]ÁòËáÄÆ­¹ýÑõ»¯Çâ¼ÓºÏÎï(xNa2SO4·yH2O2·zH2O)µÄ×é³É¿Éͨ¹ýÏÂÁÐʵÑé²â¶¨£º

¢Ù׼ȷ³ÆÈ¡1.770 0 gÑùÆ·£¬ÅäÖƳÉ100.00 mLÈÜÒºA¡£

¢Ú׼ȷÁ¿È¡25.00 mLÈÜÒºA£¬¼ÓÈëÑÎËáËữµÄBaCl2ÈÜÒºÖÁ³ÁµíÍêÈ«£¬¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïÖÁºãÖØ£¬µÃµ½°×É«¹ÌÌå0.582 5 g¡£

¢Û׼ȷÁ¿È¡25.00 mLÈÜÒºA£¬¼ÓÊÊÁ¿Ï¡ÁòËáËữºó£¬ÓÃ0.020 00 mol·L£­1 KMnO4ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄKMnO4ÈÜÒº25.00 mL¡£H2O2ÓëKMnO4·´Ó¦µÄÀë×Ó·½³ÌʽÈçÏ£º

2MnO£«5H2O2£«6H£«===2Mn2£«£«8H2O£«5O2¡ü

(2)ÉÏÊöµÎ¶¨Èô²»¼ÓÏ¡ÁòËáËữ£¬MnO±»»¹Ô­ÎªMnO2£¬ÆäÀë×Ó·½³ÌʽΪ________________________________________________________________________¡£

(3)ͨ¹ý¼ÆËãÈ·¶¨ÑùÆ·µÄ×é³É(д³ö¼ÆËã¹ý³Ì)¡£


´ð°¸¡¡(2)2MnO£«3H2O2===2MnO2¡ý£«3O2¡ü£«2OH£­£«2H2O

(3)n(Na2SO4)£½n(BaSO4)£½

£½2.50¡Á10£­3 mol

2MnO£«5H2O2£«6H£«===2Mn2£«£«8H2O£«5O2¡ü

n(H2O2)£½¡Á£½1.25¡Á10£­3 mol

m(Na2SO4)£½142 g·mol£­1¡Á2.50¡Á10£­3 mol£½0.355 g

m(H2O2)£½34 g·mol£­1¡Á1.25¡Á10£­3 mol£½0.042 5 g

n(H2O)£½£½2.50¡Á10£­3 mol

x¡Ãy¡Ãz£½n(Na2SO4)¡Ãn(H2O2)¡Ãn(H2O)£½2¡Ã1¡Ã2

¹ÊÑùÆ·µÄ×é³ÉµÄ»¯Ñ§Ê½Îª2Na2SO4·H2O2·2H2O¡£

½âÎö¡¡(2)H2O2ÓëKMnO4ÈÜÒº·´Ó¦£¬MnO±»»¹Ô­ÎªMnO2£¬¾ÝµÃʧµç×ÓÊغ㡢ÖÊÁ¿Êغã¿Éд³öÀë×Ó·½³ÌʽΪ2MnO£«3H2O2===2MnO2¡ý£«3O2¡ü£«2OH£­£«2H2O¡£

(3)25.00 mLÈÜÒºAÖк¬ÓÐNa2SO4µÄÎïÖʵÄÁ¿Îªn(Na2SO4)£½n(BaSO4)£½£½2.50¡Á10£­3 mol¡£

º¬ÓÐH2O2µÄÎïÖʵÄÁ¿Îªn(H2O2)£½n(KMnO4)£½¡Á0.020 00 mol·L£­1¡Á25.00¡Á10£­3 L£½1.25¡Á10£­3 mol¡£

ËùÈ¡25.00 mLÈÜÒºAÖÐËùº¬ÑùÆ·ÖÐË®µÄÖÊÁ¿Îªm(H2O)£½1.770 0 g¡Á£­2.50¡Á10£­3 mol¡Á142 g·mol£­1£­1.25¡Á10£­3 mol¡Á34 g·mol£­1£½0.045 00 g£¬Ôòn(H2O)£½£½2.50¡Á10£­3 mol¡£

