½«COºÍCO2µÄ»ìºÏÆøÌå¹²3.2gͨ¹ý×ãÁ¿×ÆÈȵÄÑõ»¯Í­·ÛÄ©£¬³ä·Ö·´Ó¦ºó£¬½«ÆøÌåͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖУ¨ÆøÌåÈ«²¿±»ÎüÊÕ£©£¬¹ýÂË£¬²âµÃÈÜÒºÖÊÁ¿¼õÉÙ5.6g £¬ÔòÔ­»ìºÏÆøÌåÖÐ̼ԪËصÄÖÊÁ¿·ÖÊýΪ£¨¡¡ ¡¡£©¡£

A. 21.0% B. 27.3% C. 33.3% D. 37.5%

D ¡¾½âÎö¡¿Éè·´Ó¦ºóÉú³É¶þÑõ»¯Ì¼µÄ×ÜÖÊÁ¿Îªx CO2 + Ca(OH)2 == CaCO3 ¡ý+ H2O ÈÜÒº¼õÉÙµÄÖÊÁ¿ 44 100-44 X 5.6g = x=4.4g 4.4g¶þÑõ»¯Ì¼ÖÐ̼ԪËصÄÖÊÁ¿Îª=1.2g ÔòÔ­»ìºÏÆøÌåÖÐ̼ԪËصÄÖÊÁ¿·ÖÊýΪ=37.5%£¬¹ÊÑ¡D¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º½­Î÷Ê¡¸ÓÖÝÊÐÐijÇÇøÁùУÁªÃË2018½ì¾ÅÄ꼶ÏÂѧÆÚÁª¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÎïÖÊÖУ¬ÊôÓÚ´¿¾»ÎïµÄÊÇ£¨ £©

A. ÉúÀíÑÎË® B. ¸ßÃÌËá¼Ø C. ¸»Ñõ¿ÕÆø D. ÒÒ´¼ÆûÓÍ

B ¡¾½âÎö¡¿A¡¢ÉúÀíÑÎË®ÖÐÓÐʳÑΡ¢Ë®£¬ÊôÓÚ»ìºÏÎ´íÎó£»B¡¢¸ßÃÌËá¼ØÊÇÒ»ÖÖÎïÖÊ£¬ÊôÓÚ´¿¾»ÎÕýÈ·£»C¡¢¸»Ñõ¿ÕÆø¿ÕÆøÖÐÓÐÑõÆø¡¢µªÆøµÈÎïÖÊ£¬ÊôÓÚ»ìºÏÎ´íÎó£»D¡¢ÒÒ´¼ÆûÓÍÖÐÓÐÒÒ´¼¡¢ÆûÓ͵ÈÎïÖÊ£¬ÊôÓÚ»ìºÏÎ´íÎó¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º½­ËÕÊ¡2018½ì¾ÅÄ꼶3ÔÂÄ£Ä⿼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÊµÑéÌâ

¸ù¾ÝͼʾʵÑé×°Ö㬻شðÏÂÁÐÎÊÌâ¡£

(1)ͼÖÐaÒÇÆ÷µÄÃû³Æ£ºa________________£¬b_____¡£

(2)ÓøßÃÌËá¼Ø¹ÌÌåÖÆÑõÆø£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ___________________________¡£½«×°ÖÃAºÍDÁ¬½Ó½øÐдËʵÑ飬ʵÑé½áÊø£¬Í£Ö¹¼ÓÈÈÇ°ÒªÏÈ______£¬Ä¿µÄÊÇ______¡£

(3)д³öʵÑéÊÒÖÆÈ¡¶þÑõ»¯Ì¼µÄ»¯Ñ§·½³Ìʽ________________£¬Æä·¢Éú×°ÖÿÉÓÃC(¶à¿×¸ô°åÓÃÀ´·Å¿é×´¹ÌÌå)´úÌæBµÄÓŵãÊÇ________________£¬ÈçÓÃE×°ÖÃÊÕ¼¯CO2£¬ÔòÆøÌåÓ¦´Ó____________¶ËͨÈë(Ìî¡°c¡±»ò¡°d¡±)¡£

