ij»ìºÏÈÜÒºÖк¬ÓÐÈÜÖÊNaCl¡¢KIºÍNa2S¸÷0.1 mol£¬Íù¸ÃÈÜÒºÖмÓÈëÒ»¶¨Å¨¶ÈµÄAgNO3ÈÜÒº£¬²úÉú³ÁµíµÄÖÊÁ¿Ëæ¼ÓÈëAgNO3ÈÜÒºµÄÌå»ý±ä»¯µÄ¹ØϵÈçͼËùʾ£®ÔòÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

[¡¡¡¡]

A£®

·´Ó¦¹ý³ÌÖвúÉúÒ»ÖÖºÚÉ«³Áµí£¬¸Ã³ÁµíӦΪAg2S

B£®

aµÄÊýֵΪ62.65

C£®

Ëù¼ÓAgNO3ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪmol/L

D£®

µ±¼ÓÈëAgNO3ÈÜÒºµÄÌå»ýÔÚ0¡«2V mLµÄ¹ý³ÌÖУ¬²úÉúµÄ³ÁµíÊÇAgI

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ij»ìºÏÈÜÒºÖÐÖ»º¬ÓÐÁ½ÖÖÈÜÖÊNaClºÍCuSO4£¬ÇÒn£¨NaCl£©£ºn£¨CuSO4£©=3£º2£®ÈôÒÔʯīµç¼«µç½â¸ÃÈÜÒº£¬ÏÂÁÐÍƶÏÖв»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©ÏÖÓÐmgijÆøÌ壬ËüÓÉË«Ô­×Ó·Ö×Ó¹¹³É£¬ËüµÄĦ¶ûÖÊÁ¿ÎªMg?mol-1£®Èô°¢·ü¼ÓµÂÂÞ³£ÊýÓÃNA±íʾ£¬Ôò£º
¢Ù¸ÃÆøÌåµÄÎïÖʵÄÁ¿Îª
m
M
m
M
mol£®
¢Ú¸ÃÆøÌåËùº¬Ô­×Ó×ÜÊýΪ
2mNA
M
2mNA
M
¸ö£®
¢Û¸ÃÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ
22.4m
M
22.4m
M
L£®
¢Ü¸ÃÆøÌåÈÜÓÚ1LË®ÖУ¨²»¿¼ÂÇ·´Ó¦£©£¬ÆäÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ
m
(m+1000)
¡Á100%
m
(m+1000)
¡Á100%
£®
£¨2£©ÀûÓá°»¯Ñ§¼ÆÁ¿ÔÚʵÑéÖеÄÓ¦Óá±µÄÏà¹Ø֪ʶ½øÐÐÌî¿Õ
¢Ùº¬ÓÐ6.02¡Á1023¸öÑõÔ­×ÓµÄH2SO4µÄÎïÖʵÄÁ¿ÊÇ
0.25
0.25
?mol
¢ÚÓë±ê×¼×´¿öÏÂVLCO2Ëùº¬ÑõÔ­×ÓÊýÄ¿ÏàͬµÄË®µÄÖÊÁ¿ÊÇ
18V
11.2
g
18V
11.2
g
£¨Ó÷Öʽ±íʾ£©
¢ÛÔÚÒ»¶¨µÄζȺÍѹǿÏ£¬1Ìå»ýX2 £¨g£©¸ú3Ìå»ýY2 £¨g£©»¯ºÏÉú³É2Ìå»ý»¯ºÏÎÔò¸Ã»¯ºÏÎïµÄ»¯Ñ§Ê½ÊÇ
XY3
XY3

¢ÜÈýÖÖÕýÑεĻìºÏÈÜÒºÖк¬ÓÐ0.2mol Na+¡¢0.25mol Mg2+¡¢0.4mol Cl-¡¢SO42-£¬Ôò
n£¨SO42-£© Îª
0.15
0.15
 mol
£¨3£©¼ÙÈç12CÏà¶ÔÔ­×ÓÖÊÁ¿Îª24£¬ÒÔ0.024kgËùº¬12CÔ­×ÓÊýΪ°¢·ü¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐÊýÖµ¿Ï¶¨²»±äµÄÊÇ
AC
AC
£®
A£®ÑõÆøµÄÈܽâ¶È
B£®ÆøÌåĦ¶ûÌå»ý
C£®Ò»¶¨ÖÊÁ¿µÄÆøÌåÌå»ý
D£®°¢·ü¼ÓµÂÂÞ³£Êý
E£®O2Ïà¶Ô·Ö×ÓÖÊÁ¿
F£®¸ú2mLH2 Ï໯ºÏµÄO2µÄÖÊÁ¿¡¢ÎïÖʵÄÁ¿£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

