ä·ÖÈóÏ´
B.ÒÆÒº¹ÜÈ¡²ÝËáÈÜҺʱ£¬Ð轫¼â×ì´¦µÄÒºÌå´µÈë׶ÐÎÆ¿£¬»áʹʵÑéÎó²îÆ«µÍ
C.µÎ¶¨Ê±£¬×óÊÖÇáÇáÏòÄÚ¿Ûס»îÈû£¬ÊÖÐÄ¿ÕÎÕÒÔÃâÅöµ½»îÈûʹÆäËɶ¯Â©³öÈÜÒº
D.µÎ¶¨ÖÕµã¶ÁÈ¡µÎ¶¨¹Ü¿Ì¶Èʱ£¬ÑöÊÓ±ê×¼ÒºÒºÃ棬»áʹʵÑéÎó²îÆ«¸ß
¢ÛÅжϵζ¨ÒѾ­´ïµ½ÖÕµãµÄ·½·¨ÊÇ£º                                      ¡£
¢Ü´ïµ½µÎ¶¨ÖÕµãʱ£¬ÏûºÄ¸ßÃÌËá¼ØÈÜÒº¹²20.00mL£¬Ôò²ÝËᾧÌåµÄ´¿¶ÈΪ             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

(13·Ö)ÒÒËáÒìÎìõ¥ÊÇ×é³ÉÃÛ·äÐÅÏ¢ËØÖʵijɷÖÖ®Ò»£¬¾ßÓÐÏ㽶µÄÏãζ£¬ÊµÑéÊÒÖƱ¸ÒÒËáÒìÎìõ¥µÄ·´Ó¦×°ÖÃʾÒâͼºÍÓйØÊý¾ÝÈçÏ£º

ʵÑé²½Ö裺
ÔÚAÖмÓÈë4.4 gµÄÒìÎì´¼£¬6.0 gµÄÒÒËá¡¢ÊýµÎŨÁòËáºÍ2¡«3ƬËé´ÉƬ£¬¿ªÊ¼»ºÂý¼ÓÈÈA£¬»ØÁ÷50·ÖÖÓ£¬·´Ó¦ÒºÀäÖÁÊÒκ󣬵¹Èë·ÖҺ©¶·ÖУ¬·Ö±ðÓÃÉÙÁ¿Ë®£¬±¥ºÍ̼ËáÇâÄÆÈÜÒººÍˮϴµÓ£¬·Ö³öµÄ²úÎï¼ÓÈëÉÙÁ¿ÎÞË®ÁòËáþ¹ÌÌ壬¾²ÖÃƬ¿Ì£¬¹ýÂ˳ýÈ¥ÁòËáþ¹ÌÌ壬£¬½øÐÐÕôÁó´¿»¯£¬ÊÕ¼¯140¡«143 ¡æÁó·Ö£¬µÃÒÒËáÒìÎìõ¥3.9 g¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©×°ÖÃBµÄÃû³ÆÊÇ£º                    
£¨2£©ÔÚÏ´µÓ²Ù×÷ÖУ¬µÚÒ»´ÎˮϴµÄÖ÷ҪĿµÄÊÇ£º               £» µÚ¶þ´ÎˮϴµÄÖ÷ҪĿµÄÊÇ£º                       ¡£
£¨3£©ÔÚÏ´µÓ¡¢·ÖÒº²Ù×÷ÖУ¬Ó¦³ä·ÖÕñµ´£¬È»ºó¾²Ö㬴ý·Ö²ãºó       £¨Ìî±êºÅ£©£¬
A£®Ö±½Ó½«ÒÒËáÒìÎìõ¥´Ó·ÖҺ©¶·ÉÏ¿Úµ¹³ö  
B£®Ö±½Ó½«ÒÒËáÒìÎìõ¥´Ó·ÖҺ©¶·Ï¿ڷųö
C£®ÏȽ«Ë®²ã´Ó·ÖҺ©¶·µÄÏ¿ڷųö£¬ÔÙ½«ÒÒËáÒìÎìõ¥´ÓÏ¿ڷųö
D£®ÏȽ«Ë®²ã´Ó·ÖҺ©¶·µÄÏ¿ڷųö£¬ÔÙ½«ÒÒËáÒìÎìõ¥´ÓÉϿڷųö
£¨4£©±¾ÊµÑéÖмÓÈë¹ýÁ¿ÒÒËáµÄÄ¿µÄÊÇ£º                             
£¨5£©ÊµÑéÖмÓÈëÉÙÁ¿ÎÞË®ÁòËáþµÄÄ¿µÄÊÇ£º                                        
£¨6£©ÔÚÕôÁó²Ù×÷ÖУ¬ÒÇÆ÷Ñ¡Ôñ¼°°²×°¶¼ÕýÈ·µÄÊÇ£º                   (Ìî±êºÅ)

