¡¾ÌâÄ¿¡¿ÏÂͼËùʾװÖÃÖУ¬¼×¡¢ÒÒ¡¢±ûÈý¸öÉÕ±­ÒÀ´Î·Ö±ðÊ¢·Å100 g 5.00%µÄNaOHÈÜÒº¡¢×ãÁ¿µÄCuSO4ÈÜÒººÍ100 g 10.00%µÄK2SO4ÈÜÒº£¬µç¼«¾ùΪʯīµç¼«¡£

£¨1£©½ÓͨµçÔ´£¬¾­¹ýÒ»¶Îʱ¼äºó£¬²âµÃ±ûÖÐK2SO4Ũ¶ÈΪ10.47%£¬ÒÒÖÐcµç¼«ÖÊÁ¿Ôö¼Ó¡£¾Ý´Ë»Ø´ðÎÊÌ⣺

¢ÙµçÔ´µÄN¶ËΪ£ß£ß£ß£ß£ß£ß¼«£»

¢Úµç¼«bÉÏ·¢ÉúµÄµç¼«·´Ó¦Îª£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£»

¢ÛÁÐʽ¼ÆËãµç¼«bÉÏÉú³ÉµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ý£º

£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß

¢Üµç¼«cµÄÖÊÁ¿±ä»¯Êǣߣߣߣߣߣߣߣߣßg£»

¢Ýµç½âÇ°ºó¸÷ÈÜÒºµÄËá¡¢¼îÐÔ´óСÊÇ·ñ·¢Éú±ä»¯£¬¼òÊöÆäÔ­Òò£º

¼×ÈÜÒº£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£»

ÒÒÈÜÒº£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£»

±ûÈÜÒº£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£»

