ÌâÄ¿ÄÚÈÝ

¸ù¾ÝÈç±í»Ø´ðÎÊÌâ

ζȣ¨¡æ£©

20

40

50

60

80

Èܽâ¶È

£¨g/100gË®£©

NaCl

36.0

36.6

37.0

37.3

38.4

NH4Cl

37.2

45.8

50.4

55.2

65.6

KNO3

31.6

63.9

85.5

110

169

£¨1£©50¡æÊ±£¬100gË®ÖÐ×î¶àÈܽâNaCl______ g

£¨2£©AÊÇ80¡æº¬ÓÐ120gË®µÄKNO3ÈÜÒº£¬¾­¹ýÈçϲÙ×÷£¬µÃµ½102gKNO3¹ÌÌå¡£

¢Ù AÈÜҺΪ________£¨Ñ¡Ìî¡°±¥ºÍ¡±»ò¡°²»±¥ºÍ¡±£©ÈÜÒº

¢Ú ¶ÔÒÔϹý³ÌµÄ·ÖÎö£¬ÕýÈ·µÄÊÇ_________£¨Ñ¡Ìî±àºÅ£©

a£®Aµ½BµÄ¹ý³ÌÖУ¬ÈÜÖÊÖÊÁ¿Ã»Óиıä b£®BÖÐÈÜÖÊÓëÈܼÁµÄÖÊÁ¿±ÈΪ169:100

c£®AÈÜÒºµÄÖÊÁ¿µÈÓÚ222g d£®¿ªÊ¼Îö³öKNO3¹ÌÌåµÄζÈÔÚ60¡æÖÁ80¡æÖ®¼ä

37.0 ²»±¥ºÍ a d ¡¾½âÎö¡¿±¾Ì⿼²éÁ˱¥ºÍÈÜÒººÍ²»±¥ºÍÈÜÒºÅжϺÍÈܽâ¶ÈµÄ¸ÅÄî¡£ £¨1£©50¡æÊ±£¬ÂÈ»¯ÄƵÄÈܽâ¶ÈÊÇ37.0g£¬ 100gË®ÖÐ×î¶àÈܽâNaCl37.0g£» £¨2£©¢ÙAÊÇ80¡æº¬ÓÐ120gË®µÄKNO3ÈÜÒº£¬¾­¹ý²Ù×÷£¬µÃµ½102gKNO3¹ÌÌ壬20¡æÊ±£¬ÂÈ»¯ÄƵÄÈܽâ¶ÈÊÇ37g£¬ AÈÜÒºÖк¬ÓеÄÈÜÖÊÊÇ31.6g+102g=133.6g£¬ÈܼÁÖÊÁ¿ÊÇ120g£¬80¡æÊ±£¬ÂÈ...
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ϱíÊÇCa(OH)2ºÍNaOHµÄÈܽâ¶ÈÊý¾Ý¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

ζÈ/¡æ

0

20

40

60

80

100

Èܽâ¶È/g

Ca(OH)2

0.19

0.17

0.14

0.12

0.09

0.08

NaOH

42

109

129

174

314

347

(1)ÒÀ¾ÝÉϱíÊý¾Ý£¬»æÖÆCa(OH)2ºÍNaOHµÄÈܽâ¶ÈÇúÏߣ¬ÏÂͼÖÐÄܱíʾCa(OH)2Èܽâ¶ÈÇúÏßµÄÊÇ_______(Ìî¡°A¡±»ò¡°B¡±)¡£

(2)ÒªÏë°Ñһƿ½Ó½ü±¥ºÍµÄCa(OH)2ÈÜÒº±ä³É±¥ºÍÈÜÒº£¬ÓÐÏÂÁдëÊ©£º¢Ù¼ÓÈëÇâÑõ»¯¸Æ£¬¢ÚÉý¸ßζȣ¬¢Û½µµÍζȣ¬¢Ü¼ÓÈëË®£¬¢ÝÕô·¢Ë®ºóÔÙ»Ö¸´µ½Ô­Î¶ȡ£ÆäÖоù¿ÉÐеÄÒ»×éÊÇ______(ÌîÐòºÅ)

A£®¢Ù¢Ú¢Ü B£®¢Ú¢Û¢Ü C£®¢Ù¢Û¢Ý D£®¢Ù¢Ú¢Ý

(3)20¡æÊ±£¬200g±¥ºÍNaOHÈÜÒº£¬Õô·¢10gË®ºó£¬ÔÙ»Ö¸´µ½20¡æ£¬¿ÉÎö³öNaOH¾§ÌåµÄÖÊÁ¿Îª________g¡£

(4)ÏÖÓÐ20¡æÊ±Ca(OH)2µÄ±¥ºÍÈÜÒº(¼×ÈÜÒº)£¬ÏòÆäÖмÓÈëÒ»¶¨Á¿µÄCaO¹ÌÌåºó£¬»Ö¸´µ½20¡æ£¬µÃµ½µÄÈÜÒº(ÒÒÈÜÒº)£¬ÒÑÖª£ºCaO + H2O = Ca(OH)2 ¡£ Ôò¼×ÒÒÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýµÄ¹ØÏµÎª¦Ø(ÒÒ)_____¦Ø(¼×)(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±)¡£