£¨7£©±¾ÊµÑéµÄ²úÂÊÊÇ£º        
A£®30¨G             B£®40¨G               C£®50¨G              D£®60¨G
£¨8£©ÔÚ½øÐÐÕôÁó²Ù×÷ʱ£¬Èô´Ó130 ¡æ¿ªÊ¼ÊÕ¼¯Áó·Ö£¬²úÂÊÆ«      (Ìî¸ß»òÕßµÍ)Ô­ÒòÊÇ             

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

£¨13·Ö£©´×ËáÑǸõË®ºÏÎï([Cr(CH3COO)2)]2¡¤2H2O£¬ÉîºìÉ«¾§Ì壩ÊÇÒ»ÖÖÑõÆøÎüÊÕ¼Á£¬Í¨³£ÒÔ¶þ¾ÛÌå·Ö×Ó´æÔÚ£¬²»ÈÜÓÚÀäË®ºÍÃÑ£¬Î¢ÈÜÓÚ´¼£¬Ò×ÈÜÓÚÑÎËᡣʵÑéÊÒÖƱ¸´×ËáÑǸõË®ºÏÎïµÄ×°ÖÃÈçͼËùʾ£¬Éæ¼°µÄ»¯Ñ§·½³ÌʽÈçÏ£º

Zn(s) + 2HCl(aq) = ZnCl2(aq) + H2(g)
2CrCl3(aq) + Zn(s)= 2CrCl2 (aq) + ZnCl2(aq)
2Cr2+(aq) + 4CH3COO-(aq) + 2H2O(l) = [Cr(CH3COO)2]2¡¤2H2O (s)
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¼ì²éÐé¿òÄÚ×°ÖÃÆøÃÜÐԵķ½·¨ÊÇ     ¡£
£¨2£©´×ËáÄÆÈÜÒºÓ¦·ÅÔÚ×°Öà    ÖУ¨ÌîдװÖñàºÅ£¬ÏÂͬ£©£»ÑÎËáÓ¦·ÅÔÚ×°Öà    ÖУ»
×°ÖÃ4µÄ×÷ÓÃÊÇ     ¡£
£¨3£©±¾ÊµÑéÖÐËùÓÐÅäÖÆÈÜÒºµÄË®ÐèÖó·Ð£¬ÆäÔ­ÒòÊÇ     ¡£
£¨4£©½«Éú³ÉµÄCrCl2ÈÜÒºÓëCH3COONaÈÜÒº»ìºÏʱµÄ²Ù×÷ÊÇ     ·§ÃÅA¡¢     ·§ÃÅB £¨Ìî¡°´ò¿ª¡±»ò¡°¹Ø±Õ¡±£©¡£
£¨5£©±¾ÊµÑéÖÐпÁ£Ðë¹ýÁ¿£¬ÆäÔ­ÒòÊÇ     ¡£                                             
£¨6£©ÎªÏ´µÓ[Cr(CH3COO)2)]2¡¤2H2O²úÆ·£¬ÏÂÁз½·¨ÖÐ×îÊʺϵÄÊÇ     ¡£        
A£®ÏÈÓÃÑÎËáÏ´£¬ºóÓÃÀäˮϴ        B£®ÏÈÓÃÀäˮϴ£¬ºóÓÃÒÒ´¼Ï´
C£®ÏÈÓÃÀäˮϴ£¬ºóÓÃÒÒÃÑÏ´        D£®ÏÈÓÃÒÒ´¼Ï´µÓ£¬ºóÓÃÒÒÃÑÏ´