£¨2£©Èç¹ûµç½â¹ý³ÌÖÐÍ­È«²¿Îö³ö£¬´Ëʱµç½âÄÜ·ñ¼ÌÐø½øÐУ¬ÎªÊ²Ã´£¿

£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß¡£

¡¾´ð°¸¡¿£¨1£©¢ÙÕý¼« ¢Ú4OH£­-4e£­=2H2O + O2¡ü¡£

¢ÛË®¼õÉÙµÄÖÊÁ¿£º

Éú³ÉO2Ìå»ý£º

¢Ü16g

¢Ý¼îÐÔÔö´ó£¬ÒòΪµç½âºó£¬Ë®Á¿¼õÉÙÈÜÒºÖÐNaOHŨ¶ÈÔö´ó

ËáÐÔÔö´ó£¬ÒòΪÑô¼«ÉÏOH-Éú³ÉO2£¬ÈÜÒºÖÐH+Àë×ÓŨ¶ÈÔö´ó

Ëá¼îÐÔ´óСûÓб仯£¬ÒòΪK2SO4ÊÇÇ¿ËáÇ¿¼îÑΣ¬Å¨¶ÈÔö¼Ó²»Ó°ÏìÈÜÒºµÄËá¼îÐÔ

(2)ÄܼÌÐø½øÐУ¬ÒòΪCuSO4ÈÜÒºÒÑת±äΪH2SO4ÈÜÒº£¬·´Ó¦Ò²¾Í±äΪˮµÄµç½â·´Ó¦¡£

¡¾½âÎö¡¿£¨1£©¢ÙÒÒÖÐCµç¼«ÖÊÁ¿Ôö¼Ó£¬Ôòc´¦·¢ÉúµÄ·´Ó¦Îª£ºCu2£«+2e£­=Cu£¬¼´C´¦ÎªÒõ¼«£¬ÓÉ´Ë¿ÉÍƳöbΪÑô¼«£¬aΪÒõ¼«£¬MΪ¸º¼«£¬NΪÕý¼«¡£±ûÖÐΪK2SO4£¬Ï൱ÓÚµç½âË®£¬Éèµç½âµÄË®µÄÖÊÁ¿Îªxg¡£Óɵç½âÇ°ºóÈÜÖÊÖÊÁ¿ÏàµÈÓУ¬100¡Á10%=£¨100-x£©¡Á10.47%£¬µÃx=4.5g£¬¹ÊΪ0.25mol¡£ÓÉ·½³Ìʽ2H2+O2 2H2O¿ÉÖª£¬Éú³É2molH2O£¬×ªÒÆ4molµç×Ó£¬ËùÒÔÕû¸ö·´Ó¦ÖÐת»¯0.5molµç×Ó£¬¶øÕû¸öµç·ÊÇ´®ÁªµÄ£¬¹Êÿ¸öÉÕ±­Öеĵ缫ÉÏתÒƵç×ÓÊýÊÇÏàµÈµÄ¡£¢Ú¼×ÖÐΪNaOH£¬Ï൱ÓÚµç½âH2O£¬Ñô¼«b´¦ÎªÒõÀë×ÓOH£­·Åµç£¬¼´4OH£­-4e£­=2H2O + O2¡ü¡£¢ÛתÒÆ0.5molµç×Ó£¬ÔòÉú³ÉO2Ϊ0.5/4=0.125mol£¬±ê¿öϵÄÌå»ýΪ0.125¡Á22.4=2.8L¡£¢ÜCu2£«+2e£­=Cu£¬×ªÒÆ0.5molµç×Ó£¬ÔòÉú³ÉµÄm(Cu)=0.5/2 ¡Á64 =16g¡£¢Ý¼×ÖÐÏ൱ÓÚµç½âË®£¬¹ÊNaOHµÄŨ¶ÈÔö´ó£¬pH±ä´ó¡£ÒÒÖÐÒõ¼«ÎªCu2£«·Åµç£¬Ñô¼«ÎªOH£­·Åµç£¬ËùÒÔH£«Ôö¶à£¬¹ÊpH¼õС¡£±ûÖÐΪµç½âË®£¬¶ÔÓÚK2SO4¶øÑÔ£¬ÆäpH¼¸ºõ²»±ä¡££¨2£©Í­È«²¿Îö³ö£¬¿ÉÒÔ¼ÌÐøµç½âH2SO4£¬Óеç½âÒº¼´¿Éµç½â¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿A¡¢B¡¢C¡¢D¡¢E¡¢FÊǺ˵çºÉÊýÒÀ´ÎÔö´óµÄÁùÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬AÔªËصÄÔ­×ÓºËÄÚÖ»ÓÐ1¸öÖÊ×Ó£»BÔªËصÄÔ­×Ӱ뾶ÊÇÆäËùÔÚÖ÷×åÖÐ×îСµÄ£¬BµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½ÎªHBO3£»CÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊý±È´ÎÍâ²ã¶à4£»CµÄÒõÀë×ÓÓëDµÄÑôÀë×Ó¾ßÓÐÏàͬµÄµç×ÓÅŲ¼£¬Á½ÔªËØ¿ÉÐγɻ¯ºÏÎïD2C£»C¡¢EͬÖ÷×å¡£

(1)BÔÚÖÜÆÚ±íÖеÄλÖÃ______________________________________________

(2)FÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïµÄ»¯Ñ§Ê½Îª___________________________________¡£

(3)ÔªËØC¡¢D¡¢EÐγɵļòµ¥Àë×Ӱ뾶ÓÉСµ½´óµÄ˳Ðò________________________(ÓÃÀë×Ó·ûºÅ±íʾ)¡£

(4)Óõç×Óʽ±íʾ»¯ºÏÎïD2CµÄÐγɹý³Ì£º__________________________________________________¡£

C¡¢D»¹¿ÉÐγɻ¯ºÏÎïD2C2£¬D2C2Öк¬ÓеĻ¯Ñ§¼üÊÇ_________________________________________¡£

