´ÎÁ×Ëá(H3PO2)ÊÇÒ»ÖÖ¾«Ï¸Á×»¯¹¤²úÆ·£¬¾ßÓнÏÇ¿»¹Ô­ÐÔ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)H3PO2ÊÇÒ»ÖÖÖÐÇ¿Ëᣬд³öÆäµçÀë·½³Ìʽ£º_____________________________

________________________________________________________________________¡£

(2)H3PO2¼°NaH2PO2¾ù¿É½«ÈÜÒºÖеÄAg£«»¹Ô­ÎªÒø£¬´Ó¶ø¿ÉÓÃÓÚ»¯Ñ§¶ÆÒø¡£

¢ÙH3PO2ÖУ¬PÔªËصĻ¯ºÏ¼ÛΪ________¡£

¢ÚÀûÓÃH3PO2½øÐл¯Ñ§¶ÆÒø·´Ó¦ÖУ¬Ñõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ4¡Ã1£¬ÔòÑõ»¯²úÎïΪ________(Ìѧʽ)¡£

¢ÛNaH2PO2Ϊ________(Ìî¡°ÕýÑΡ±»ò¡°ËáʽÑΡ±)£¬ÆäÈÜÒºÏÔ________(Ìî¡°ÈõËáÐÔ¡±¡°ÖÐÐÔ¡±»ò¡°Èõ¼îÐÔ¡±)¡£

(3)H3PO2µÄ¹¤ÒµÖÆ·¨ÊÇ£º½«°×Á×(P4)ÓëBa(OH)2ÈÜÒº·´Ó¦Éú³ÉPH3ÆøÌåºÍBa(H2PO2)2£¬ºóÕßÔÙÓëH2SO4·´Ó¦¡£Ð´³ö°×Á×ÓëBa(OH)2ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º____________________________________________¡£

(4)H3PO2Ò²¿ÉÓõçÉøÎö·¨ÖƱ¸£¬¡°ËÄÊÒµçÉøÎö·¨¡±¹¤×÷Ô­ÀíÈçͼËùʾ(ÑôĤºÍÒõĤ·Ö±ðÖ»ÔÊÐíÑôÀë×Ó¡¢ÒõÀë×Óͨ¹ý)£º

¢Ùд³öÑô¼«µÄµç¼«·´Ó¦Ê½£º_________________________________________¡£

¢Ú·ÖÎö²úÆ·Êҿɵõ½H3PO2µÄÔ­Òò£º_____________________________________

________________________________________________________________________¡£

¢ÛÔçÆÚ²ÉÓá°ÈýÊÒµçÉøÎö·¨¡±ÖƱ¸H3PO2£º½«¡°ËÄÊÒµçÉøÎö·¨¡±ÖÐÑô¼«ÊÒµÄÏ¡ÁòËáÓÃH3PO2Ï¡ÈÜÒº´úÌ棬²¢³·È¥Ñô¼«ÊÒÓë²úÆ·ÊÒÖ®¼äµÄÑôĤ£¬´Ó¶øºÏ²¢ÁËÑô¼«ÊÒÓë²úÆ·ÊÒ¡£ÆäȱµãÊDzúÆ·ÖлìÓÐ________ÔÓÖÊ£¬¸ÃÔÓÖʲúÉúµÄÔ­ÒòÊÇ________________________________¡£


(1)H3PO2H2PO£«H£«¡¡(2)¢Ù£«1¡¡¢ÚH3PO4¡¡¢ÛÕýÑΡ¡Èõ¼îÐÔ

(3)2P4£«3Ba(OH)2£«6H2O£½3Ba(H2PO2)2£«2PH3¡ü

(4)¢Ù2H2O£­4e£­===O2¡ü£«4H£«

¢ÚÑô¼«ÊÒµÄH£«´©¹ýÑôĤÀ©É¢ÖÁ²úÆ·ÊÒ£¬Ô­ÁÏÊÒµÄH2PO´©¹ýÒõĤÀ©É¢ÖÁ²úÆ·ÊÒ£¬¶þÕß·´Ó¦Éú³ÉH3PO2¡¡¢ÛPO¡¡H2PO»òH3PO2±»Ñõ»¯