×ÛÉÏ¿ÉÖª£¬x¡Ãy¡Ãz£½n(Na2SO4)¡Ãn(H2O2)¡Ãn(H2O)£½2¡Ã1¡Ã2£¬¹ÊÁòËáÄÆ­¹ýÑõ»¯Çâ¼ÓºÏÎïµÄ»¯Ñ§Ê½Îª2Na2SO4·H2O2·2H2O¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÔÏÂÐðÊöÖУ¬´íÎóµÄÊÇ(¡¡¡¡)

A£®ÄÆÔ­×ÓºÍÂÈÔ­×Ó×÷ÓÃÉú³ÉNaClºó£¬Æä½á¹¹µÄÎȶ¨ÐÔÔöÇ¿

B£®ÔÚÂÈ»¯ÄÆÖУ¬³ýÂÈÀë×ÓºÍÄÆÀë×ӵľ²µçÎüÒý×÷ÓÃÍ⣬»¹´æÔÚµç×ÓÓëµç×Ó¡¢Ô­×ÓºËÓëÔ­×ÓºËÖ®¼äµÄÅųâ×÷ÓÃ

C£®ÈκÎÀë×Ó¼üÔÚÐγɵĹý³ÌÖбض¨Óеç×ӵĵÃÓëʧ

D£®½ðÊôÄÆÓëÂÈÆø·´Ó¦Éú³ÉÂÈ»¯Äƺó£¬ÌåϵÄÜÁ¿½µµÍ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁи÷×éÎïÖʵÄÈÛ¡¢·Ðµã¸ßµÍÖ»Óë·¶µÂ»ªÁ¦ÓйصÄÊÇ(¡¡¡¡)

A£®Li¡¢Na¡¢K¡¢Rb

B£®HF¡¢HCl¡¢HBr¡¢HI

C£®LiCl¡¢NaCl¡¢KCl¡¢RbCl

D£®F2¡¢Cl2¡¢Br2¡¢I2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÅжÏÏÂÁÐÈÜÒºÔÚ³£ÎÂϵÄËá¡¢¼îÐÔ(ÔÚÀ¨ºÅÖÐÌî¡°ËáÐÔ¡±¡¢¡°¼îÐÔ¡±»ò¡°ÖÐÐÔ¡±)¡£

(1)ÏàͬŨ¶ÈµÄHClºÍNaOHÈÜÒºµÈÌå»ý»ìºÏ(¡¡¡¡)

(2)ÏàͬŨ¶ÈµÄCH3COOHºÍNaOHÈÜÒºµÈÌå»ý»ìºÏ(¡¡¡¡)

(3)ÏàͬŨ¶ÈNH3·H2OºÍHClÈÜÒºµÈÌå»ý»ìºÏ(¡¡¡¡)

(4)pH£½2µÄHClºÍpH£½12µÄNaOHÈÜÒºµÈÌå»ý»ìºÏ(¡¡¡¡)

(5)pH£½3µÄHClºÍpH£½10µÄNaOHÈÜÒºµÈÌå»ý»ìºÏ(¡¡¡¡)

(6)pH£½3µÄHClºÍpH£½12µÄNaOHÈÜÒºµÈÌå»ý»ìºÏ(¡¡¡¡)

(7)pH£½2µÄCH3COOHºÍpH£½12µÄNaOHÈÜÒºµÈÌå»ý»ìºÏ(¡¡¡¡)

(8)pH£½2µÄHClºÍpH£½12µÄNH3·H2OµÈÌå»ý»ìºÏ(¡¡¡¡)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÑ֪ijζÈÏÂCH3COOHµÄµçÀë³£ÊýK£½1.6¡Á10£­5¡£¸ÃζÈÏ£¬Ïò20 mL 0.01 mol·L£­1 CH3COOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.01 mol·L£­1 KOHÈÜÒº£¬ÆäpH±ä»¯ÇúÏßÈçͼËùʾ(ºöÂÔζȱ仯)¡£Çë»Ø´ðÏÂÁÐÓйØÎÊÌ⣺

