初三化学题，要详细过程，今晚发的加分
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\u65b9\u7a0b\u5f0f\uff1aCaO+H20=Ca\uff08OH\uff092 Ca\uff08OH\uff092+Na2CO3=CaCO3+2NaOH
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3.\u52a0\u5165\u78b3\u9178\u94a0\uff0c\u6709\u6c89\u6dc0\u4ea7\u751f\uff0c\u5219\u8bf4\u660e\u8fd8\u6709\u591a\u4f59\u7684\u6c22\u6c27\u5316\u9499\u5b58\u5728\uff0c\u6c22\u6c27\u5316\u9499\u4e0e\u78b3\u9178\u94a0\u53cd\u5e94\u751f\u6210\u6c89\u6389\uff0c\u6240\u4ee5\u6ee4\u6db2\u662f\u6c22\u6c27\u5316\u94a0\u548c\u6c22\u6c27\u5316\u9499
\uff081\uff09\u4ea7\u751f\u6c27\u6c14\u7684\u8d28\u91cf 78g-74.8g=3.2g
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2H2O2=MnO2=2H2O+O2
68 32
x 3.2g
68/x=32/3.2g
x=6.8g
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y*8.72%=100g*3% y=34.4g
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(1000g*30%-z)/(1000g-32z/68)*100%=8.72%
z=221.9g
所以S元素的质量m(S)=10t
S~SO2
32 64
故m(SO2):m(S)=64:32=2:1
所以产生so2 20t
S~H2SO4
32 98
所以产生m（H2SO4）=98/32*10=30.625t
m(0.005%的硫酸溶液)=30.625/0.00005=612500t
S → SO2 → H2SO4
32 64 98
1000×1% X Y×0.005%
列等式 得 X=20t Y=512500t
1000t原煤中含硫1000*0.01=10t
s——so2
32 64
10 x
x=20
s——h2so4
32 98
10 y
y=30.625
30.625/0.00005=612500
绛旓細C 璁炬婚噺涓 100 鍒 姘㈢殑璐ㄩ噺涓1 CaO---Ca(OH)2---2H 56 74 2 X Y 1 X=28 Y=37 28/[100-(37-28)]*100%=30.8
绛旓細锛1.锛夊弽搴旂敓鎴鐨娌夋穩璐ㄩ噺涓簒 鏍规嵁璐ㄩ噺瀹堟亽瀹氬緥锛60 + 45=x+100 x=5g锛圕aCO3锛夛紙2.锛夋墍寰楁憾娑蹭腑婧惰川鐨勮川閲忓垎鏁帮細Na2CO3 + Ca锛圢O3锛2= CaCO3 + 2NaNO3 106---100---174 x---5---y x=5銆03g锛圢a2CO3锛塏aNO3鍜孨a2CO3鐨勬贩鍚堢墿10g涓殑NaNO3鐨勮川閲=10-5.03=4.97g y=8.7g锛圢aNO3锛...
绛旓細鐐圭噧 2H2+O2===2H2O 4銆3236 x銆y9g x/4=y/32=9g/36 瑙ｅ緱 x = 1g锛寉=8g 鏁呭弬涓庡弽搴旂殑姘㈡皵鍜屾哀姘旀槸1g鍜8g锛屾贩鍚堟皵浣撶殑缁勬垚鍙兘鏄疉(杩樻湁鍙兘杩10g涓湁2gH2+8gO2).
绛旓細鍒嗘瀽杩囩▼ 锛1锛夊彇灏戦噺婧舵恫浜庤瘯绠′腑锛屽姞鍏ヨ冻閲忔隘鍖栭挕婧舵恫锛屼骇鐢熺櫧鑹叉矇娣锛岃鏄庡彲鑳芥湁CO32-銆丼O42-锛屽師鍥狅細CO32- +Ba2+===BaCO3鈫擄紱SO42- +Ba2+===BaSO4鈫 锛2锛夊悜婊ゆ福涓姞鍏ョ█纭濋吀锛屾护娓ｅ叏閮ㄦ憾瑙ｅ苟浜х敓姘旀场锛岃鏄庤繖涓矇娣鏄疊aCO3锛屾皵娉℃槸CO2锛岀‘瀹氫竴瀹氬惈鏈塁O32-锛屼竴瀹氭病鏈塖O42- 鍘熷洜锛欱aCO3...