[½âÎö] (1)H3PO2ΪһԪÈõËᣬÆäµçÀë·½³ÌʽΪH3PO2H£«£«H2PO¡£(2)ÓÉ»¯ºÏ¼Û´úÊýºÍΪ0¿ÉÈ·¶¨PΪ£«1¼Û£»¢Ú¸ù¾ÝÌâÖÐÐÅϢд³ö»¯Ñ§·½³ÌʽΪ4Ag£«£«H3PO2£«2H2O===4Ag£«H3PO4£«4H£«£¬¼´Ñõ»¯²úÎïΪH3PO4£»¢ÛNaH2PO2Ϊǿ¼îÈõËáÑΣ¬ÈÜÒº³ÊÈõ¼îÐÔ¡£(3)¸ù¾ÝÌâÖÐÐÅÏ¢ºÍ·´Ó¦Ç°ºóÔªËØ»¯ºÏ¼Û±ä»¯Ð´³ö»¯Ñ§·½³ÌʽΪ2P4£«3Ba(OH)2£«6H2O===2PH3¡ü£«3Ba(H2PO2)2¡£(4)¢ÙÑô¼«ÊÇË®µçÀë³öµÄOH£­·Åµç£¬Æ䷴ӦʽΪ2H2O£­4e£­===O2¡ü£«4H£«£»¢ÚÑô¼«ÊÒÖеÄH£«´©¹ýÑôĤ½øÈë²úÆ·ÊÒ£¬Ô­ÁÏÊÒµÄH2PO´©¹ýÒõĤÀ©É¢ÖÁ²úÆ·ÊÒ£¬¶þÕß·´Ó¦Éú³ÉH3PO2£»¢ÛÑô¼«ÊÒÄÚ¿ÉÄÜÓв¿·ÖH2PO»òH3PO2ʧµç×Ó·¢ÉúÑõ»¯·´Ó¦£¬µ¼ÖÂÉú³ÉÎïÖлìÓÐPO¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


½«15 mL 2 mol/L Na2CO3ÈÜÒºÖðµÎ¼ÓÈëµ½40 mL 0.5 mol/L MClnÑÎÈÜÒºÖУ¬Ç¡ºÃ½«ÈÜÒºÖеÄMn£«Àë×ÓÍêÈ«³ÁµíΪ̼ËáÑΣ¬ÔòMClnÖÐnÖµÊÇ(¡¡¡¡)

A£®4        B£®3

C£®2        D£®1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


½«±¥ºÍFeCl3ÈÜÒºµÎÈë·ÐË®²¢Öó·ÐÒ»¶Îʱ¼ä£¬¿ÉµÃµ½ºìºÖÉ«ÒºÌ壬´ËÒºÌå²»¾ßÓеÄÐÔÖÊÊÇ(¡¡¡¡)

A£®¹âÊøͨ¹ý¸ÃÒºÌåʱÐγɹâÁÁµÄ¡°Í¨Â·¡±

B£®²åÈëʯīµç¼«Í¨Ö±Á÷µçºó£¬ÓÐÒ»¼«¸½½üÒºÌåÑÕÉ«¼ÓÉî

C£®Ïò¸ÃÒºÌåÖмÓÈëÏõËáÒøÈÜÒº£¬ÎÞ³Áµí²úÉú

D£®½«¸ÃÒºÌå¼ÓÈÈ¡¢Õô¸É¡¢×ÆÉÕºó£¬ÓÐÑõ»¯ÎïÉú³É

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ë®ÈÜÒºÖÐÄÜ´óÁ¿¹²´æµÄÒ»×éÀë×ÓÊÇ(¡¡¡¡)

A£®Na£«¡¢Ca2£«¡¢Cl£­¡¢SO

B£®Fe2£«¡¢H£«¡¢SO¡¢ClO£­

C£®K£«¡¢Fe3£«¡¢NO¡¢SCN£­

D£®Mg2£«¡¢NH¡¢Cl£­¡¢SO

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Áò»¯ÇâµÄת»¯ÊÇ×ÊÔ´ÀûÓúͻ·¾³±£»¤µÄÖØÒªÑо¿¿ÎÌâ¡£ÓÉÁò»¯Çâ»ñµÃÁòµ¥ÖÊÓжàÖÖ·½·¨¡£

(1)½«ÉÕ¼îÎüÊÕH2SºóµÄÈÜÒº¼ÓÈëµ½ÈçͼËùʾµÄµç½â³ØµÄÑô¼«Çø½øÐеç½â¡£µç½â¹ý³ÌÖÐÑô¼«Çø·¢ÉúÈçÏ·´Ó¦£º