(1)aµãÈÜÒºÖÐc(H£«)Ϊ________£¬pHԼΪ________¡£

(2)a¡¢b¡¢c¡¢dËĵãÖÐË®µÄµçÀë³Ì¶È×î´óµÄÊÇ__________£¬µÎ¶¨¹ý³ÌÖÐÒËÑ¡ÓÃ____________×÷ָʾ¼Á£¬µÎ¶¨ÖÕµãÔÚ________(Ìî¡°cµãÒÔÉÏ¡±»ò¡°cµãÒÔÏ¡±)¡£

(3)ÈôÏò20 mLÏ¡°±Ë®ÖÐÖðµÎ¼ÓÈëµÈŨ¶ÈµÄÑÎËᣬÔòÏÂÁб仯Ç÷ÊÆÕýÈ·µÄÊÇ________(Ìî×Öĸ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂͼ±íʾˮÖÐc(H£«)ºÍc(OH£­)µÄ¹Øϵ£¬ÏÂÁÐÅжϴíÎóµÄÊÇ (¡¡¡¡)

A£®Á½ÌõÇúÏß¼äÈÎÒâµã¾ùÓÐc(H£«)¡Ác(OH£­)£½Kw

B£®MÇøÓòÄÚÈÎÒâµã¾ùÓÐc(H£«)£¼c(OH£­)

C£®Í¼ÖÐT1£¼T2

D£®XZÏßÉÏÈÎÒâµã¾ùÓÐpH£½7

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÖÓÐÁ½Æ¿Î¶ȷֱðΪ15 ¡æºÍ45 ¡æ£¬pH¾ùΪ1µÄÁòËáÈÜÒº£¬ÏÂÁÐÓйØ˵·¨²»ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®Á½ÈÜÒºÖеÄc(OH£­)ÏàµÈ

B£®Á½ÈÜÒºÖеÄc(H£«)Ïàͬ

C£®µÈÌå»ýÁ½ÖÖÈÜÒºÖкͼîµÄÄÜÁ¦Ïàͬ

D£®Á½ÈÜÒºÖеÄc(H2SO4)»ù±¾Ïàͬ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÃæÊÇÓйØÎïÖʵÄת»¯¹Øϵͼ(ÓÐЩÎïÖÊÒÑÊ¡ÂÔ)¡£

ÈôAΪµ¥ÖÊ£¬EÔÚ³£ÎÂÏÂΪҺÌ壬CµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª78¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)»­³öAµÄÔ­×ӽṹʾÒâͼ________£¬FµÄµç×ÓʽÊÇ________¡£

(2)ÏÂÃæ¶ÔCÎïÖʽṹ¡¢ÐÔÖʵÄÍƶÏÖУ¬²»ÕýÈ·µÄÊÇ________¡£

A£®¾ÃÖÃÓÚ¿ÕÆøÖлá±ä³É°×É«

B£®¾ßÓÐÇ¿Ñõ»¯ÐÔ

C£®¾§ÌåÖдæÔÚÀë×Ó¼üºÍ¹²¼Û¼ü

D£®ÓöʪÈóµÄ×ÏɫʯÈïÊÔÖ½Ö»ÄÜʹÆä±äÀ¶É«

(3)ÈôCÊǺ¬Ñõ»¯ºÏÎïÇÒÑõΪ18Oʱ£¬ÔòCÓëD·´Ó¦ËùµÃ²úÎïµÄĦ¶ûÖÊÁ¿·Ö±ðΪ__________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Na2CO3¹ÌÌå·ÛÄ©ÖлìÓÐÉÙÁ¿NaHCO3£¬ÓÃʲô·½·¨³ýÔÓ£¿Na2CO3ÈÜÒºÖлìÓÐÉÙÁ¿NaHCO3£¬ÓÃʲô·½·¨³ýÔÓ£¿NaHCO3ÈÜÒºÖлìÓÐÉÙÁ¿Na2CO3£¬ÓÃʲô·½·¨³ýÔÓ£¿

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