¾Æ¾«µÆ ³¤¾±Â©¶· 2KMnO4K2MnO4+MnO2+O2¡ü ½«µ¼¹ÜÒƳöË®Ãæ ·Àֹˮµ¹ÎüÔì³ÉÊÔ¹ÜÕ¨ÁÑ CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü ¿ØÖÆ·´Ó¦µÄ·¢ÉúÓëÍ£Ö¹£¬½ÚÔ¼Ò©Æ· c ¡¾½âÎö¡¿±¾ÌâÖ÷Òª¿¼²éÁËÒÇÆ÷µÄÃû³Æ¡¢ÆøÌåµÄÖÆȡװÖúÍÊÕ¼¯×°ÖõÄÑ¡Ôñ£¬Í¬Ê±Ò²¿¼²éÁË»¯Ñ§·½³ÌʽµÄÊéдÔӵȣ¬×ÛºÏÐԱȽÏÇ¿£®ÆøÌåµÄÖÆȡװÖõÄÑ¡ÔñÓë·´Ó¦ÎïµÄ״̬ºÍ·´Ó¦µÄÌõ¼þÓйأ»ÆøÌåµÄÊÕ¼¯×°ÖõÄÑ¡ÔñÓëÆøÌåµÄÃܶȺÍÈܽâ...

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º½­ËÕÊ¡2018½ì¾ÅÄ꼶3ÔÂÄ£Ä⿼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÈËÌåÄÚÔªËØʧºâÊÇÖØÒªµÄÖ²¡ÒòËØ¡£ÏÂÁм²²¡¿ÉÄÜÓëȱ¸ÆÓйصÄÊÇ

A¡¢ØþÙͲ¡ B¡¢´ó²±×Ó²¡ C¡¢Æ¶ÑªÖ¢ D¡¢ÖÇÁ¦µÍÏÂ

A ¡¾½âÎö¡¿ ÊÔÌâ·ÖÎö£ºÈËÌåÄÚÔªËØʧºâÊÇÖØÒªµÄÖ²¡ÒòËØ¡£È±¸Æ»áÒýÆðØþÙͲ¡¡£¹ÊÑ¡A.

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£ºÉ½¶«Ê¡2018½ì¾ÅÄ꼶һģ»¯Ñ§ÊÔ¾í ÌâÐÍ£º×ÛºÏÌâ

½ðÊô²ÄÁϾßÓÐÓÅÁ¼µÄÐÔÄÜ£¬±»¹ã·ºÓ¦ÓÃÓÚÉú²ú¡¢Éú»îÖС£

¢ÅÏÂÁнðÊôÖÆÆ·ÖУ¬Ö÷ÒªÀûÓÃÁ˽ðÊôµ¼µçÐÔµÄÊÇ_________¡££¨ÌîÐòºÅ£©

A£®»Æ½ðÊÎÆ·   B£®Ìú¹ø   C£®Í­µ¼Ïß D£®²»Ðâ¸Öµ¶¾ß

¢Æ¸ÖÌúÐâÊ´»áÔì³ÉÑÏÖصÄ×ÊÔ´ÀË·Ñ£¬·ÀÖ¹»ò¼õ»º¸ÖÌúÐâÊ´µÄ³£Ó÷½·¨ÓÐ__________¡££¨Ð´³öÒ»Ìõ¼´¿É£©

¢Ç¡°ÔøÇàµÃÌúÔò»¯ÎªÍ­¡±£¬ÕâÊÇÊÀ½çʪ·¨Ò±½ðµÄÏÈÇý¡£ÊÔд³öÓÃÌúºÍÁòËáÍ­ÈÜҺΪԭÁϽøÐÐʪ·¨Á¶Í­µÄ»¯Ñ§·½³Ìʽ____________£¬ËüÊôÓÚ__________ ·´Ó¦¡££¨Ìî¡°»¯ºÏ¡±¡¢¡°·Ö½â¡±¡¢¡°¸´·Ö½â¡±¡¢¡°Öû»¡±Ö®Ò»£©

¢ÈΪÁ˲ⶨij»ÆÍ­£¨Í­Ð¿ºÏ½ð£©ÑùÆ·µÄ×é³É£¬Ä³Ñо¿ÐÔѧϰС×é³ÆÈ¡Á˸ÃÑùÆ·20g£¬ÏòÆäÖÐÖðµÎ¼ÓÈë9.8%µÄÏ¡ÁòËáÖÁ¸ÕºÃ²»ÔÙ²úÉúÆøÌåΪֹ¡£·´Ó¦¹ý³ÌÖУ¬Éú³ÉÆøÌåÓëËùÓÃÁòËáÈÜÒºµÄÖÊÁ¿¹ØϵÈçÏÂͼËùʾ¡£ÊÔ¼ÆË㣺¸Ã»ÆÍ­ÑùÆ·ÖÐÍ­µÄÖÊÁ¿Îª______________£¿

