4£®ÒÑ֪ij¡°84Ïû¶¾Òº¡±Æ¿Ì岿·Ö±êÇ©Èçͼ1Ëùʾ£¬¸Ã¡°84Ïû¶¾Òº¡±Í¨³£Ï¡ÊÍ100±¶£¨Ìå»ýÖ®±È£©ºóʹÓã®Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ä³Í¬Ñ§È¡100mL¸Ã¡°84Ïû¶¾Òº¡±£¬Ï¡ÊͺóÓÃÓÚÏû¶¾£¬Ï¡ÊͺóµÄÈÜÒºÖÐc£¨Na+£©=0.04 mol•L-1£®
£¨2£©¸Ãͬѧ²ÎÔĸá°84Ïû¶¾Òº¡±µÄÅä·½£¬ÓûÓÃNaClO¹ÌÌåÅäÖÆ480mL¸ÃÏû¶¾Òº£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇC£¨Ìî×Öĸ£©£®
A£®Èçͼ2ËùʾµÄÒÇÆ÷ÖУ¬ÓÐÈýÖÖÊDz»ÐèÒªµÄ£¬»¹ÐèÒªÒ»ÖÖ²£Á§ÒÇÆ÷
B£®ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴ¾»ºó£¬Ó¦ºæ¸Éºó²ÅÄÜÓÃÓÚÈÜÒºÅäÖÆ
C£®ÅäÖƹý³ÌÖУ¬Î´ÓÃÕôÁóˮϴµÓÉÕ±­ºÍ²£Á§°ô¿ÉÄܵ¼Ö½á¹ûÆ«µÍ
D£®ÐèÒª³ÆÁ¿NaClO¹ÌÌåµÄÖÊÁ¿Îª143.0g
£¨3£©ÅäÖÆÈÜÒºµÄ¹ý³ÌÈçͼ3£¬¸ÃͬѧµÄ´íÎó²½ÖèÓÐB
A.1´¦        B.2´¦        C.3´¦      D.4´¦
£¨4£©¡°84Ïû¶¾Òº¡±ÓëÏ¡ÁòËá»ìºÏʹÓÿÉÔöÇ¿Ïû¶¾ÄÜÁ¦£¬Ä³Ïû¶¾Ð¡×éÈËÔ±ÓÃ98%£¨ÃܶÈΪ1.84g•cm-3£©µÄŨÁòËáÅäÖÆ2L 2.3mol•L-1µÄÏ¡ÁòËáÓÃÓÚÔöÇ¿¡°84Ïû¶¾Òº¡±µÄÏû¶¾ÄÜÁ¦£®ÐèÓÃŨÁòËáµÄÌå»ýΪ250mL£®

·ÖÎö £¨1£©¸ù¾Ýº¬25%NaClO¡¢1000mL¡¢ÃܶÈ1.19g•cm-3£¬½áºÏc=$\frac{1000¦Ñw}{M}$À´¼ÆËã´ÎÂÈËáÄƵÄÎïÖʵÄÁ¿Å¨¶È£¬ÒÀ¾ÝÏ¡ÊÍÇ°ºóÈÜÖʵÄÎïÖʵÄÁ¿²»±äÀ´¼ÆË㣻
£¨2£©¸ù¾ÝÈÜÒºµÄÅäÖƼ°c=$\frac{n}{V}$¡¢m=nMÀ´·ÖÎö£»
£¨3£©ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÒ»°ã²½Ö裺¼ÆËã¡¢³ÆÁ¿¡¢ÈܽâÀäÈ´¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔÈ£¬¾Ý´Ë½â´ð£»
£¨4£©¸ù¾ÝÈÜҺϡÊÍÇ°ºóÈÜÖʵÄÎïÖʵÄÁ¿²»±ä¼ÆË㣮

½â´ð ½â£º£¨1£©c£¨NaClO£©=c=$\frac{1000¦Ñw}{M}$=$\frac{1000¡Á1.19¡Á25%}{74.5}$=4.0 mol•L-1£¬Ï¡Êͺóc£¨NaClO£©=$\frac{1}{100}$¡Á4.0 mol•L-1=0.04 mol•L-1£¬c£¨Na+£©=c£¨NaClO£©=0.04 mol•L-1£¬
¹Ê´ð°¸Îª£º0.04£»
£¨2£©A£®ÐèÓÃÍÐÅÌÌìƽ³ÆÁ¿NaClO¹ÌÌ壬ÐèÓÃÉÕ±­À´ÈܽâNaClO£¬ÐèÓò£Á§°ô½øÐнÁ°èºÍÒýÁ÷£¬ÐèÓÃÈÝÁ¿Æ¿ºÍ½ºÍ·µÎ¹ÜÀ´¶¨ÈÝ£¬Í¼Ê¾µÄA¡¢B¡¢²»ÐèÒª£¬µ«»¹Ðè²£Á§°ôºÍ½ºÍ·µÎ¹Ü£¬¹ÊA´íÎó£»
B£®ÅäÖƹý³ÌÖÐÐèÒª¼ÓÈëË®£¬ËùÒÔ¾­Ï´µÓ¸É¾»µÄÈÝÁ¿Æ¿²»±Øºæ¸ÉºóÔÙʹÓ㬹ÊB´íÎó£»
C£®Öƹý³ÌÖУ¬Î´ÓÃÕôÁóˮϴµÓÉÕ±­ºÍ²£Á§°ô£¬µ¼ÖÂÅäÖƵÄÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿¼õС£¬½á¹ûÆ«µÍ£¬¹ÊCÕýÈ·£»
D£®Ó¦Ñ¡È¡500 mLµÄÈÝÁ¿Æ¿½øÐÐÅäÖÆ£¬È»ºóÈ¡³ö480 mL¼´¿É£¬ËùÒÔÐèÒªNaClOµÄÖÊÁ¿£º0.5 L¡Á4.0 mol•L-1¡Á74.5 g•mol-1=149 g£¬¹ÊD´íÎó£»
¹Ê´ð°¸Îª£ºC£»
£¨3£©ÒÆÒººóδϴµÓÉÕ±­ºÍ²£Á§°ô£¬¶¨ÈÝʱ¶ÁÊýӦƽÊӿ̶ÈÏߣ¬
¹ÊÑ¡£ºB£º
£¨4£©ÈÜҺϡÊÍÇ°ºóÈÜÖʵÄÎïÖʵÄÁ¿²»±ä£¬Å¨ÁòËáµÄŨ¶ÈΪc=$\frac{1000¡Á1.84g/L¡Á98%}{98g/mol}$=18.4mol/L£¬¼ÙÉèÐèҪŨÁòËáµÄÌå»ýΪV£¬
ÔòV¡Á18.4mol/L=2L¡Á2.3mol/L£¬V=0.25L=250mL£¬
¹Ê´ð°¸Îª£º250£®

µãÆÀ ±¾Ì⿼²éÎïÖʵÄÁ¿Å¨¶ÈµÄ¼ÆËãÒÔ¼°ÈÜÒºµÄÅäÖÆ£¬Îª¸ßƵ¿¼µã£¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦¡¢ÊµÑéÄÜÁ¦ºÍ¼ÆËãÄÜÁ¦µÄ¿¼²é£¬×¢Òâ°ÑÎÕÏà¹Ø¼ÆË㹫ʽµÄÔËÓã¬ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

13£®ÏÂÁÐÃüÃûÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®3-¼×»ù¶¡ÍéB£®2£¬2£¬4£¬4-Ëļ׻ùÐÁÍé
C£®2-¼×»ù-3-ÒÒÏ©»ùÒÒÍéD£®2-¼×»ù-3-ÎìÏ©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

15£®³ýÈ¥À¨ºÅÄÚÔÓÖÊËùÓÃÊÔ¼ÁºÍ·½·¨£¬ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÒÒ´¼£¨ÒÒËᣩ   Éúʯ»Ò          ¹ýÂË
B£®ÒÒ´¼£¨±½·Ó£©  NaOHÈÜÒº       ·ÖÒº
C£®äåÒÒÍ飨ÒÒ´¼£©  Ë®              ·ÖÒº
D£®±½¼×Ëᣨ±½¼×ËáÄÆ£©ÁòËá           ÕôÁó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

12£®ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®1molNa2O2Óë×ãÁ¿Ë®·´Ó¦×ªÒƵç×ÓÊýΪ2NA
B£®28g N2¡¢COµÄ»ìºÏÆøÌåÖк¬ÓÐÔ­×ÓµÄÊýÄ¿µÈÓÚ2NA
C£®³£Î³£Ñ¹Ï£¬22.4L NH3·Ö×ӵĵç×ÓÊý¡¢ÖÊ×ÓÊý¾ùΪ10NA
D£®1mol N2Óë3mol H2»ìºÏ£¬³ä·Ö·´Ó¦ºóÉú³É°±Æø·Ö×ÓÊýΪ2NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

19£®°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣮
£¨1£©¸ù¾ÝϵͳÃüÃû·¨£¬µÄÃû³ÆÊÇ2-¼×»ù¶¡Í飻
£¨2£©ôÇ»ùµÄµç×ÓʽÊÇ£»
£¨3£©Ë³-2-¶¡Ï©µÄ½á¼òÊÇ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

9£®£¨1£©Ð´³öÏÂÁл¯ºÏÎïµÄÃû³Æ»ò½á¹¹¼òʽ£º
¢Ù¼×ËáÒÒõ¥£¬
¢Ú2£¬5-¶þ¼×»ù-2£¬4-¼º¶þÏ©µÄ½á¹¹¼òʽ£º£®
£¨2£©Âé»ÆËØÓֳƻƼÊÇÎÒ¹úÌض¨µÄÖÐÒ©²ÄÂé»ÆÖÐËùº¬ÓеÄÒ»ÖÖÉúÎï¼î£¬¾­ÎÒ¹ú¿Æѧ¼ÒÑо¿·¢ÏÖÆä½á¹¹Èçͼ£º

Âé»ÆËØÖк¬Ñõ¹ÙÄÜÍŵÄÃû³ÆÊÇôÇ»ù£¬ÊôÓÚ´¼ÀࣨÌî¡°´¼¡±»ò¡°·Ó¡±£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

16£®ÏÂÁи÷×éÎïÖÊÖУ¬ÎïÖÊÖ®¼äͨ¹ýÒ»²½·´Ó¦¾ÍÄÜʵÏÖÈçͼËùʾת»¯µÄÊÇ£¨¡¡¡¡£©
abc
AAlAlCl3Al£¨OH£©3
BNONO2HNO3
CSiSiO2H2SiO3
DFeS2SO2H2SO4
A£®AB£®BC£®CD£®D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

13£®¿×ȸʯµÄÖ÷Òª³É·ÖÊÇCu2£¨OH£©2CO3£¨º¬Fe2O3¡¢FeCO3¡¢Al2O3¡¢SiO2ÔÓÖÊ£©£¬¹¤ÒµÉÏÓÿ×ȸʯÖƱ¸ÁòËáÍ­µÄµÚÒ»²½ÐèÓùýÁ¿µÄÁòËáÈܽⲢ¹ýÂË£®³£ÎÂÏ£¬·Ö±ðÈ¡ÂËÒº²¢ÏòÆäÖмÓÈëÖ¸¶¨ÎïÖÊ£¬·´Ó¦ºóµÄÈÜÒºÖÐÖ÷Òª´æÔÚµÄÒ»×éÀë×ÓÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¼ÓÈë¹ýÁ¿°±Ë®£ºFe3+¡¢NH4+¡¢SO42-¡¢OH-
B£®¼ÓÈë¹ýÁ¿NaOHÈÜÒº£ºNa+¡¢AlO2-¡¢SO42-¡¢OH-
C£®¼ÓÈë¹ýÁ¿H2O2ÈÜÒº£ºFe2+¡¢H+¡¢SO42-¡¢Cu2+
D£®¼ÓÈë¹ýÁ¿NaHCO3ÈÜÒº£ºNa+¡¢Al3+¡¢SO42-¡¢HCO3-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

1£®°´ÒªÇóÌî¿Õ£º
£¨1£©Ô­×ÓÖÖÀàÓÉÖÊ×ÓÊýºÍÖÐ×ÓÊý£¬¾ö¶¨£»µÚÈýÖÜÆڰ뾶×îСµÄÔªËØÐγɵļòµ¥Àë×ÓµÄÀë×ӽṹʾÒâͼÊÇ£»
£¨2£©ÒÔ»ÆÌú¿óΪԭÁÏÉú²úÁòËáµÄ¹¤ÒÕÁ÷³ÌÈçͼ1Ëùʾ£¬É豸BµÄÃû³ÆΪ½Ó´¥Ê½£»É豸AÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ4FeS2+11O2$\frac{\underline{\;¸ßÎÂ\;}}{\;}$2Fe2O3+8SO2£»
£¨3£©Í¼2ÊÇʵÑéÊÒÖг£ÓÃÓڲⶨÑÎËáºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦ÈȵÄ×°Ö㬴Ë×°ÖÃÃû³Æ½Ð¼òÒ×Á¿Èȼƣ»
£¨4£©0.5mol CH3OH£¨l£©ÍêȫȼÉÕÉú³É¶þÑõ»¯Ì¼ÆøÌåºÍҺ̬ˮʱ·Å³öÈÈÁ¿Îª363.2kJ£¬Ð´³ö±íʾCH3OH£¨l£©È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽCH3OH£¨l£©+$\frac{3}{2}$O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-726.4KJ/mol£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