µç¶Æ³§Ôø²ÉÓÃÓÐÇèµç¶Æ¹¤ÒÕ£¬´¦ÀíÓÐÇèµç¶ÆµÄ·Ïˮʱ£¬¿ÉÔÚ´ß»¯¼ÁTiO2×÷ÓÃÏ£¬ÏÈÓÃNaClO½«CN-Àë×ÓÑõ»¯³ÉCNO-£¬ÔÚËáÐÔÌõ¼þÏÂCNO-¼ÌÐø±»NaClOÑõ»¯³ÉN2ºÍCO2¡£»·±£¹¤×÷ÈËÔ±ÔÚÃܱÕϵͳÖÐÓÃÏÂͼװÖýøÐÐʵÑ飬ÒÔÖ¤Ã÷´¦Àí·½·¨µÄÓÐЧÐÔ£¬²¢Í¨¹ý²â¶¨¶þÑõ»¯Ì¼µÄÁ¿È·¶¨CN-±»´¦ÀíµÄ°Ù·ÖÂÊ¡£
½«Å¨Ëõºóº¬CN-Àë×ÓµÄÎÛË®Óë¹ýÁ¿NaClOÈÜÒºµÄ»ìºÏÒº¹²200 mL(ÆäÖÐCN-µÄŨ¶ÈΪ0.05 mol•L-1µ¹Èë¼×ÖУ¬ÈûÉÏÏðƤÈû£¬Ò»¶Îʱ¼äºó£¬´ò¿ªÏðƤÈûºÍ»îÈû£¬Ê¹ÈÜҺȫ²¿·ÅÈëÒÒÖУ¬¹Ø±Õ»îÈû¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¼×Öз´Ó¦µÄÀë×Ó·½³ÌʽΪ________________________£¬ÒÒÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ________________________¡£
£¨2£©ÒÒÖÐÉú³ÉµÄÆøÌå³ýN2ºÍCO2Í⣬»¹ÓÐHCl¼°¸±²úÎïCl2µÈ¡£±ûÖмÓÈëµÄ³ýÔÓÊÔ¼ÁÊDZ¥ºÍʳÑÎË®£¬Æä×÷ÓÃÊÇ_____________________£¬¶¡ÔÚʵÑéÖеÄ×÷ÓÃÊÇ______________£¬×°Óмîʯ»ÒµÄ¸ÉÔï¹ÜµÄ×÷ÓÃÊÇ______________________________¡£
£¨3£©ÎìÖÐÊ¢Óк¬Ca(OH)2 0.02molµÄʯ»ÒË®£¬ÈôʵÑéÖÐÎìÖй²Éú³É0.82 g³Áµí£¬Ôò¸ÃʵÑéÖвâµÃCN-±»´¦ÀíµÄ°Ù·ÖÂʵÈÓÚ__________¡£¸Ã²âµÃÖµÓ빤ҵʵ¼Ê´¦ÀíµÄ°Ù·ÖÂÊÏà±È×ÜÊÇÆ«µÍ£¬¼òҪ˵Ã÷¿ÉÄÜÔÒòÖ®Ò»_______________________¡£
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´£ººÓÄÏÊ¡ÄÏÑôÊÐ2016-2017ѧÄê¸ß¶þÏÂѧÆÚµÚÒ»´ÎÔ¿¼£¨3Ô£©»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÑ¡ÔñÌâ
¸ßÎÂϳ¬Ñõ»¯¼Ø¾§Ìå³ÊÁ¢·½Ìå½á¹¹£¬¾§ÌåÖÐÑõµÄ»¯ºÏ¼Û²¿·ÖΪ0¼Û£¬²¿·ÖΪ£2¼Û¡£ÈçÓÒͼËùʾΪ³¬Ñõ»¯¼Ø¾§ÌåµÄÒ»¸ö¾§°û(¾§ÌåÖÐ×îСµÄÖظ´µ¥Ôª)£¬ÔòÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ
A£®³¬Ñõ»¯¼ØµÄ»¯Ñ§Ê½ÎªKO2£¬Ã¿¸ö¾§°ûº¬ÓÐ4¸öK£«ºÍ4¸öO2-
B£®¾§ÌåÖÐÿ¸öK£«ÖÜΧÓÐ8¸öO2-£¬Ã¿¸öOÖÜΧÓÐ8¸öK£«
C£®¾§ÌåÖÐÓëÿ¸öK£«¾àÀë×î½üµÄK£«ÓÐ8¸ö
D£®¾§ÌåÖУ¬0¼ÛÑõÓë£2¼ÛÑõµÄÊýÄ¿±ÈΪ2:1
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016-2017ѧÄê¹óÖÝÊ¡¸ßÒ»£¨3Ô£©¿ªÑ§Öʼ컯ѧÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