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绛旓細2H2O2= MnO2= 2H2O+ O2鈫 68 32 x 3.2g 68/x=32/3.2g x=6.8g 6.8g/100g*100%=6.8 2KClO3 =MnO2鈻= 2KCl + 3O2 鈫 245 149 96 m n 19.2g 245/m=149/n=96/19.2g m=49g n=29.8g
绛旓細1.璁惧彇娴撶～閰竂ml,鍒欐湁濡備笅绛夊紡 X锛坢l锛*1.84g/ml*0.98=100*0.098,X=10/1.84 娴撶～閰镐綋绉*瀵嗗害=鎬昏川閲*鍚噺=纭吀鐨閲忥紝鎬昏川閲*鍚噺=纭吀鐨勯噺锛屼袱鑰呯浉绛 2.璁惧彇30%鐨勭洂閰竂ml,鍒欐湁濡備笅绛夊紡 (X(ml)*1.15g/ml*0.3)/(115+X*1.15)=0.15 鐩愰吀浣撶Н*瀵嗗害*鐩愰吀鐧惧垎鏁=鐩愰吀鐨勮川閲忥紝...
绛旓細鍒濅笁鍖栧璇曢 鍙兘鐢ㄥ埌鐨勭浉瀵瑰師瀛愯川閲:H鈥1 C鈥12 N鈥14 O鈥16 Cu鈥64 绗厾鍗(閫夋嫨棰 鍏36鍒) 涓銆侀夋嫨棰(姣忓皬棰2鍒,鍏36鍒,姣忓皬棰樺垪鍑虹殑鍥涗釜閫夐」涓,鍙湁涓椤圭鍚堥鐩姹) 1銆佷笅鍒楀彉鍖栦竴瀹氬彂鐢熶簡鍖栧鍙樺寲鐨勬槸( ) A銆佸啺闆瀺鍖 B銆侀鐗╄厫鐑 C銆侀噾妗傞棣 D銆佸皹鍦熼鎵 2銆佷笅鍒楃墿璐ㄤ腑灞炰簬姘у寲鐗╃殑鏄...
绛旓細瑙:(1)鐢遍鎰忕煡,鍦ㄥ姞鐑殑鏉′欢涓婯MnO4瀹屽叏鍒嗚В鐢熸垚K2MnO4銆丮nO2鍜孫2锛孠ClO3瀹屽叏鍒嗚В鐢熸垚 KCl鍜孫2,鑰孠2MnO4銆並Cl閮借兘婧朵簬姘达紝MnO2涓嶈兘婧朵簬姘淬傚洜姝わ紝杩囨护鍚庢墍寰楀浐浣撴槸MnO2锛屽叾璐ㄩ噺涓1.30g銆傝娣峰悎鐗╀腑KMnO4鐨勮川閲忎负X锛孠MnO4瀹屽叏鍒嗚В鐢熸垚O2鐨勮川閲忎负Y 锛屽垯:2KMnO4=螖=K2MnO4+MnO2+O2鈫 2*158~~~...
绛旓細纰抽吀閽鐨璐ㄩ噺鍒嗘暟=8.48g/10g=84.8 (2)鏍峰搧涓挔鍏冪礌鐨勮川閲=8.48g*46/106+锛10g-8.48g)*23/58.5 =3.68g+0.6g =4.28g 鏍峰搧涓挔鍏冪礌鐨勮川閲忓垎鏁=4.28g/10g=42.8 棰樹簩锛150t娴撶缉娴锋按涓隘鍖栭晛鐨勮川閲=150t*9.5%=14.25t 璁鹃渶瑕佹阿姘у寲閽欑殑璐ㄩ噺涓簓 MgCl2+Ca(OH)2=CaCl2+Mg(OH)2 ...