CH3COOHÊÇÖÐѧ»¯Ñ§Öг£ÓõÄÒ»ÔªÈõËᣬÇë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Èô·Ö±ð½«pH=2µÄÑÎËáºÍ´×ËáÏ¡ÊÍ100±¶£¬ÔòÏ¡ÊͺóÈÜÒºµÄpH£ºÑÎËá
£¾
£¾
´×ËᣨÌî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±£©£®
£¨2£©½«100mL 0.1mol?L-1µÄCH3COOHÈÜÒºÓë50mL 0.2mol?L-1µÄNaOHÈÜÒº»ìºÏ£¬ËùµÃÈÜÒº³Ê
¼î
¼î
ÐÔ£¬Ô­Òò
CH3COO-+H2O?CH3COOH+OH-
CH3COO-+H2O?CH3COOH+OH-
£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£®
£¨3£©ÒÑ֪ij»ìºÏÈÜÒºÖÐÖ»º¬ÓÐCH3COO-¡¢H+¡¢Na+¡¢OH-ËÄÖÖÀë×Ó£¬ÇÒÀë×ÓŨ¶È´óС¹ØϵΪ£ºc£¨CH3COO-£©£¾c£¨H+£©£¾c£¨Na+£©£¾c£¨OH-£©£¬Ôò¸ÃÈÜÒºÖк¬ÓеÄÈÜÖÊΪ
CH3COOHºÍCH3COONa
CH3COOHºÍCH3COONa
£®
£¨4£©ÒÑÖªKa£¨CH3COOH£©=1.76¡Á10-5£¬Ka£¨HNO2£©=4.6¡Á10-4£¬ÈôÓÃͬŨ¶ÈµÄNaOHÈÜÒº·Ö±ðÖк͵ÈÌå»ýÇÒpHÏàµÈµÄCH3COOHºÍHNO2£¬ÔòÏûºÄNaOHÈÜÒºµÄÌå»ý¹ØϵΪ£ºÇ°Õß
£¾
£¾
ºóÕߣ¨Ìî¡°£¾£¬£¼»ò=¡±£©
£¨5£©ÒÑÖª25¡æʱ£¬0.1mol?L-1´×ËáÈÜÒºµÄpHԼΪ3£¬ÏòÆäÖмÓÈëÉÙÁ¿´×ËáÄƾ§Ì壬·¢ÏÖÈÜÒºµÄpHÔö´ó£®¶ÔÉÏÊöÏÖÏóÓÐÁ½ÖÖ²»Í¬µÄ½âÊÍ£º¼×ͬѧÈÏΪ´×ËáÄƳʼîÐÔ£¬ËùÒÔÈÜÒºµÄpHÔö´ó£»ÒÒͬѧ¸ø³öÁíÍâÒ»ÖÖ²»Í¬ÓÚ¼×ͬѧµÄ½âÊÍ£¬ÇëÄãд³öÒÒͬѧ¿ÉÄܵÄÀíÓÉ
CH3COOH?CH3COO-+H+£¬Òò´×ËáÄƵçÀ룬ʹCH3COO-Ũ¶ÈÔö´ó£¬´×ËáµÄµçÀëƽºâÄæÏòÒƶ¯£¬H+Ũ¶ÈϽµ£¬pHÔö´ó
CH3COOH?CH3COO-+H+£¬Òò´×ËáÄƵçÀ룬ʹCH3COO-Ũ¶ÈÔö´ó£¬´×ËáµÄµçÀëƽºâÄæÏòÒƶ¯£¬H+Ũ¶ÈϽµ£¬pHÔö´ó
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÎÊ´ðÌâ

£¨1£©ÏÖÓÐmgijÆøÌ壬ËüÓÉË«Ô­×Ó·Ö×Ó¹¹³É£¬ËüµÄĦ¶ûÖÊÁ¿ÎªMg?mol-1£®Èô°¢·ü¼ÓµÂÂÞ³£ÊýÓÃNA±íʾ£¬Ôò£º
¢Ù¸ÃÆøÌåµÄÎïÖʵÄÁ¿Îª mol£®
¢Ú¸ÃÆøÌåËùº¬Ô­×Ó×ÜÊýΪ ¸ö£®
¢Û¸ÃÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ L£®
¢Ü¸ÃÆøÌåÈÜÓÚ1LË®ÖУ¨²»¿¼ÂÇ·´Ó¦£©£¬ÆäÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ £®
£¨2£©ÀûÓá°»¯Ñ§¼ÆÁ¿ÔÚʵÑéÖеÄÓ¦Óá±µÄÏà¹Ø֪ʶ½øÐÐÌî¿Õ
¢Ùº¬ÓÐ6.02¡Á1023¸öÑõÔ­×ÓµÄH2SO4µÄÎïÖʵÄÁ¿ÊÇ ?mol
¢ÚÓë±ê×¼×´¿öÏÂVLCO2Ëùº¬ÑõÔ­×ÓÊýÄ¿ÏàͬµÄË®µÄÖÊÁ¿ÊÇ £¨Ó÷Öʽ±íʾ£©
¢ÛÔÚÒ»¶¨µÄζȺÍѹǿÏ£¬1Ìå»ýX2 £¨g£©¸ú3Ìå»ýY2 £¨g£©»¯ºÏÉú³É2Ìå»ý»¯ºÏÎÔò¸Ã»¯ºÏÎïµÄ»¯Ñ§Ê½ÊÇ 
¢ÜÈýÖÖÕýÑεĻìºÏÈÜÒºÖк¬ÓÐ0.2mol Na+¡¢0.25mol Mg2+¡¢0.4mol Cl-¡¢SO42-£¬Ôò
n£¨SO42-£© Îª  mol
£¨3£©¼ÙÈç12CÏà¶ÔÔ­×ÓÖÊÁ¿Îª24£¬ÒÔ0.024kgËùº¬12CÔ­×ÓÊýΪ°¢·ü¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐÊýÖµ¿Ï¶¨²»±äµÄÊÇ £®
A£®ÑõÆøµÄÈܽâ¶È
B£®ÆøÌåĦ¶ûÌå»ý
C£®Ò»¶¨ÖÊÁ¿µÄÆøÌåÌå»ý
D£®°¢·ü¼ÓµÂÂÞ³£Êý
E£®O2Ïà¶Ô·Ö×ÓÖÊÁ¿
F£®¸ú2mLH2 Ï໯ºÏµÄO2µÄÖÊÁ¿¡¢ÎïÖʵÄÁ¿£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