X¡¢Y¡¢Z¡¢W¾ùΪ¶ÌÖÜÆÚÔªËØ£¬ËüÃÇÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÈçͼËùʾ¡£ÒÑÖªYÔ×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµÄ3±¶¡£ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ
A. Ô×Ӱ뾶£ºW>Z>Y>X
B. ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄËáÐÔ£ºZ>W>X
C. Yµ¥Öʵķеã±ÈZµ¥Öʵķеã¸ß
D. Wµ¥ÖʸúË®·´Ó¦ÐγɵÄÈÜÒº¾ßÓÐƯ°×ÐÔ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ì½Î÷Ê¡¸ßÈý3ÔÂÁª¿¼Àí¿Æ×ۺϻ¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º¼ò´ðÌâ
Îíö²ÌìÆøƵ·±³öÏÖ£¬ÑÏÖØÓ°ÏìÈËÃǵÄÉú»îºÍ½¡¿µ¡£ÆäÖÐÊ×ÒªÎÛȾÎïΪ¿ÉÎüÈë¿ÅÁ£ÎïPM2.5£¬ÆäÖ÷ÒªÀ´Ô´ÎªÈ¼Ãº¡¢»ú¶¯³µÎ²ÆøµÈ¡£Òò´Ë£¬¶ÔPM2.5¡¢SO2¡¢NOxµÈ½øÐÐÑо¿¾ßÓÐÖØÒªÒâÒå¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©½«PM2.5Ñù±¾ÓÃÕôÁóË®´¦ÀíÖƳɴý²âÊÔÑù¡£
Èô²âµÃ¸ÃÊÔÑùËùº¬Ë®ÈÜÐÔÎÞ»úÀë×ӵĻ¯Ñ§×é·Ö¼°Æäƽ¾ùŨ¶ÈÈçÏÂ±í£º
Àë×Ó | K+ | Na+ | NH4+ | SO42- | NO3- | Cl- |
Ũ¶Èmol/L | 4¡Á10-6 | 6¡Á10-6 | 2¡Á10-5 | 4¡Á10-5 | 3¡Á10-5 | 2¡Á10-5 |
¸ù¾Ý±íÖÐÊý¾ÝÅжÏÊÔÑùµÄpH=_________¡£
£¨2£©Æû³µÎ²ÆøÖÐNOxºÍCOµÄÉú³É£º
¢ÙÒÑÖªÆû¸×ÖÐÉú³ÉNOµÄ·´Ó¦Îª£ºN2(g)+O2(g)2NO(g) ¡÷H>0ºãΣ¬ºãÈÝÃܱÕÈÝÆ÷ÖУ¬ÏÂÁÐ˵·¨ÖÐÄÜ˵Ã÷¸Ã·´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ____
A.»ìºÏÆøÌåµÄÃܶȲ»Ôٱ仯 B.»ìºÏÆøÌåµÄƽ¾ù·Ö×ÓÁ¿²»Ôٱ仯
C.N2¡¢O2¡¢NOµÄÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã1¡Ã2 D.ÑõÆøµÄ°Ù·Öº¬Á¿²»Ôٱ仯
¢ÚÆû³µÈ¼ÓͲ»ÍêȫȼÉÕʱ²úÉúCO£¬ÓÐÈËÉèÏë°´ÏÂÁз´Ó¦³ýÈ¥CO£¬2CO(g)=2C(s)+O2(g)£¬ÒÑÖª¸Ã·´Ó¦µÄ¡÷H£¾0£¬Ôò¸ÃÉèÏëÄÜ·ñʵÏÖ______________£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©
£¨3£©Îª¼õÉÙSO2µÄÅÅ·Å£¬³£²ÉÈ¡µÄ´ëÊ©ÓУº
¢Ù½«Ãº×ª»¯ÎªÇå½àÆøÌåȼÁÏ¡£ÒÑÖª£º
H2(g)+ 1/2O2(g) =H2O(g) ¡÷H=£241.8kJ¡¤mol-1
C(s)+1/2O2(g) =CO(g)¡÷H =-110.5kJ¡¤mol-1
д³ö½¹Ì¿ÓëË®ÕôÆø·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º___________________¡£
¢ÚÏ´µÓº¬SO2µÄÑÌÆø¡£
£¨4£©Æû³µÎ²Æø¾»»¯µÄÖ÷ÒªÔÀí£º2NO(g)+2CO(g)2CO2(g)+N2(g)£»¡÷H£¼0£¬Èô¸Ã·´Ó¦ÔÚ¾øÈÈ¡¢ºãÈݵÄÃܱÕÌåϵÖнøÐУ¬ÏÂÁÐʾÒâͼÕýÈ·ÇÒÄÜ˵Ã÷·´Ó¦ÔÚ½øÐе½t1ʱ¿Ì´ïµ½Æ½ºâ״̬µÄÊÇ_________£¨ÌîÐòºÅ£©¡££¨ÈçͼÖÐvÕý¡¢K¡¢n¡¢w·Ö±ð±íʾÕý·´Ó¦ËÙÂÊ¡¢Æ½ºâ³£Êý¡¢ÎïÖʵÄÁ¿¡¢ÖÊÁ¿·ÖÊý£©
³µÁ¾ÅŷŵĵªÑõ»¯ÎúȼÉÕ²úÉúµÄ¶þÑõ»¯ÁòÊǵ¼ÖÂÎíö²ÌìÆøµÄ¡°×ï¿ý»öÊס±Ö®Ò»¡£»îÐÔÌ¿¿É´¦Àí´óÆøÎÛȾÎïNO¡£ÔÚ5LÃܱÕÈÝÆ÷ÖмÓÈëNOºÍ»îÐÔÌ¿£¨¼ÙÉèÎÞÔÓÖÊ£©¡£Ò»¶¨Ìõ¼þÏÂÉú³ÉÆøÌåEºÍF¡£µ±Î¶ȷֱðÔÚT1¡æºÍT2¡æʱ£¬²âµÃ¸÷ÎïÖÊƽºâʱÎïÖʵÄÁ¿£¨n/mol£©ÈçÏÂ±í£º
ÎïÖÊ Î¶È/¡æ | »îÐÔÌ¿ | NO | E | F |
³õʼ | 3.000 | 0.10 | 0 | 0 |
T1 | 2.960 | 0.020 | 0.040 | 0.040 |
T2 | 2.975 | 0.050 | 0.025 | 0.025 |
£¨1£©Ð´³öNOÓë»îÐÔÌ¿·´Ó¦µÄ»¯Ñ§·½³Ìʽ______________________£»
£¨2£©¼ÆËãÉÏÊö·´Ó¦T1¡æʱµÄƽºâ³£ÊýK1=__________________£»ÈôT1£¼T2£¬Ôò¸Ã·´Ó¦µÄ¡÷H__________________0(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£
£¨3£©ÉÏÊö·´Ó¦T1¡æʱ´ïµ½»¯Ñ§Æ½ºâºóÔÙͨÈë0.1molNOÆøÌ壬Ôò´ïµ½Ð»¯Ñ§Æ½ºâʱNOµÄת»¯ÂÊΪ________£»
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ì½Î÷Ê¡¸ßÈý3ÔÂÁª¿¼Àí¿Æ×ۺϻ¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ÏÂÁÐʵÑé²Ù×÷¼°ÏÖÏóÄܹ»´ïµ½¶ÔӦʵÑéÄ¿µÄµÄÊÇ£¨ ¡¡£©
ʵÑéÄÚÈÝ | ʵÑéÄ¿µÄ | |
A | ½«SO2ͨÈëÆ·ºìÈÜÒºÖУ¬Æ·ºìÍÊÉ«£»¼ÓÈÈÍÊÉ«ºóÆ·ºìÈÜÒº£¬ÈÜÒº»Ö¸´ºìÉ« | Ö¤Ã÷ÑÇÁòËáµÄ²»Îȶ¨ÐÔ |
B | ³£ÎÂÏ£¬Ïò±¥ºÍNa2CO3ÈÜÒºÖмÓÉÙÁ¿BaSO4·ÛÄ©£¬¹ýÂË£¬ÏòÏ´¾»µÄ³ÁµíÖмÓÏ¡ÑÎËᣬÓÐÉÙÁ¿ÆøÅݲúÉú | Ö¤Ã÷³£ÎÂÏ KSP(BaSO4)£¾KSP(BaCO3) |
C | ³£ÎÂϲⶨÎïÖʵÄÁ¿Å¨¶ÈÏàͬµÄÑÎËáºÍ´×ËáÈÜÒºµÄpH£ºÑÎËápHСÓÚ´×ËápH | Ö¤Ã÷ÏàͬÌõ¼þÏ£¬ÔÚË®ÖÐHClµçÀë³Ì¶È´óÓÚCH3COOH |
D | ½«Å¨ÁòËáºÍ̼µ¥ÖÊ»ìºÏ¼ÓÈÈ£¬Ö±½Ó½«Éú³ÉµÄÆøÌåͨÈë×ãÁ¿µÄ³ÎÇåʯ»ÒË®£¬Ê¯»ÒË®±ä»ë×Ç | ¼ìÑéÆøÌå²úÎïÖÐCO2µÄ´æÔÚ |
