³£ÎÂÏ£¬Ksp(AgCl)=1.8¡Á10-10, Ksp(AgI)=1.0¡Á10-14,½«µÈÌå»ýµÄAgClºÍAgIµÄ±¥ºÍÈÜÒºµÄÇå»ìºÏ£¬ÔÙÏòÆäÖмÓÈëÒ»¶¨Á¿µÄAgNO3¹ÌÌ壬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A. ÏòAgIÇåÒºÖмÓÈëAgNO3£¬c(Ag+)Ôö´ó£¬Ksp(AgI)Ò²Ôö´ó
B. Á½ÇåÒº»ìºÏ£¬AgClºÍAgI¶¼³Áµí
C. ÈôÈ¡0.1435gAgCl¹ÌÌå¼ÓÈë100mLË®£¨ºöÂÔÌå»ý±ä»¯£©£¬c(Cl-)Ϊ0.01mol/L
D. ÈôAgNO3×ãÁ¿£¬AgClºÍAgI¶¼¿É³Áµí£¬µ«ÒÔAgClΪÖ÷
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´£º2017½ìºÓ±±Ê¡¸ßÈýÏÂѧÆÚÖÜÁ·£¨2£©»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
½«×ãÁ¿µÄXÆøÌåͨÈëYÈÜÒºÖУ¬ÊµÑé½á¹ûÓëÔ¤²âµÄÏÖÏóÒ»ÖµÄ×éºÏÊÇ
A. Ö»ÓÐ¢Ù¢Ú¢Û B. Ö»ÓÐ¢Ù¢Ú C. Ö»ÓÐ¢Ú D. ¢Ú¢Ü
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìºþÄÏÊ¡ÉÛÑôÊиßÈýµÚ¶þ´Î´óÁª¿¼Àí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º¼ò´ðÌâ
Ç軯¼Ø£¨KCN£©ÊÇÒ»ÖÖÓо綾µÄÎïÖÊ£¬Öü´æºÍʹÓÃʱ±ØÐë×¢Òⰲȫ¡£ÒÑÖª£ºKCN+H2O2=KOCN+H2O¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©OCN-ÖÐËùº¬ÈýÖÖÔªËصĵÚÒ»µçÀëÄÜ´Ó´óµ½Ð¡µÄ˳ÐòΪ_________£¨ÓÃÔªËØ·ûºÅ±íʾ£¬ÏÂͬ£©£¬µç¸ºÐÔ´Ó´óµ½Ð¡µÄ˳ÐòΪ________£»»ù̬µªÔ×ÓÍâΧµç×ÓÅŲ¼Ê½Îª__________¡£
£¨2£©H2O2ÖеĹ²¼Û¼üÀàÐÍΪ_______£¨Ìî¡°¦Ò¼ü¡±»ò¡°¦Ð¼ü¡±) £¬ÆäÖÐÑõÔ×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪ_________£»·Ö×ÓÖÐ4¸öÔ×Ó______£¨Ìî¡°ÔÚ¡±»ò¡°²»ÔÚ¡±£©Í¬Ò»ÌõÖ±ÏßÉÏ£»H2O2Ò×ÈÜÓÚË®³ýËüÃǶ¼ÊǼ«ÐÔ·Ö×ÓÍ⣬»¹ÒòΪ____________________¡£
£¨3£©ÓëOCN-¼üºÏ·½Ê½ÏàͬÇÒ»¥ÎªµÈµç×ÓÌåµÄ·Ö×ÓΪ________£¨ÈξÙÒ»Àý£©£»ÔÚÓëOCN-»¥ÎªµÈµç×ÓÌåµÄ΢Á£ÖУ¬ÓÉÒ»ÖÖÔªËØ×é³ÉµÄÒõÀë×ÓÊÇ____________¡£