(5)ÏÖÓÐ60¡æÊ±º¬Ca(OH)2ºÍNaOHÁ½ÖÖÈÜÖʵı¥ºÍÈÜÒº£¬ÈôÒªµÃµ½½Ï´¿¾»µÄNaOH¾§Ì壬Ӧ²ÉÈ¡µÄÎïÀí·½·¨ÊÇ________¡£

ÂÔ ÂÔ ÂÔ ÂÔ ÂÔ ¡¾½âÎö¡¿£¨1£©Óɱí¿ÉÖª£¬ÇâÑõ»¯ÄƵÄÈܽâ¶ÈËæÎ¶ȵÄÉý¸ß¶øÉý¸ß£¬ÇâÑõ»¯¸ÆµÄÈܽâ¶ÈËæÎ¶ȵÄÉý¸ß¶ø½µµÍ£¬¹ÊCa(OH)2Èܽâ¶ÈÇúÏßµÄÊÇA; £¨2£©ÇâÑõ»¯¸ÆµÄÈܽâ¶ÈËæÎ¶ȵÄÉý¸ß¶ø½µµÍ£¬¹Ê¿ÉÒÔ¼ÓÈëÈÜÖÊ£¬½µµÍζÈÊDz»±¥ºÍµÄÇâÑõ»¯¸ÆÈÜÒº±äΪ±¥ºÍÈÜÒº£¬¢Ù¢Ú¢ÝÕýÈ·£¬¹ÊÑ¡D£» £¨3£©20¡æÊ±£¬ÇâÑõ»¯ÄƵÄÈܽâ¶ÈΪ109g£¬¹Ê10gË®ÖÐÄÜÈܽâÇâÑõ»¯ÄƵÄÖÊÁ¿ÊÇ10.9g£» £¨4£©Ïò...

»¯Ñ§ÊÇÒ»ÃÅÒÔʵÑéΪ»ù´¡µÄѧ¿Æ¡£¸ù¾ÝÏÂͼ»Ø´ðÎÊÌ⣺

£¨1£©ÒÇÆ÷¢ÙµÄÃû³ÆÊÇ______________¡£

£¨2£©ÊµÑéÊÒÓÃÂÈËá¼ØºÍ¶þÑõ»¯ÃÌÖÆÈ¡ÑõÆø£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________£¬ÒªÊÕ¼¯Ò»Æ¿½Ï´¿¾»µÄÆøÌ壬ѡÓõķ¢ÉúºÍÊÕ¼¯×°ÖÃÊÇ______£¨Ìî×ÖĸÐòºÅ£¬ÏÂͬ£©¡£

£¨3£©ÊµÑéÊÒÖÆÈ¡¶þÑõ»¯Ì¼£¬·¢Éú×°ÖÃÒªÄÜ´ïµ½¡°¿ØÖÆ·´Ó¦ËÙ¶È¡±£¬ÐèҪѡÓõÄÒÇÆ÷ºÍÓÃÆ·ÓÐ_________£»·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________£»Ñ¡ÓõÄÊÕ¼¯×°ÖÃÊÇC£¬Ôò¶þÑõ»¯Ì¼´Ó____£¨Ìîµ¼¹ÜµÄÐòºÅ£©½øÈë¡£

£¨4£©×°ÖÃHÖУ¬Ïò¸ÉÔïʯÈïÊÔÖ½µÄa¶ËµÎÈëÕôÁóË®£¬¹Û²ìµ½µÄʵÑéÏÖÏóΪ_______£¬¸ÃʵÑéµÃ³öµÄ½áÂÛΪ___________¡£

£¨5£©Ð¡Ã÷ͬѧÒÔ·ÏÆúµÄ¡°°ô°ô±ù¡±ËÜÁϹÜΪ²ÄÁÏ£¬ÖÆ³ÉµÄÆøÌå·¢Éú×°Ö㨸÷²¿·ÖÁ¬½Ó½ôÃܺó£¬ÈçÏÂͼËùʾ£©¡£¸ÃʵÑéÉè¼ÆµÄÓŵãÓУºÄÜ¿ØÖÆ·´Ó¦ËÙÂʺÍ____________¡££¨Ð´Ò»µã¼´¿É£©

×¶ÐÎÆ¿ 2KClO32KCl + 3O2¡ü AD BEG CaCO3 +2HCl==CaCl2 + CO2¡ü+H2O m ÊÔÖ½a¶Ë±ä³ÉºìÉ« ¶þÑõ»¯Ì¼ÓëË®·´Ó¦Éú³ÉÁË̼Ëá ¿ØÖÆ·´Ó¦µÄÍ£Ö¹ºÍ·¢Éú ¡¾½âÎö¡¿±¾ÌâÖ÷Òª¿¼²éÁËÒÇÆ÷µÄÃû³Æ¡¢ÆøÌåµÄÖÆÈ¡×°ÖúÍÊÕ¼¯×°ÖõÄÑ¡Ôñ£¬Í¬Ê±Ò²¿¼²éÁË»¯Ñ§·½³ÌʽµÄÊéд¡¢ÆøÌåµÄÐÔÖʵȣ¬×ÛºÏÐԱȽÏÇ¿¡£ÆøÌåµÄÖÆÈ¡×°ÖõÄÑ¡ÔñÓë·´Ó¦ÎïµÄ״̬ºÍ·´Ó¦µÄÌõ¼þÓйأ»ÆøÌåµÄÊÕ¼¯×°ÖõÄÑ¡ÔñÓëÆøÌåµÄ...

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø