ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÓйØÎïÖʵÄÁ¿µÄ¼ÆËãÊÇÖÐѧ»¯Ñ§µÄÖØÒª²¿·Ö£¬Çë»Ø´ðÏÂÁÐÓйØÎïÖʵÄÁ¿µÄ¼ÆËãÎÊÌâ¡£

£¨1£©ÔÚ±ê×¼×´¿öÏ£¬67.2 L CO2ÊÇ__________mol£¬ÖÊÁ¿Îª_______g£¬º¬ÓÐ__________¸öCO2·Ö×Ó£¬ÆäÖк¬ÓÐ__________molÑõÔ­×Ó¡£

£¨2£©ÔÚ±ê×¼×´¿öÏ£¬1.7 g°±ÆøËùÕ¼µÄÌå»ýԼΪ_________L£¬ËüÓëͬÌõ¼þÏÂ_____mol H2Sº¬ÓÐÏàͬµÄÇâÔ­×ÓÊý¡£

£¨3£©ÏÖÓÐCO¡¢CO2¡¢O3ÈýÖÖÆøÌå,ËüÃǷֱ𶼺¬ÓÐ1 mol O,ÔòÈýÖÖÆøÌåµÄÎïÖʵÄÁ¿Ö®±ÈΪ_________¡£

£¨4£©±ê×¼×´¿öÏÂ,11.2 L XÆøÌå·Ö×ÓµÄÖÊÁ¿Îª16 g,ÔòXÆøÌåµÄĦ¶ûÖÊÁ¿ÊÇ___________¡£

¡¾´ð°¸¡¿3.0 132 3NA 6 2.24 0.15 6£º3£º2 32 g/mol

¡¾½âÎö¡¿

(1)ÔÚ±ê×¼×´¿öÏ£¬67.2 L CO2µÄÎïÖʵÄÁ¿Îª=3mol£¬ÖÊÁ¿Îª44g/mol¡Á3mol

=132g£¬º¬ÓÐ3NA¸öCO2·Ö×Ó£¬ÆäÖк¬ÓÐ3mol¡Á2=6molÑõÔ­×Ó£»

(2)n(NH3)==0.1mol£¬V(NH3)=0.1mol¡Á22.4L/mol=2.24L£¬n(H)=0.3mol£¬Ôòn(H2S)=0.15mol£»

(3)CO¡¢CO2¡¢O3ÈýÖÖÆøÌ壬ËüÃǺ¬ÓеÄÑõÔ­×Ó¸öÊýÖ®±ÈΪ1:2:3£¬ÔòÏÖÈýÖÖÆøÌåÖÐÑõÔ­×ÓµÄÎïÖʵÄÁ¿¶¼Îª1mol£¬Ôòn(CO)=1mol£¬n(CO2)=mol£¬n(O3)=mol£¬ÔòÕâÈýÖÖÆøÌåµÄÎïÖʵÄÁ¿Ö®±ÈΪ1::=6:3:2£»

(4)±ê×¼×´¿öÏ£¬11.2 L XÆøÌåµÄÎïÖʵÄÁ¿Îª=0.5mol£¬ÆøÌåµÄÖÊÁ¿Îª16 g£¬ÔòXÆøÌåµÄĦ¶ûÖÊÁ¿Îª=32g/mol¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿îëÍ­ÊÇÁ¦Ñ§¡¢»¯Ñ§×ÛºÏÐÔÄÜÁ¼ºÃµÄºÏ½ð£¬¹ã·ºÓ¦ÓÃÓÚÖÆÔì¸ß¼¶µ¯ÐÔÔª¼þ¡£ÒÔÏÂÊÇ´Óij·Ï¾ÉîëÍ­Ôª¼þ(º¬BeO¡¢CuS¡¢ÉÙÁ¿FeSºÍSiO2)ÖлØÊÕîëºÍÍ­Á½ÖÖ½ðÊôµÄÁ÷³Ì¡£

ÒÑÖª£º¢ñ£®îë¡¢ÂÁÔªËØ´¦ÓÚÖÜÆÚ±íÖеĶԽÇÏßλÖ㬻¯Ñ§ÐÔÖÊÏàËÆ

¢ò£®³£ÎÂÏ£ºKsp[Cu(OH)2]¡¢=2.2¡Á10£­20¡¢Ksp[Fe(OH)3]=4.0¡Á10£­38¡¢Ksp[Mn(OH)2]=2.l¡Á10£­13

(1)д³öîëÍ­Ôª¼þÖÐSiO2ÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ______________________¡£

(2)ÂËÔüBµÄÖ÷Òª³É·ÖΪ___________________(Ìѧʽ)¡£Ð´³ö·´Ó¦¢ñÖк¬î뻯ºÏÎïÓë¹ýÁ¿ÑÎËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ___________________________________________¡£

(3)¢ÙÈÜÒºCÖк¬NaCl¡¢BeCl2ºÍÉÙÁ¿HCl£¬ÎªÌá´¿BeCl2£¬Ñ¡ÔñºÏÀí²½Öè²¢ÅÅÐò________¡£

a£®¼ÓÈë¹ýÁ¿µÄNaOH b£®¹ýÂË c£®¼ÓÈëÊÊÁ¿µÄHCl

d£®¼ÓÈë¹ýÁ¿µÄ°±Ë® e£®Í¨Èë¹ýÁ¿µÄCO2 f£®Ï´µÓ

¢Ú´ÓBeCl2ÈÜÒºÖеõ½BeCl2¹ÌÌåµÄ²Ù×÷ÊÇ___________________________________¡£

(4)MnO2Äܽ«½ðÊôÁò»¯ÎïÖеÄÁòÔªËØÑõ»¯Îªµ¥ÖÊÁò£¬Ð´³ö·´Ó¦¢òÖÐCuS·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ________________________________¡£