A. A B. B C. C D. D
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìÁÉÄþÊ¡°°É½ÊиßÈý3ÔÂÔ¿¼Àí¿Æ×ۺϻ¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
¶ÔÏÂÁи÷ÓлúÎïͬ·ÖÒì¹¹ÌåÊýÄ¿£¨²»¿¼ÂÇÁ¢ÌåÒì¹¹£©µÄÅжÏÖÐÕýÈ·µÄÊÇ
A. ·Ö×ÓʽΪC4H8£¬ÄÜ·¢Éú¼Ó³É·´Ó¦µÄͬ·ÖÒì¹¹ÌåÓÐ2ÖÖ
B. ·Ö×ÓʽΪC8H10µÄ¶þ¼×±½£¬±½»·ÉÏ.µÄÒ»¸öÇâÔ×Ó±»äåÔ×ÓÈ¡´ú£¬ËùµÃ²úÎïÓÐ6ÖÖ
C. ·Ö×ÓʽΪC4H8Cl2£¬Ö»ÓÐÒ»¸ö¼×»ùµÄͬ·ÖÒì¹¹ÌåÓÐ5ÖÖ
D. ·Ö×ÓʽΪC5H10O2£¬Ö»ÓÐÒ»¸ö¹ÙÄÜÍÅ£¬ÇÒÄÜÈÜÓÚË®µÄͬ·ÖÒì¹¹ÌåÓÐ2ÖÖ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016-2017ѧÄê½ËÕÊ¡¸ßÒ»ÏÂѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
¶ÌÖÜÆÚÖ÷×åÔªËØA¡¢B¡¢C¡¢D¡¢EµÄÔ×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐAÓëCͬÖ÷×壬AÓëÆäËüÔªËز»ÔÚͬһÖÜÆÚ£¬BÓëDͬÖ÷×壬³£ÎÂÏÂDµÄµ¥ÖÊΪµ»ÆÉ«¹ÌÌ壮ÏÂÁÐÍƶÏÖÐÕýÈ·µÄÊÇ£¨ £©
A. Ô×Ӱ뾶ÓÉСµ½´óµÄ˳Ðò£ºr£¨C£©£¼r£¨D£©£¼r£¨E£©
B. ÔªËØD¡¢E·Ö±ðÓëAÐγɵĻ¯ºÏÎïµÄÈÈÎȶ¨ÐÔ£ºE£¾D
C. ÔªËØDµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄËáÐÔ±ÈEµÄÇ¿
D. ÔªËØB·Ö±ðÓëA¡¢CÐγɵĻ¯ºÏÎïÖл¯Ñ§¼üµÄÀàÐÍÍêÈ«Ïàͬ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016-2017ѧÄê½ËÕÊ¡¸ßÒ»ÏÂѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ÏÂÁÐÎïÖʵĵç×ÓʽÊéдÕýÈ·µÄÊÇ£¨ £©
A. B. C. D.
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016-2017ѧÄêºÓÄÏÊ¡ÂåÑôÊиßÒ»3ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ÎåÖÖ¶ÌÖÜÆÚÔªËصÄijЩÐÔÖÊÈç±íËùʾ£¨ÆäÖÐÖ»ÓÐW¡¢Y¡¢ZΪͬÖÜÆÚÔªËØ£©ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ( )
A. ÓÉQÓëYÐγɵĻ¯ºÏÎïÖÐÖ»´æÔÚÀë×Ó¼ü B. ZÓëXÖ®¼äÐγɵĻ¯ºÏÎï¾ßÓл¹ÔÐÔ
C. ÓÉX¡¢Y¡¢ZÈýÖÖÔªËØÐγɵĻ¯ºÏÎһ¶¨Êǹ²¼Û»¯ºÏÎï D. YÓëWÐγɵĻ¯ºÏÎïÖУ¬YÏÔ¸º¼Û
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