ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿¢ñ¡¢°´ÒªÇóÌî¿Õ£º

£¨1£©32 g CH4£¬ÆäĦ¶ûÖÊÁ¿Îª_______£¬Ô¼º¬ÓÐ____Ħµç×Ó£¬ÔÚ±ê×¼×´¿öÏÂËùÕ¼µÄÌå»ýԼΪ____L¡£

£¨2£©µÈÌå»ýµÈÎïÖʵÄÁ¿Å¨¶ÈµÄCa(OH)2ÈÜÒºÓëNaHCO3ÈÜÒº»ìºÏ£¬Àë×Ó·½³ÌʽΪ£º______¡£

£¨3£©H++HCO3-£½H2O+CO2¡ü¶ÔÓ¦µÄÒ»¸ö»¯Ñ§·½³Ìʽ________________________¡£

£¨4£©ÖƱ¸ÇâÑõ»¯Ìú½ºÌåµÄÀë×Ó·½³Ìʽ________________¡£

¢ò¡¢ÏÖÓÐÒÔÏÂÎïÖÊ£º

¢ÙNaCl¹ÌÌå ¢ÚҺ̬SO3 ¢ÛҺ̬µÄ´×Ëá ¢Ü¹¯ ¢ÝBaSO4¹ÌÌå ¢ÞÕáÌÇ(C12H22O11) ¢ß¾Æ¾«(C2H5OH) ¢àÈÛ»¯µÄKNO3 ¢áÑÎËá¡£Çë»Ø´ðÏÂÁÐÎÊÌâ(ÌîÐòºÅ)¡£

£¨1£©ÒÔÉÏÎïÖÊÖÐÄܵ¼µçµÄÊÇ__________________¡£

£¨2£©ÒÔÉÏÎïÖÊÊôÓÚµç½âÖʵÄÊÇ_________________¡£

£¨3£©ÒÔÉÏÎïÖÊÖÐÊôÓڷǵç½âÖʵÄÊÇ______________¡£

¡¾´ð°¸¡¿16g/mol 20 44.8 Ca2++OH£­+HCO3£­£½CaCO3¡ý+H2O HCl+NaHCO3£½NaCl+CO2¡ü+H2O Fe3++3H2OFe(OH)3(½ºÌå)+3H+ ¢Ü¢à¢á ¢Ù¢Û¢Ý¢à ¢Ú¢Þ¢ß

¡¾½âÎö¡¿

¢ñ¡¢£¨1£©¼×ÍéµÄÏà¶Ô·Ö×ÓÖÊÁ¿ÊÇ16£¬Ôò¼×ÍéµÄĦ¶ûÖÊÁ¿ÊÇ16g/mol£¬32 g CH4µÄÎïÖʵÄÁ¿ÊÇ32g¡Â16g/mol£½2mol£¬1·Ö×Ó¼×Í麬ÓÐ10¸öµç×Ó£¬Òò´Ëº¬ÓÐ20molµç×Ó¡£¸ù¾ÝV=nVm¿ÉÖªÔÚ±ê×¼×´¿öÏÂËùÕ¼µÄÌå»ýԼΪ2mol¡Á22.4L/mol£½44.8L¡£

£¨2£©µÈÌå»ýµÈÎïÖʵÄÁ¿Å¨¶ÈµÄCa(OH)2ÈÜÒºÓëNaHCO3ÈÜÒº»ìºÏ·´Ó¦Éú³É̼Ëá¸Æ¡¢ÇâÑõ»¯ÄƺÍË®£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪCa2++OH£­+HCO3£­£½CaCO3¡ý+H2O¡£

£¨3£©H++HCO3-£½H2O+CO2¡ü¶ÔÓ¦µÄÒ»¸ö»¯Ñ§·½³Ìʽ¿ÉÒÔÊÇHCl+NaHCO3£½NaCl+CO2¡ü+H2O¡£

£¨4£©±¥ºÍÂÈ»¯ÌúÈÜÒºµÎÈë·ÐË®ÖмÌÐø¼ÓÈÈÒ»¶Îʱ¼äºó¼´¿ÉµÃµ½ÇâÑõ»¯Ìú½ºÌ壬ÔòÖƱ¸ÇâÑõ»¯Ìú½ºÌåµÄÀë×Ó·½³ÌʽΪFe3++3H2OFe(OH)3(½ºÌå)+3H+¡£

¢ò¡¢ÈÜÓÚË®»òÈÛÈÚ״̬ÏÂÄܹ»µ¼µçµÄ»¯ºÏÎïÊǵç½âÖÊ£¬ÈÜÓÚË®ºÍÔÚÈÛÈÚ״̬Ͼù²»Äܵ¼µçµÄ»¯ºÏÎïÊǷǵç½âÖÊ£¬º¬ÓÐ×ÔÓÉÒƶ¯µç×Ó»òÀë×ÓµÄÎïÖÊ¿ÉÒÔµ¼µç¡£

¢ÙNaCl¹ÌÌå²»µ¼µç£¬Êǵç½âÖÊ£»

¢ÚҺ̬SO3²»µ¼µç£¬ÊǷǵç½âÖÊ£»

¢ÛҺ̬µÄ´×Ëá²»µ¼µç£¬Êǵç½âÖÊ£»

¢Ü¹¯ÊǽðÊô£¬Äܵ¼µç£¬²»Êǵç½âÖÊÒ²²»ÊǷǵç½âÖÊ£»

¢ÝBaSO4¹ÌÌå²»µ¼µç£¬Êǵç½âÖÊ£»

¢ÞÕáÌÇ(C12H22O11)²»µ¼µç£¬ÊǷǵç½âÖÊ£»