(5)ÈÜÒºDÖк¬c(Cu2+)=2.2mol¡¤L£­1¡¢c(Fe3+)=0.008mol¡¤L£­1¡¢c(Mn2+)=0.01mol¡¤L£­1£¬ÖðµÎ¼ÓÈëÏ¡°±Ë®µ÷½ÚpH¿ÉÒÀ´Î·ÖÀ룬Ê×ÏȳÁµíµÄÊÇ___________(ÌîÀë×Ó·ûºÅ)¡£

¡¾ÌâÄ¿¡¿³¤Õ÷ÈýºÅ¼×ÔËÔØ»ð¼ý(CZ¡ª3A)Óë¡°æ϶𹤳̡°Ì«¿Õ°Ú¶É³µ¡±¡°±±¶·ÎÀÐÇ¡±¡°·çÔÆÎÀÐÇ¡±µÈÖйúº½Ìì´óʼþ½ôÃÜÏàÁ¬¡£ËüÊÇ´óÐÍÈý¼¶ÒºÌåÍƽø¼Á»ð¼ý£¬Ò»×Ó¼¶ºÍ¶þ×Ó¼¶¾ùʹÓÃÆ«¶þ¼×ëÂ(UDMH)-ËÄÑõ»¯¶þµª(NTO)Íƽø¼Á£¬·´Ó¦²úÎïÂÌÉ«ÎÞÎÛȾ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÒÑÖª(CH3)2NNH2(l)+4O2(g)= 2CO2(g)+4H2O(g)+N2(g) ¡÷H=a kJ/mol

N2(g)+O2(g)= 2NO(g) ¡÷H=b kJ/mol

2NO(g)+O2(g)= N2O4(l) ¡÷H=c kJ/mol

ÔòUDMH-NTOÍƽø¼Á·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ________________________¡£

(2)Æ«¶þ¼×ëÂÒ×ÈÜÓÚË®£¬ÆäһˮºÏÎïµÄµçÀ뷽ʽÓëһˮºÏ°±(Kb=1.7¡Á10-5)ÏàËƵ«µçÀë³£Êý¸üС£¬Æ«¶þ¼×ëÂһˮºÏÎïµÄµçÀë·½³ÌʽΪ___________________£¬ÏòÆ«¶þ¼×ëÂÈÜÒºÖмÓÈëµÈÎïÖʵÄÁ¿µÄ´×Ëá(Ka=-1.7¡Á10-5)£¬³ä·Ö·´Ó¦ºóÈÜÒº³Ê______(Ìî¡°ËáÐÔ¡±¡°¼îÐÔ¡±»ò¡°ÖÐÐÔ¡±)¡£¼îÐÔÆ«¶þ¼×롪¿ÕÆøȼÁϵç³ØµÄµç½âÖÊÈÜÒºÊÇ20¡«30£¥µÄKOHÈÜÒº£¬µç³Ø¹¤×÷ʱÕý¼«µÄµç¼«·´Ó¦Ê½Îª__________________________¡£

(3)N2O4ÊÇNO2µÄ¶þ¾Û²úÎNO¡¢NO2µÈµªÑõ»¯ÎïÊÇÖ÷ÒªµÄ´óÆøÎÛȾÎµªÑõ»¯ÎïÓëÐü¸¡ÔÚ´óÆøÖеÄ΢Á£Ï໥×÷ÓÃʱ£¬Éæ¼°ÈçÏ·´Ó¦£º

(I)2NO2(g)+NaCl(s)NaNO3(s)+ClNO(g) ¡÷H1<0 K1

(II)2NO(g)+Cl2(g)2ClNO(g) ¡÷H2<0 K2

Ôò4NO2(g)+2NaCl(s) 2NaNO3(s)+2NO(g)+C12(g)µÄƽºâ³£ÊýΪ_______(ÓÃK1¡¢K2±íʾ)£»ÎªÑо¿²»Í¬Ìõ¼þ¶Ô·´Ó¦(¢ò)µÄÓ°Ï죬T¡æʱ£¬Ïò2LºãÈÝÃܱÕÈÝÆ÷ÖмÓÈë0.2mol NOºÍ0.2mol C12£¬5minʱ´ïµ½Æ½ºâ£¬·´Ó¦¹ý³ÌÖÐÈÝÆ÷ÄÚµÄѹǿ¼õСÁË10%£¬Ôò5minÄÚ·´Ó¦µÄƽ¾ùËÙÂÊv(C1NO)=_____mol/(L¡¤min)£¬NOµÄƽºâת»¯ÂʦÁ1=_____£¥£»ÈôÆäËûÌõ¼þ±£³Ö²»±ä£¬Ê¹·´Ó¦(¢ò)ÔÚ³õʼÈÝ»ýΪ2LµÄºãѹÃܱÕÈÝÆ÷ÖнøÐУ¬ÔòNOµÄƽºâת»¯ÂʦÁ2_______¦Á1(Ìî¡°>¡±¡°<¡±»ò¡°=¡±)¡£

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

¾«Ó¢¼Ò½ÌÍø