£¨4£©KCNµÄ¾§°û½á¹¹ÈçͼËùʾ¡£¾§ÌåÖÐK£«µÄÅäλÊýΪ_______£¬ÈôÆ侧°û²ÎÊýa=0.648nm£¬ÔòKCN ¾§ÌåµÄÃܶÈΪ_______g/cm3£¨Áгö¼ÆËãʽ£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ì¼ªÁÖÊ¡³¤´ºÊиßÈýÏÂѧÆÚµÚ¶þ´ÎÖÊÁ¿¼à²âÀí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ʹÓÃÐÂÐ͵缫²ÄÁÏ£¬ÒÔN2¡¢H2Ϊµç¼«·´Ó¦ÎÒÔHC1-NH4C1Ϊµç½âÖÊÈÜÒº£¬¿ÉÖÆÔì³öÒ»ÖÖ¼ÈÄÜÌṩµçÄÜ£¬ÓÖÄÜʵÏÖµª¹Ì¶¨µÄÐÂÐÍȼÁϵç³Ø£¬ÔÀíÈçÏÂͼËùʾ¡£ÏÂÁÐÓйطÖÎöÕýÈ·µÄÊÇ£¨ £©
A. ͨÈëH2¡ª¼«ÎªÕý¼« B. ·ÖÀë³öµÄÎïÖÊXΪHC1
C. µç³Ø¹¤×÷Ò»¶Îʱ¼äºó£¬ÈÜÒºpH¼õС D. ͨÈëN2Ò»¼«µÄµç¼«·´Ó¦Ê½Îª£ºN2+6e-+8H+=2NH4+
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìºþÄÏÊ¡¡¢ºâÑô°ËÖеÈÊ®ÈýУÖصãÖÐѧ¸ßÈýµÚÒ»´ÎÁª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ
µª»¯ÂÁ(AlN)ÊÇÒ»ÖÖÐÂÐÍÎÞ»ú·Ç½ðÊô²ÄÁÏ¡£ÎªÁË·ÖÎöijAlNÑùÆ·£¨ÑùÆ·ÖеÄÔÓÖʲ»ÓëNaOHÈÜÒº·´Ó¦£©ÖÐ AlNµÄº¬Á¿£¬Ä³ÊµÑéС×éÉè¼ÆÁËÈçÏÂÁ½ÖÖʵÑé·½°¸¡£ÒÑÖª£ºAlN+NaOH+H2O£½NaAlO2+NH3¡ü
¡¾·½°¸1¡¿È¡Ò»¶¨Á¿µÄÑùÆ·£¬ÓÃÒÔÏÂ×°ÖòⶨÑùÆ·ÖÐAlNµÄ´¿¶È(¼Ð³Ö×°ÖÃÒÑÂÔÈ¥)¡£
£¨1£©ÉÏͼװÖÃÖУ¬UÐιÜBÖÐËù×°¹ÌÌåΪ________£¬CÖÐÇòÐθÉÔï¹ÜµÄ×÷ÓÃÊÇ_______________________¡£
£¨2£©¹Ø±ÕK1´ò¿ªK2£¬ÔÙ´ò¿ª·ÖҺ©¶·»îÈû£¬¼ÓÈëNaOHŨÈÜÒº£¬ÖÁ²»ÔÙ²úÉúÆøÌå¡£´ò¿ªK1£¬Í¨È뵪ÆøÒ»¶Îʱ¼ä£¬²â¶¨C×°Ö÷´Ó¦Ç°ºóµÄÖÊÁ¿±ä»¯¡£Í¨È뵪ÆøµÄÄ¿µÄÊÇ_______________________________________¡£
£¨3£©ÓÉÓÚÉÏÊö×°Öû¹´æÔÚ____________ȱÏÝ£¬µ¼Ö²ⶨ½á¹ûÆ«¸ß¡£
¡¾·½°¸2¡¿°´ÒÔϲ½Öè²â¶¨ÑùÆ·ÖÐAlNµÄ´¿¶È£º
£¨4£©²½Öè¢ÚÉú³É³ÁµíµÄÀë×Ó·½³ÌʽΪ___________________¡£
£¨5£©²½Öè¢ÛµÄ²Ù×÷ÖÐÓõ½µÄÖ÷Òª²£Á§ÒÇÆ÷ÊÇ_________¡£AlNµÄ´¿¶ÈÊÇ__________£¨ÓÃm1¡¢m2±íʾ£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìºþÄÏÊ¡¡¢ºâÑô°ËÖеÈÊ®ÈýУÖصãÖÐѧ¸ßÈýµÚÒ»´ÎÁª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ÒÑÖªÓлúÎïA(C10H20O2)ÔÚËáÐÔÌõ¼þÏÂÄܹ»Ë®½âµÃBºÍC, BÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔÑõ»¯µÃµ½C£¬Ôò·ûºÏÌõ¼þµÄÓлúÎïAÓÐ
A. 4ÖÖ B. 8ÖÖ C. 16ÖÖ D. 32ÖÖ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìºþÄÏÊ¡¡¢ºâÑô°ËÖеÈÊ®ÈýУÖصãÖÐѧ¸ßÈýµÚÒ»´ÎÁª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ÏÂͼËùʾÒÇÆ÷¿ÉÓÃÓÚʵÑéÊÒÖƱ¸ÉÙÁ¿ÎÞË®FeCl3£¬ÒÇÆ÷Á¬½Ó˳ÐòÕýÈ·µÄÊÇ
A. a-b-c-d-e-e-f-g-h B. a-e-d-c-b-h-i-g
C. a-d-e-c-b-h-i-g D. a-c-b-d-e-h-i-f
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016-2017ѧÄê½ËÕÊ¡¸ß¶þ3ÔÂѧҵˮƽ²âÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
Ò»¶¨Î¶ÈÏ£¬ÔÚÈÝ»ý²»±äµÄÃܱÕÈÝÆ÷Öз¢ÉúÈçÏ·´Ó¦£ºI2(g) + H2(g)2HI(g)¡£¸Ã·´Ó¦´ïµ½Æ½ºâ״̬µÄ±êÖ¾ÊÇ£¨ £©
A. ÈÝÆ÷ÖÐÆøÌåµÄÑÕÉ«²»Ôٸıä
B. I2¡¢H2¡¢HIµÄ·Ö×ÓÊýÖ®±ÈΪ1¡Ã1¡Ã2
C. I2(g) ºÍH2(g)Íêȫת»¯Îª HI(g)
D. µ¥Î»Ê±¼äÄÚÉú³Én mol I2µÄͬʱÉú³Én mol H2
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìËÄ´¨Ê¡¸ßÈýÉÏѧÆÚµÚËÄ´ÎÔ¿¼Àí¿Æ×ۺϻ¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º¼ò´ðÌâ
ÇâÄܵĴ洢ÊÇÇâÄÜÓ¦ÓõÄÖ÷Ҫƿ¾±£¬ÅäλÇ⻯Îï¡¢¸»ÇâÔØÌ廯ºÏÈÍÊÇÄ¿Ç°Ëù²ÉÓõÄÖ÷Òª´¢Çâ²ÄÁÏ¡£
£¨1£©Ti£¨BH4£©2ÊÇÒ»ÖÖ¹ý¶ÉÔªËØÅðÇ⻯Îï´¢Çâ²ÄÁÏ¡£ÔÚ»ù̬TiÖУ¬¼Ûµç×ÓÅŲ¼Ê½Îª ____£¬¼Ûµç×ÓÅŲ¼Í¼Îª____
£¨2£©Òº°±ÊǸ»ÇâÎïÖÊ£¬ÊÇÇâÄܵÄÀíÏëÔØÌ壬ÀûÓÃN2+3H22NH3,ʵÏÖ´¢ÇâºÍÊäÇâ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_____ £»
a£®NH3·Ö×ÓÖеªÔ×ӵĹìµÀÔÓ»¯·½Ê½Îªsp2ÔÓ»¯
b£®NH+4ÓëPH+4¡¢CH4¡¢BH-4¡¢ClO¡ª4»¥ÎªµÈµç×ÓÌå
c£®Ïàͬѹǿʱ£¬NH3µÄ·Ðµã±ÈPH3µÄ·Ðµã¸ß
d£®[Cu£¨NH3£©4]2+Àë×ÓÖУ¬NÔ×ÓÊÇÅäλÔ×Ó
£¨3£©Óü۲ãµç×Ó¶Ô»¥³âÀíÂÛÍƶÏSnBr2·Ö×ÓÖУ¬SnÔ×ӵĹìµÀÔÓ»¯·½Ê½Îª__________£¬
SnBr2·Ö×ÓÖÐ Sn-BrµÄ¼ü½Ç______120¡ã(Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±)¡£
(4) NiO µÄ¾§Ìå½á¹¹ÓëÂÈ»¯ÄÆÏàͬ£¬ ÔÚ¾§°ûÖÐÄøÀë×ÓµÄÅäλÊýÊÇ_______¡£
ÒÑÖª¾§°ûµÄ±ß³¤Îª a nm£¬ NiO µÄĦ¶ûÖÊÁ¿Îª b g¡¤mol-1£¬ NAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ ÔòNiO ¾§ÌåµÄÃܶÈΪ_________g¡¤cm-3¡£
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