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

K3[Fe(C2O4)3]¡¤3H2O[Èý²ÝËáºÏÌú(¢ó)Ëá¼Ø¾§Ìå]Ò×ÈÜÓÚË®,ÄÑÈÜÓÚÒÒ´¼,¿É×÷ΪÓлú·´Ó¦µÄ´ß»¯¼Á¡£ÊµÑéÊÒ¿ÉÓÃÌúмΪԭÁÏÖƱ¸,Ïà¹Ø·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ:Fe+H2SO4FeSO4+H2¡ü
FeSO4+H2C2O4+2H2OFeC2O4¡¤2H2O¡ý+H2SO4
2FeC2O4¡¤2H2O+H2O2+H2C2O4+3K2C2O42K3[Fe(C2O4)3]+6H2O
2Mn+5C2+16H+2Mn2++10CO2¡ü+8H2O
»Ø´ðÏÂÁÐÎÊÌâ:
(1)ÌúмÖг£º¬ÁòÔªËØ,Òò¶øÔÚÖƱ¸FeSO4ʱ»á²úÉúÓж¾µÄH2SÆøÌå,¸ÃÆøÌå¿ÉÓÃÇâÑõ»¯ÄÆÈÜÒºÎüÊÕ¡£ÏÂÁÐÎüÊÕ×°ÖÃÕýÈ·µÄÊÇ¡¡¡¡¡¡¡¡¡¡¡£ 

(2)ÔÚ½«Fe2+Ñõ»¯µÄ¹ý³ÌÖÐ,Ðè¿ØÖÆÈÜҺζȲ»¸ßÓÚ40 ¡æ,ÀíÓÉÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡;µÃµ½K3[Fe(C2O4)3]ÈÜÒººó,¼ÓÈëÒÒ´¼µÄÀíÓÉÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
(3)¾§ÌåÖÐËùº¬½á¾§Ë®¿Éͨ¹ýÖØÁ¿·ÖÎö·¨²â¶¨,Ö÷Òª²½ÖèÓÐ:¢Ù³ÆÁ¿,¢ÚÖÃÓÚºæÏäÖÐÍѽᾧˮ,¢ÛÀäÈ´,¢Ü³ÆÁ¿,¢Ý¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡(ÐðÊö´Ë²½²Ù×÷),¢Þ¼ÆËã¡£²½Öè¢ÛÈôδÔÚ¸ÉÔïÆ÷ÖнøÐÐ,²âµÃµÄ¾§ÌåÖÐËùº¬½á¾§Ë®º¬Á¿¡¡¡¡¡¡¡¡(Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족);²½Öè¢ÝµÄÄ¿µÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
(4)¾§ÌåÖÐC2º¬Á¿µÄ²â¶¨¿ÉÓÃËáÐÔKMnO4±ê×¼ÈÜÒºµÎ¶¨¡£³ÆÈ¡Èý²ÝËáºÏÌú(¢ó)Ëá¼Ø¾§Ìåm gÈÜÓÚË®Åä³É250 mLÈÜÒº,È¡³ö20.00 mL·ÅÈë׶ÐÎÆ¿ÖÐ,ÓÃ0.010 0 mol¡¤L-1ËữµÄ¸ßÃÌËá¼ØÈÜÒº½øÐеζ¨¡£
¢ÙÏÂÁвÙ×÷¼°Ëµ·¨ÕýÈ·µÄÊÇ¡¡¡¡¡¡¡£ 
A.µÎ¶¨¹ÜÓÃÕôÁóˮϴµÓºó,¼´¿É×°Èë±ê×¼ÈÜÒº
B.×°Èë±ê×¼ÈÜÒººó,°ÑµÎ¶¨¹Ü¼ÐÔڵζ¨¹Ü¼ÐÉÏ,ÇáÇáת¶¯»îÈû,·Å³öÉÙÁ¿±ê×¼Òº,ʹ¼â×ì³äÂúÒºÌå
C.½Ó½üÖÕµãʱ,ÐèÓÃÕôÁóË®³åÏ´Æ¿±ÚºÍµÎ¶¨¹Ü¼â¶ËÐü¹ÒµÄÒºµÎ
¢ÚÓÐͬѧÈÏΪ¸ÃµÎ¶¨¹ý³Ì²»ÐèҪָʾ¼Á,ÄÇôµÎ¶¨ÖÕµãµÄÏÖÏóΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡,Èô´ïµ½µÎ¶¨ÖÕµãÏûºÄ¸ßÃÌËá¼ØÈÜÒºV mL,ÄÇô¾§ÌåÖÐËùº¬C2µÄÖÊÁ¿·ÖÊýΪ¡¡¡¡¡¡¡¡(Óú¬V¡¢mµÄʽ×Ó±íʾ)¡£ 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