(5)C¡¢EµÄÇ⻯Î·ÐµãÓɸߵ½µÍ˳ÐòÊÇ£º_______________________________¡£

(6)д³ö̼µ¥ÖÊÓëEµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïŨÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¬²¢Óõ¥ÏßÇűêÃ÷µç×ÓµÄתÒÆ·½Ïò_______________¡£µ±×ªÒƵç×ÓΪ0.2molʱ£¬±ê×¼×´¿öÏ·´Ó¦²úÉúÆøÌå_______________L¡£

(7)ÒÑÖªEµ¥ÖʺÍFµ¥ÖʵÄË®ÈÜÒº·´Ó¦»áÉú³ÉÁ½ÖÖÇ¿ËᣬÆäÀë×Ó·½³ÌʽΪ_________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ô­×ÓÐòÊýΪ 31 µÄÔªËØ R£¬ÔÚÖÜÆÚ±íÖеÄλÖÃΪ

A.µÚÈýÖÜÆÚµÚVA×åB.µÚËÄÖÜÆÚµÚIIIA ×å

C.µÚÎåÖÜÆÚµÚIIIA ×åD.µÚËÄÖÜÆÚµÚVA ×å

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖªA¡¢B¡¢CºÍ¼×¡¢ÒÒ¡¢±û¾ùÊÇÓɶÌÖÜÆÚÔªËØÐγɵÄÎïÖÊ£¬DÊǹý¶ÉÔªËØÐγɵij£¼ûµ¥ÖÊ£¬ËüÃÇÖ®¼äÄÜ·¢ÉúÈçÏ·´Ó¦¡££¨Í¼ÖÐÓÐЩ·´Ó¦µÄ²úÎïºÍ·´Ó¦Ìõ¼þûÓбê³ö£©

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©±ûµÄµç×ÓʽΪ_____£»×é³ÉÆøÌåÒÒµÄÔªËØÔÚÖÜÆÚ±íµÄλÖÃ____£»ÎïÖÊBº¬ÓеĻ¯Ñ§¼üÀàÐÍÊÇ______£»

£¨2£©Ð´³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ£º

¢Ü _________________________£»

¢Ý _____________________£»

£¨3£©½ðÊôDÓëÏ¡ÏõËá·´Ó¦£¬²úÉú±ê×¼×´¿öÏÂ1.12LµÄNOÆøÌ壬Ôò²Î¼Ó·´Ó¦µÄÏõËáΪ_______mol¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ò»¶¨Ìõ¼þÏ£¬ÔÚÃܱպãÈÝÈÝÆ÷ÖУ¬Äܱíʾ·´Ó¦X(g)+2Y(g)2Z(g)Ò»¶¨´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ(¡¡¡¡)

¢ÙX¡¢Y¡¢ZµÄÃܶȲ»ÔÙ·¢Éú±ä»¯ ¢ÚvÕý£¨Y£©£½ 2vÄ棨X£© ¢ÛÈÝÆ÷ÖеÄѹǿ²»ÔÙ·¢Éú±ä»¯ ¢Üµ¥Î»Ê±¼äÄÚÉú³Énmol Z£¬Í¬Ê±Éú³É2nmolY

A. ¢Ù¢Ú B. ¢Ù¢Ü C. ¢Ú¢Û D. ¢Û¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÈçͼËùʾ£¬¸Ü¸Ë AB Á½¶Ë·Ö±ð¹ÒÓÐÌå»ýÏàͬ£®ÖÊÁ¿ÏàµÈµÄ¿ÕÐÄÍ­ÇòºÍ¿ÕÐÄÌúÇò£¬µ÷½Ú¸Ü¸Ë²¢Ê¹ÆäÔÚË®Öб£³Öƽºâ£¬È»ºóСÐĵØÏòÉÕ±­ÖÐÑëµÎÈë M µÄŨÈÜÒº£¬Ò»¶Îʱ¼äºó£¬ ÏÂÁÐÓйظܸ˵ÄÆ«ÏòÅжÏÕýÈ·µÄÊÇ£¨ÊµÑé¹ý³ÌÖУ¬²»¿¼ÂÇÁ½ÇòµÄ¸¡Á¦±ä»¯£©£¨ £©