C ÖƳɲ»Ðâ¸Ö£¨ Fe + CuSO4 FeSO4 + Cu Öû» 13.5g ¡¾½âÎö¡¿¢Å ½ðÊô×öµ¼ÏßÊÇÀûÓÃÁ˽ðÊôµÄµ¼µçÐÔ£¬¹ÊÑ¡C ¢Æ ·ÀÖ¹½ðÊôÐâÊ´µÄ·½·¨Óн«¸ÖÌúÖƳɲ»Ðâ¸Ö»ò ¸ÖÌú±íÃæ½à¾»ºó£¬¸²¸Ç±£»¤²ã£¬ÀýÈçÍ¿ÓÍ¡¢Æᣬ¶Æп£¬¿¾À¶¹¤ÒÕ µÈ£© ¢Ç ÌúÓëÁòËáÍ­·´Ó¦Éú³ÉÍ­ºÍÁòËáÑÇÌú£¬·´Ó¦·½³ÌʽΪFe + CuSO4 FeSO4 + Cu £¬¸Ã·´Ó¦Êǵ¥ÖÊÓ뻯ºÏÎï·´Ó¦Éú³Éеĵ¥ÖʺÍеĻ¯ºÏÎï...

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£ºÉ½¶«Ê¡2018½ì¾ÅÄ꼶һģ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÓйØÇâÑõ»¯ÄÆÐÔÖÊ̽¾¿ÊµÑéµÄÐðÊöÖУ¬´íÎóµÄÊÇ£¨ £©¡£

A. ¼×Ïò½Ó½ü·ÐÌÚµÄË®ÖмÓÈëÒ»¶¨Á¿NaOH¹ÌÌ壬ҺÌåѸËÙ·ÐÌÚ

B. ÒÒ½«ÉÙÁ¿Í··¢¼ÓÈëµ½ÈȵÄŨNaOHÈÜÒºÖУ¬Í··¢Öð½¥ÈܽâÏûʧ

C. ±ûÏò¾ÃÖÿÕÆøÀïµÄÇâÑõ»¯ÄÆÈÜÒºÖеμÓÎÞÉ«·Ó̪ÊÔÒº£¬¼ìÑéÆäÊÇ·ñ±äÖÊ

D. ¶¡ÏòÊ¢ÂúCO2ÆøÌåµÄ¼¯ÆøÆ¿ÖмÓÈëÊÊÁ¿Å¨ÉÕ¼îÈÜÒº£¬¼¦µ°±»¡°ÍÌ¡±ÈëÆ¿ÖÐ

C ¡¾½âÎö¡¿A¡¢ÇâÑõ»¯ÄÆÈÜÓÚË®·Å³öÈÈÁ¿£¬¹ÊÏò½Ó½ü·ÐÌÚµÄË®ÖмÓÈëÒ»¶¨Á¿NaOH¹ÌÌ壬ҺÌåѸËÙ·ÐÌÚ£¬ÕýÈ·£» B¡¢Í··¢Êǵ°°×ÖÊ£¬ÓëÈȵÄÇâÑõ»¯ÄÆÈÜÓÚ·´Ó¦£¬¹ÊÍ··¢Öð½¥ÈܽâÏûʧ£¬ÕýÈ·£» C¡¢ÇâÑõ»¯ÄƱäÖʺóÉú³ÉµÄ̼ËáÄÆÈÜÒºÈÔÈ»ÏÔ¼îÐÔ£¬¹Ê²»ÄÜÓÃÎÞÉ«·Ó̪ÊÔÒº£¬¼ìÑéÆäÊÇ·ñ±äÖÊ£¬´íÎó£» D¡¢ÏòÊ¢ÂúCO2ÆøÌåµÄ¼¯ÆøÆ¿ÖмÓÈëÊÊÁ¿Å¨ÉÕ¼îÈÜÒº£¬ÇâÑõ»¯ÄÆÓë¶þÑõ»¯Ì¼·´Ó¦£¬Æ¿ÄÚѹǿ½µµÍ£¬ÔÚ´óÆøѹµÄ×÷ÓÃÏ£¬¼¦µ°...