¢ß¾Æ¾«(C2H5OH)²»µ¼µç£¬ÊǷǵç½âÖÊ£»

¢àÈÛ»¯µÄKNO3Äܵ¼µç£¬Êǵç½âÖÊ£»

¢áÑÎËáÄܵ¼µç£¬ÊôÓÚ»ìºÏÎ²»Êǵç½âÖÊÒ²²»ÊǷǵç½âÖÊ¡£Ôò

£¨1£©ÒÔÉÏÎïÖÊÖÐÄܵ¼µçµÄÊǢܢà¢á¡£

£¨2£©ÒÔÉÏÎïÖÊÊôÓÚµç½âÖʵÄÊǢ٢ۢݢࡣ

£¨3£©ÒÔÉÏÎïÖÊÖÐÊôÓڷǵç½âÖʵÄÊǢڢޢߡ£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Áò»¯¼î·¨Êǹ¤ÒµÉÏÖƱ¸Na2S2O3µÄ·½·¨Ö®Ò»£¬·´Ó¦Ô­ÀíΪ£º2Na2S+Na2CO3+4SO2=3Na2S2O3+CO2£¨¸Ã·´Ó¦¡÷H£¾0£©¡£Ä³Ñо¿Ð¡×éÔÚʵÑéÊÒÓÃÁò»¯¼î·¨ÖƱ¸Na2S2O3¡¤5H2OÁ÷³ÌÈçÏ¡£

£¨1£©ÎüÁò×°ÖÃÈçͼËùʾ¡£

¢Ù×°ÖÃBµÄ×÷ÓÃÊǼìÑé×°ÖÃAÖÐSO2µÄÎüÊÕЧÂÊ£¬BÖÐÊÔ¼ÁÊÇ_____________£¬±íÃ÷SO2ÎüÊÕЧÂʵ͵ÄʵÑéÏÖÏóÊÇBÖÐÈÜÒº______________¡£

¢ÚΪÁËʹSO2¾¡¿ÉÄÜÎüÊÕÍêÈ«£¬ÔÚ²»¸Ä±äAÖÐÈÜҺŨ¶È¡¢Ìå»ýµÄÌõ¼þÏ£¬³ýÁ˼°Ê±½Á°è·´Ó¦ÎïÍ⣬»¹¿É²ÉÈ¡µÄºÏÀí´ëÊ©ÊÇ______________¡££¨ÈÎдһÌõ£©

£¨2£©¼ÙÉ豾ʵÑéËùÓõÄNa2CO3º¬ÉÙÁ¿NaCl¡¢NaOH£¬Éè¼ÆʵÑé·½°¸½øÐмìÑé¡££¨ÊÒÎÂʱCaCO3±¥ºÍÈÜÒºµÄpH£½10.2£©£¬ÏÞÑ¡ÊÔ¼Á¼°ÒÇÆ÷£ºÏ¡ÏõËá¡¢AgNO3ÈÜÒº¡¢CaCl2ÈÜÒº¡¢·Ó̪ÈÜÒº¡¢ÕôÁóË®¡¢pH¼Æ¡¢ÉÕ±­¡¢ÊԹܡ¢µÎ¹Ü

ÐòºÅ

ʵÑé²Ù×÷

Ô¤ÆÚÏÖÏó

½áÂÛ

¢Ù

È¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®£¬³ä·ÖÕñµ´Èܽ⣬____________ ¡£

Óа×É«³ÁµíÉú³É

ÑùÆ·º¬NaCl

¢Ú

ÁíÈ¡ÉÙÁ¿ÑùÆ·ÓÚÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®£¬³ä·Ö½Á°èÈܽ⣬_______¡£

Óë°×É«³ÁµíÉú³É£¬ÉϲãÇåÒºpH>10.2

ÑùÆ·º¬NaOH

£¨3£©Na2S2O3ÈÜÒºÊǶ¨Á¿ÊµÑéÖеij£ÓÃÊÔ¼Á£¬²â¶¨ÆäŨ¶ÈµÄ¹ý³ÌÈçÏ£º

µÚÒ»²½£º×¼È·³ÆÈ¡a g KIO3(Ïà¶Ô·Ö×ÓÖÊÁ¿£º214)¹ÌÌåÅä³ÉÈÜÒº£¬

µÚ¶þ²½£º¼ÓÈë¹ýÁ¿KI¹ÌÌåºÍH2SO4ÈÜÒº£¬µÎ¼Óָʾ¼Á£¬

µÚÈý²½£ºÓÃNa2S2O3ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄNa2S2O3ÈÜÒºµÄÌå»ýΪv mL¡£Ôòc(Na2S2O3)£½______mol¡¤L-1¡££¨Ö»ÁгöËãʽ£¬²»×÷ÔËË㣩

ÒÑÖª£ºIO3-+I-+6H+=3I2+3H2O 2S2O32-+I2=S4O62-+2I-¡£¼×ͬѧʢװNa2S2O3ÈÜҺ֮ǰδÈóÏ´£¬ÕâÑù²âµÃµÄNa2S2O3µÄŨ¶È¿ÉÄÜ________£¨Ìî¡°ÎÞÓ°Ï족¡¢¡°Æ«µÍ¡±»ò¡°Æ«¸ß¡±)£»ÒÒͬѧµÚÒ»²½ºÍµÚ¶þ²½µÄ²Ù×÷¶¼ºÜ¹æ·¶£¬µÚÈý²½µÎËÙÌ«Âý£¬ÕâÑù²âµÃµÄNa2S2O3µÄŨ¶È¿ÉÄÜ________(Ìî¡°ÎÞÓ°Ï족¡¢¡° Æ«µÍ¡±»ò¡°Æ«¸ß¡±)¡£

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

¾«Ó¢¼Ò½ÌÍø