谁能给我高一全部的化学公式
\u9ad8\u4e00\u5fc5\u4fee1\u6240\u6709\u5316\u5b66\u516c\u5f0f\u4e2a\u57fa\u672c\u516c\u5f0f\uff081\uff09 n = N / NA \uff082\uff09 n = m / M \uff083\uff09 Vm = V / n \uff084\uff09 cB = nB / V \u5176\u4ed6\u63a8\u5bfc\u516c\u5f0f\uff085\uff09cB=1000\u03c1w/M \uff086\uff09M=22.4\u03c1\uff08\u6807\u51b5\u4e0b\uff09 \uff087\uff09c\u7a00\u00b7v\u7a00=c\u6d53\u00b7v\u6d53 \uff088\uff09m\u7a00\u00b7w\u7a00=m\u6d53\u00b7w\u6d53 \uff089\uff09\u7b49\u6e29\u7b49\u538b\u4e0b\uff1aV1/V2=n1/n2
\u57fa\u672c\u5c31\u8fd9\u4e9b\u591f\u7528\u4e86
N1：N2=n1：n2(微粒数之比=物质的量之比）
密度p=M/Vm p1：p2=M1：M2（相同条件下，密度之比=相对分子质量之比）
c=1000pw%/M
绛旓細绗竴绔鍖栧鍙嶅簲鍙婂叾鑳介噺鍙樺寲 (1)Fe2O3锛3CO 2Fe锛3CO2(閲戝睘姘у寲鐗╃儹杩樺師)(2)CuO锛婬2 Cu锛婬2O (3)WO3锛3H2 W锛3H2O (4)C锛婬2O H2锛婥O(姘寸叅姘)(5)C锛2CuO 2Cu锛婥O2鈫 (6)4P锛5O2 2P2O5 (7)2KMnO4 K2MnO4锛婱nO2锛婳2鈫(鐑垎瑙)(8)2HgO 2Hg锛婳2鈫 (9)2KCl O3 2KCl锛3...
绛旓細鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 楂樹竴鍖栧鏂圭▼寮 涓銆 纰遍噾灞: 1. 鏂板垏鐨勯挔鏈夐摱鐧借壊鍏夋辰,浣嗗緢蹇彂鏆;鏂圭▼寮:4Na+O2=2Na2O;璇ヤ骇鐗╀笉绋冲畾銆傞挔鍦ㄧ┖姘斾腑鐕冪儳鏃,鍙戝嚭榛勮壊鐨勭伀鐒;鍚屾椂鐢熸垚娣￠粍鑹茬殑鍥轰綋,鏂圭▼寮:2Na+O2鐐圭噧=== Na2O2銆傞攤鐕冪儳鏂圭▼寮:4Li+O2鐐圭噧=== 2Li2O;閽剧噧鐑ф柟绋嬪紡:K+O2鐐圭噧=== KO2...
绛旓細鐭ラ亾澶ф湁鍙涓绛斾富 鍥炵瓟閲:1.8涓 閲囩撼鐜:82% 甯姪鐨勪汉:2707涓 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 鍖栧鏂圭▼寮忓ぇ鍏2007-11-19 21:44闈為噾灞炲崟璐(F2 ,Cl2 , O2 , S, N2 , P , C , Si) 1, 姘у寲鎬: F2 + H2 === 2HF F2 +Xe(杩囬噺)===XeF2 2F2(杩囬噺)+Xe===XeF4 nF2 +2M===...
绛旓細2NaOH = 2NaNO2 + H2O 65銆丼i + 2F 2 = SiF4 66銆丼i + 2NaOH + H2O = NaSiO3 +2H2鈫 67銆佺鍗曡川鐨勫疄楠屽鍒舵硶锛氱矖纭呯殑鍒跺彇锛歋iO2 + 2C 楂樻俯鐢电倝 Si + 2CO 锛堢煶鑻辨矙锛夛紙鐒︾⒊锛 锛堢矖纭咃級绮楃杞彉涓虹函纭咃細Si锛堢矖锛 + 2Cl2 鈻 SiCl4 SiCl4 + 2H2 楂樻俯 Si锛堢函锛+ 4HCl ...