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£ºÉ½¶«Ê¡2018½ì¾ÅÄ꼶һģ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂͼΪ³õÖл¯Ñ§ÊµÑ鿼²éÖг£¼ûµÄ²¿·Ö²Ù×÷£¬ÆäÖÐÕýÈ·µÄÊÇ£¨¡¡ ¡¡£©¡£

A. A B. B C. C D. D

D ¡¾½âÎö¡¿A¡¢²â¶¨ÈÜÒºµÄpHʱ²»Äܽ«pHÊÔÖ½Ö±½Ó²åÈë´ý²âÒºÌåÖУ¬´íÎó£» B¡¢ÏòÊÔ¹ÜÖмÓÈë¹ÌÌåÌú¶¤Ê±£¬Òª½«ÊÔ¹Üƽ·Å£¬´íÎó£» C¡¢¸Ã¹ýÂË×°ÖÃÖÐȱÉÙ²£Á§°ô£¬´íÎó£» D¡¢¼ì²é×°ÖõÄÆøÃÜÐԵķ½·¨ÊÇÕýÈ·µÄ¡£¹ÊÑ¡D¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£ºÖØÇìÊУ¨½­½ò¶þÖеȣ©°ËУ2018½ì¾ÅÄ꼶ÏÂѧÆÚµÚÒ»½×¶Î¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓÃÊʵ±µÄ·½·¨°ÑÏõËá¼ØµÄ²»±¥ºÍÈÜҺת±äΪ±¥ºÍÈÜÒº,ÆäÈÜÒºÖÊÁ¿( )

A. ±ä´ó B. ±äС C. ²»±ä D. ±ä´ó¡¢±äС¡¢²»±ä¶¼ÓпÉÄÜ

D ¡¾½âÎö¡¿½«ÏõËá¼ØµÄ²»±¥ºÍÈÜÒº±äΪ±¥ºÍÈÜÒºµÄ·½·¨ÓмÓÈëÏõËá¼Ø£¬¼õÉÙË®£¬½µµÍζȣ¬µ±¼ÓÈëÏõËá¼Øʱ£¬ÈÜÒºµÄÖÊÁ¿Ôö¼Ó£¬µ±¼õÉÙˮʱ£¬ÈÜÒºµÄÖÊÁ¿¼õС£¬µ±½µµÍζÈʱ£¬ÈÜÒºµÄÖÊÁ¿²»±ä£¬¹ÊÑ¡D¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£ººÓ±±Ê¡2018½ì¾ÅÄ꼶ÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚÉú²úÉú»îÖг£Óõ½ÏÂÁÐÎïÖÊ£¬ÊôÓÚ´¿¾»ÎïµÄÊÇ

A. ʯ»ÒË® B. ²»Ðâ¸Ö C. ¸É±ù D. ÑÀ¸à

C ¡¾½âÎö¡¿A¡¢Ê¯»ÒË®Öк¬ÓÐË®¡¢ÇâÑõ»¯¸ÆµÈÎïÖÊ£¬ÊôÓÚ»ìºÏÎ¹ÊÑ¡Ïî´íÎó£» B¡¢²»Ðâ¸ÖÊÇÌúµÄºÏ½ð£¬²»Ðâ¸ÖÖгýº¬ÌúÍ⣬»¹ÓÐC¡¢Cr¡¢Ni µÈ£¬ÊôÓÚ»ìºÏÎ¹ÊÑ¡Ïî´íÎó£» C¡¢¸É±ùÖ»º¬ÓÐÒ»ÖÖÎïÖÊ£¬ÊôÓÚ´¿¾»Î¹ÊÑ¡ÏîÕýÈ·£» D¡¢ÑÀ¸àÖк¬ÓÐĦ²Á¼Á£¨Ì¼Ëá¸ÆµÈ£©¡¢±£Êª¼Á£¨¸ÊÓ͵ȣ©¡¢Ìðζ¼ÁµÈ¶àÖÖÎïÖÊ£¬ÊôÓÚ»ìºÏÎ¹ÊÑ¡Ïî´íÎó£® ¹ÊÑ¡C£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