S2£­£­2e£­===S¡¡(n£­1)S£«S2£­===S

¢Ùд³öµç½âʱÒõ¼«µÄµç¼«·´Ó¦Ê½£º________________¡£

¢Úµç½âºóÑô¼«ÇøµÄÈÜÒºÓÃÏ¡ÁòËáËữµÃµ½Áòµ¥ÖÊ£¬ÆäÀë×Ó·½³Ìʽ¿Éд³É__________________________¡£

(2)½«H2SºÍ¿ÕÆøµÄ»ìºÏÆøÌåͨÈëFeCl3¡¢FeCl2¡¢CuCl2µÄ»ìºÏÈÜÒºÖз´Ó¦»ØÊÕS£¬ÆäÎïÖÊת»¯ÈçͼËùʾ¡£

¢ÙÔÚͼʾµÄת»¯ÖУ¬»¯ºÏ¼Û²»±äµÄÔªËØÊÇ________¡£

¢Ú·´Ó¦Öе±ÓÐ1 mol H2Sת»¯ÎªÁòµ¥ÖÊʱ£¬±£³ÖÈÜÒºÖÐFe3£«µÄÎïÖʵÄÁ¿²»±ä£¬ÐèÏûºÄO2µÄÎïÖʵÄÁ¿Îª________¡£

¢ÛÔÚζÈÒ»¶¨ºÍ²»²¹¼ÓÈÜÒºµÄÌõ¼þÏ£¬»ºÂýͨÈë»ìºÏÆøÌ壬²¢³ä·Ö½Á°è¡£ÓûʹÉú³ÉµÄÁòµ¥ÖÊÖв»º¬CuS£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ________________¡£

(3)H2SÔÚ¸ßÎÂÏ·ֽâÉú³ÉÁòÕôÆøºÍH2¡£Èô·´Ó¦ÔÚ²»Í¬Î¶ÈÏ´ﵽƽºâʱ£¬»ìºÏÆøÌåÖи÷×é·ÖµÄÌå»ý·ÖÊýÈçͼËùʾ£¬H2SÔÚ¸ßÎÂÏ·ֽⷴӦµÄ»¯Ñ§·½³ÌʽΪ__________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


H2O2ÊÇÒ»ÖÖÂÌÉ«Ñõ»¯»¹Ô­ÊÔ¼Á£¬ÔÚ»¯Ñ§Ñо¿ÖÐÓ¦Óù㷺¡£

(1)ijС×éÄâÔÚͬŨ¶ÈFe3£«µÄ´ß»¯Ï£¬Ì½¾¿H2O2Ũ¶È¶ÔH2O2·Ö½â·´Ó¦ËÙÂʵÄÓ°Ïì¡£ÏÞÑ¡ÊÔ¼ÁÓëÒÇÆ÷£º30%H2O2ÈÜÒº¡¢0.1 mol¡¤L£­1Fe2(SO4)3ÈÜÒº¡¢ÕôÁóË®¡¢×¶ÐÎÆ¿¡¢Ë«¿×Èû¡¢Ë®²Û¡¢½º¹Ü¡¢²£Á§µ¼¹Ü¡¢Á¿Í²¡¢Ãë±í¡¢ºãÎÂˮԡ²Û¡¢×¢ÉäÆ÷¡£

¢Ùд³ö±¾ÊµÑéH2O2·Ö½â·´Ó¦·½³Ìʽ²¢±êÃ÷µç×ÓתÒƵķ½ÏòºÍÊýÄ¿£º______________________________¡£

¢ÚÉè¼ÆʵÑé·½°¸£ºÔÚ²»Í¬H2O2Ũ¶ÈÏ£¬²â¶¨________(ÒªÇóËù²âµÃµÄÊý¾ÝÄÜÖ±½ÓÌåÏÖ·´Ó¦ËÙÂÊ´óС)¡£

¢ÛÉè¼ÆʵÑé×°Öã¬Íê³ÉͼÖеÄ×°ÖÃʾÒâͼ¡£

¢Ü²ÎÕÕϱí¸ñʽ£¬ÄⶨʵÑé±í¸ñ£¬ÍêÕûÌåÏÖʵÑé·½°¸(ÁгöËùÑ¡ÊÔ¼ÁÌå»ý¡¢Ðè¼Ç¼µÄ´ý²âÎïÀíÁ¿ºÍËùÄⶨµÄÊý¾Ý£»Êý¾ÝÓÃ×Öĸ±íʾ)¡£