绛旓細璋佽兘鎬荤粨涓涓楂樹竴鍖栧鏂圭▼寮鍏ㄩ儴鐨 3涓洖绛 #鐑# 涓轰粈涔堢幇鍦ㄦ儏鏅枩鍓ц秺鏉ヨ秺灏戜簡?are23 2012-08-08 路 TA鑾峰緱瓒呰繃2628涓禐 鐭ラ亾灏忔湁寤烘爲绛斾富 鍥炵瓟閲:487 閲囩撼鐜:0% 甯姪鐨勪汉:341涓 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 甯屾湜鏈夊府鍔1銆侀潪閲戝睘鍗曡川(F2 ,Cl2 , O2 , S, N2 , P , C , ...
绛旓細楂樹竴瀛﹁繃鐨勫寲瀛鏂圭▼寮!!!璋佹湁??鎬ラ渶!! 鍙鏄楂樹竴鐨鐨勫氨琛!!... 鍙鏄珮涓鐨勭殑灏辫!! 灞曞紑...浠ヤ笂涓変釜鍙嶅簲涓鈥滈浄闆ㄥ彂搴勭鈥濆師鐞 鎵╁睍鍙嶅簲鏈:4NO2+O2+2H2O = 4HNO3 4NO+3O2+2H2O = 4HNO3...灞曞紑鍏ㄩ儴 鐓х墖鐓х殑涓嶆槸寰堟竻鏅,鏈変粈涔堢湅涓嶆噦鐨勫彲浠ラ棶鎴戙傝繖浜涢兘鏄垜浠佸笀鎬荤粨鐨勯珮涓鏈熸湯鐨勬柟绋嬪紡,鍊煎緱...
绛旓細灞曞紑鍏ㄩ儴 楂樹竴鐨勫寲瀛鎯宠瀛﹀ソ,瀛︿範濂藉寲瀛︾殑鏂圭▼寮忔槸涓嶅彲閬垮厤鐨,涓嬮潰鏄垜缁欏ぇ瀹跺甫鏉ョ殑鏈夊叧浜庨珮涓鐨勫寲瀛︽柟绋嬪紡浠嬬粛,甯屾湜鑳藉甯姪鍒板ぇ瀹躲 楂樹竴鍖栧鏂圭▼寮忕殑璇︾粏浠嬬粛(涓) 1銆侀噾灞為挔鎶曞埌纭吀閾滄憾娑蹭腑鐨勫寲瀛︽柟绋嬪紡: 2Na+2H2O=2NaOH+H2鈫 CuSO4+2NaOH=Cu(OH) 2鈫+ Na2SO4 (鍏堝啋姘旀场鍐嶈摑鑹叉矇娣) 2銆侀噾灞為挔涓庣洂閰...
绛旓細涓囪兘鎭掔瓑寮忥細n=N/NA=m/M=V/Vm=cV(aq)N1锛歂2=n1锛歯2(寰矑鏁颁箣姣=鐗╄川鐨勯噺涔嬫瘮锛夊瘑搴=M/Vm p1锛歱2=M1锛歁2锛堢浉鍚屾潯浠朵笅锛屽瘑搴︿箣姣=鐩稿鍒嗗瓙璐ㄩ噺涔嬫瘮锛塩=1000pw%/M
绛旓細锛35锛 (宸ヤ笟鍒跺HNO3鐨勪簲涓弽搴)锛堝師鏂欎负锛氭按鍜岀┖姘旓級锛氣憼2H2O 2H2鈫+O2鈫 鈶 N2+3H2 2NH3 鈶4NH3+5O2 4NO+6H2O 鈶2NO+O2锛2NO2 鈶3NO2+ H2O 锛2HNO3+NO (36) 4HNO3锛4NO2鈫+O2鈫+2H2O (瑙佸厜鎴栧彈鐑垎瑙)(37) (宸ヤ笟涓婂埗澶囩幓鐠冪殑涓や釜涓昏鍙嶅簲)锛歋iO2+Na2CO3 Na2SiO3+CO2 S...
绛旓細n=N/NA n=m/M n=V/22.4(鏍囧喌涓嬬殑姘斾綋)C=n/V C=1000dw/M(d琛ㄧず婧舵恫鐨勫瘑搴︼紝w琛ㄧず婧惰川鐨勮川閲忓垎鏁) d=m/V m锛堟憾娑诧級=m锛堟憾璐)+m锛堟憾鍓傦級 w=m锛堟憾璐級/m锛堟憾娑诧級 n=CV