A. µ± MΪÑÎËá¡¢¸Ü¸ËΪµ¼Ìåʱ£¬A ¶Ë¸ß£¬B ¶ËµÍ

B. µ± MΪ AgNO3¡¢¸Ü¸ËΪµ¼Ìåʱ£¬A ¶Ë¸ß£¬B ¶ËµÍ

C. µ± MΪ CuSO4¡¢¸Ü¸ËΪµ¼Ìåʱ£¬A ¶ËµÍ£¬B ¶Ë¸ß

D. µ± MΪ CuSO4¡¢¸Ü¸ËΪ¾øÔµÌåʱ£¬A ¶ËµÍ£¬B ¶Ë¸ß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¢ñ.»¯ºÏÎï Mg5Al3(OH)19(H2O)4 ¿É×÷»·±£ÐÍ×èȼ²ÄÁÏ£¬ÊÜÈÈʱ°´ÈçÏ»¯Ñ§·½³Ìʽ·Ö½â£º2Mg5Al3(OH)19(H2O)427H2O¡ü+10MgO+3Al2O3

(1)д³ö¸Ã»¯ºÏÎï×÷×èȼ¼ÁµÄÁ½ÌõÒÀ¾Ý___________¡£

(2)ÓÃÀë×Ó·½³Ìʽ±íʾ³ýÈ¥¹ÌÌå²úÎïÖÐ Al2O3 µÄÔ­Àí___________

(3)ÒÑÖª MgO¿ÉÈÜÓÚNH4ClµÄË®ÈÜÒº£¬Óû¯Ñ§·½³Ìʽ±íʾÆäÔ­Àí_____________¡£

¢ò.´ÅÐÔ²ÄÁÏ A(M=296g/mol)ÊÇÓÉÁ½ÖÖÔªËØ×é³ÉµÄ»¯ºÏÎijÑо¿Ð¡×é°´ÈçͼÁ÷³Ì̽ ¾¿Æä×é³É£º

Çë»Ø´ð£º

(4)AµÄ»¯Ñ§Ê½Îª_____¡£C ÖгʻÆÉ«µÄÀë×Ó¶ÔÓ¦ÔªËØÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃΪ__________¡£

(5)ÒÑÖª»¯ºÏÎï A ÄÜÓëÏ¡ÁòËá·´Ó¦£¬Éú³ÉÒ»ÖÖµ­»ÆÉ«²»ÈÜÎïºÍÒ»ÖÖÆøÌå(±ê¿öÏ£¬¸ÃÆøÌå¶Ô°±ÆøµÄÏà¶ÔÃܶÈΪ2)£¬¸ÃÆøÌå·Ö×ӵĵç×ÓʽΪ_____¡£Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ_________¡£

(6)д³öF¡úG·´Ó¦µÄ»¯Ñ§·½³Ìʽ____________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿1L1mol/LAl2(SO4)3ÈÜÒºÖк¬ÓÐSO42-µÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ£¨ £©

A.1mol/LB.20mol/L

C.3mol/LD.2mol/L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚÒ»ÃܱÕÈÝÆ÷ÖУ¬µÈÎïÖʵÄÁ¿µÄAºÍB·¢Éú·´Ó¦£ºA£¨g£©+2B(g) 2C(g),·´Ó¦´ïƽºâʱ£¬Èô»ìºÏÆøÌåÖÐAºÍBµÄÎïÖʵÄÁ¿Ö®ºÍÓëCµÄÎïÖʵÄÁ¿ÏàµÈ£¬ÔòÕâʱAµÄת»¯ÂÊΪ£¨ £©

A. 40% B. 50% C. 60% D. 70%

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