¡¡¡¡ÎïÀíÁ¿

ʵÑéÐòºÅ ¡¡¡¡

V[0.1 mol¡¤L£­1

Fe2(SO4)3]/mL

¡­¡­

1

a

¡­¡­

2

a

¡­¡­

(2)ÀûÓÃͼ(a)ºÍ(b)ÖеÄÐÅÏ¢£¬°´Í¼(c)×°ÖÃ(Á¬Í¨µÄA¡¢BÆ¿ÖÐÒѳäÓÐNO2ÆøÌå)½øÐÐʵÑé¡£¿É¹Û²ìµ½BÆ¿ÖÐÆøÌåÑÕÉ«±ÈAÆ¿ÖеÄ__________(Ìî¡°É»ò¡°Ç³¡±)£¬ÆäÔ­ÒòÊÇ____________________________¡£

¡¡¡¡¡¡¡¡¡¡(a)¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡(b)

(c)

ͼ21

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¼îʽ̼ËáÂÁþ

[MgaAlb(OH)c(CO3)d¡¤xH2O]³£ÓÃ×÷ËÜÁÏ×èȼ¼Á¡£

(1)¼îʽ̼ËáÂÁþ¾ßÓÐ×èȼ×÷Óã¬ÊÇÓÉÓÚÆäÊÜÈÈ·Ö½âÐèÎüÊÕ´óÁ¿ÈÈÁ¿ºÍ________________________________¡£

(2)MgaAlb(OH)c(CO3)d¡¤xH2OÖÐa¡¢b¡¢c¡¢dµÄ´úÊý¹ØϵʽΪ________¡£

(3)Ϊȷ¶¨¼îʽ̼ËáÂÁþµÄ×é³É£¬½øÐÐÈçÏÂʵÑ飺

¢Ù׼ȷ³ÆÈ¡3.390 g ÑùÆ·Óë×ãÁ¿Ï¡ÑÎËá³ä·Ö·´Ó¦£¬Éú³ÉCO2 0.560 L(ÒÑ»»Ëã³É±ê×¼×´¿öÏÂ)¡£

¢ÚÁíÈ¡Ò»¶¨Á¿ÑùÆ·ÔÚ¿ÕÆøÖмÓÈÈ£¬ÑùÆ·µÄ¹ÌÌå²ÐÁôÂÊ(¡Á100%)Ëæζȵı仯ÈçͼËùʾ(ÑùÆ·ÔÚ270 ¡æʱÒÑÍêȫʧȥ½á¾§Ë®£¬600 ¡æÒÔÉϲÐÁô¹ÌÌåΪ½ðÊôÑõ»¯ÎïµÄ»ìºÏÎï)¡£

¸ù¾ÝÒÔÉÏʵÑéÊý¾Ý¼ÆËã¼îʽ̼ËáÂÁþÑùÆ·ÖеÄn(OH£­)¡Ãn(CO)(д³ö¼ÆËã¹ý³Ì)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


±È½ÏÒÒÍéºÍÒÒ´¼µÄ½á¹¹£¬ÏÂÁÐ˵·¨´íÎóµÄÊÇ(¡¡¡¡)

A£®Á½¸ö̼ԭ×ÓÒÔµ¥¼üÏàÁ¬

B£®·Ö×ÓÀﶼº¬ÓÐ6¸öÏàͬµÄÇâÔ­×Ó

C£®ÒÒ»ùÓëÒ»¸öÇâÔ­×ÓÏàÁ¬¾ÍÊÇÒÒÍé·Ö×Ó

D£®ÒÒ»ùÓëÒ»¸öôÇ»ùÏàÁ¬¾ÍÊÇÒÒ´¼·Ö×Ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ò»¶¨ÄÜÔÚÏÂÁÐÈÜÒºÖдóÁ¿¹²´æµÄÀë×Ó×éÊÇ(¡¡¡¡)

A£®º¬ÓдóÁ¿Al3£«µÄÈÜÒº£ºNa£«¡¢NH¡¢SO¡¢Cl£­

B£®c(H£«)£½1¡Á10£­13 mol/LµÄÈÜÒº£ºNa£«¡¢Ca2£«¡¢SO¡¢CO

C£®º¬ÓдóÁ¿Fe3£«µÄÈÜÒº£ºNa£«¡¢Mg2£«¡¢NO¡¢SCN£­

D£®º¬ÓдóÁ¿NOµÄÈÜÒº£ºH£«¡¢Fe2£«¡¢SO¡¢Cl£­

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