高一化学习题
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B\uff1a\u9519\u8bef\uff0c\u5411\u9971\u548c\u98df\u76d0\u6c34\u4e2d\u6ef4\u52a0\u7a00\u786b\u9178\u4e0d\u4f1a\u4ea7\u751fHCl\u7684\uff0c\u53ef\u4ee5\u50cfNaCl\u56fa\u4f53\u4e0a\u6ef4\u52a0\u6d53\u786b\u9178\u6765\u5236\u53d6HCl
C\uff1a\u6b63\u786e
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2C 注意纯硫酸还可以吸收三氧化硫
3D 看似数学的排列组合题,其实不然。仅碳和氢两种就能组成成千上万种。
4B 氧气会把氢硫酸氧化成硫和水;氯气使其变成硫和强酸——盐酸;硫酸铜可以把硫离子沉淀掉,形成不溶于强酸的硫化铜,同时生成强酸——硫酸;二氧化硫与氧气作用相同。
(先写这四道)
1.CH2Sio3
2.C2.22Kg
3.D多于12种
4.B(2)(3)
5.Am+n=2
6.A当X。Y都是费劲舒时,X一定Y比活泼
C当X是金属时,Y可能是金属,也可能是非金属
7.共有0.15 mol气体.
其中O2为0.05 mol,1.6 g.
则剩余氢气为0.1 mol,0.25 g.
即H2为0.075 mol.
所以H2和O2的物质的量之比为3:1.
1.C,只有C是分子晶体,
2.B,3.D,4.B,5.A,6.AC,7.3:2
1.C,只有C是分子晶体,
2.B
3.D
4.B
5.A
6.AC
7.3:2
题目太多,分数太少.你自己答.
绛旓細楂樹竴鍖栧鏂圭▼寮忓皬缁:(1) 4Na+O2=2Na2O (甯告俯) 鈽(2) Na2O+O2=鍔犵儹= 2Na2O2 (3) 2Na+O2= Na2O2(鐐圭噧)(娉:鍙嶅簲鏉′欢涓嶅悓;鐢熸垚鐗╀篃涓嶅悓銆)(4)...姝ょ被闂璇疯缁嗗湴鍋氬競闈笂鍙傝冧功涓瀛愬弽搴旈儴鍒嗙殑涔犻鍙﹀涓涓噸瑕佺殑鏂圭▼寮忓氨鏄疌L2鍜孎EBR2鐨勫弽搴斻傛垜浠煡閬,BR2鍙互鎶奆E姘у寲鎴愪笁浠,鎵浠ヨ繖涓弽搴斾腑CL2瑕...
绛旓細x=81.07t 60*6g/95%z=478g/31*10t x=245.76t 鍗虫瘡鏈堟墍闇Na2CO3銆丆aCO3鍜孲iO2鐨勮川閲忓垎鍒负72.36t銆81.07t銆245.76t 娌¤閿欑殑璇 杩欓亾棰樻槸楂樹竴蹇呬慨涓绗洓绔犵涓鑺傛棤鏈洪噾灞炴潗鏂欑殑涓昏鈥斺旂鐨勮鍚缁冧範棰11棰榽 鍢垮樋 鎴戝仛浜嗗ソ鍑犻亶浜 鏈変粈涔堥棶棰樺彲浠ョ暀涓嬫潵~
绛旓細1.A锛孊eF2锛2.C锛2NO+O2===2NO2 6NO +3O2 ===6NO2 3NO2+H2O===2HNO3+NO 6NO2+2H2O ===4HNO3+2NO 4NO+3O2+2H2O===4HNO3 鎵浠ヨ鎶婂墿涓嬬殑5mlNO杞寲涓篐NO3锛岄渶瑕5*3/4=3.75ml鐨勬哀姘旓紝鍐嶅姞涓5ml锛屽緱鍒8.75ml銆3.C,鐗╄川鐨勯噺娴撳害=p*w/M锛岀幇鍦╝<b,M=M,鍗硃1*...
绛旓細鎴戠粰浣犲啓鍑犱釜姣旇緝绠鍗曠殑棰橈細棰樼洰锛24.5gH2SO4鐨勭墿璐ㄧ殑閲忔槸澶氬皯锛1.50molNa2CO3鐨勮川閲忔槸澶氬皯锛熻В锛欻2SO4鐨勬懇灏旇川閲廙=98g/mol銆侶2SO4鐨勭墿璐ㄧ殑閲 n=m/M=24.5/98g/mol=0.25mol.Na2CO3鐨勭浉瀵瑰垎瀛愯川閲廙=106g/mol銆侼a2CO3鐨勮川閲弇 =nM=106g/mol*1.5mol=159g .绛斻傘傘傞鐩細鎽╁皵璐ㄩ噺鏈...
绛旓細閲戝睘鎶曞叆纭吀涓彲鑳戒細浜х敓浠ヤ笅涓夌鎯呭喌锛氣懘纭吀杩囬噺锛屼袱绉嶉噾灞炲潎瀹屽叏鍙嶅簲锛涒懙閲戝睘杩囬噺锛岀~閰哥殑閲忓潎涓嶈冻锛屼袱绉嶉噾灞為兘鏈畬鍏ㄥ弽搴旓紱鈶剁~閰哥殑鐢ㄩ噺瀵逛竴绉嶉噾灞炴潵璇存槸瓒抽噺鐨勶紝瀵逛簬鍙﹀涓绉嶉噾灞炴槸涓嶈冻閲忕殑銆俷(Al)=10.8/27=0.4mol n(Mg)=10.8/24=0.45mol 涓嶢l瀹屽叏鍙嶅簲闇纭吀:(0.4*3/2)/3=0.2L...
绛旓細鏌愮┖鐡堕噸51G锛岃婊℃煇姘斾綋A鍚庣殑璐ㄩ噺涓51.27G銆傚悓娓┿佸悓鍘嬩笅锛岃婊℃阿姘旀椂鐨勮川閲忎负51.08G锛堢┖鐡跺師鐪熺┖锛夛紝鍒欐姘斾綋鐨勬懇灏旇川閲忎负澶氬皯锛2.鎶11浣撶Н鐨勬阿姘旓紝5浣撶Н鐨勬哀姘斿拰1浣撶Н姘皵鍦ㄥ瘑闂鍣ㄤ腑鐢ㄧ數鐏姳鐐圭噧銆傛伆瀹屽叏鍙嶅簲锛屾墍寰楁憾娑叉憾璐ㄧ殑璐ㄩ噺鍒嗘暟涓哄灏戯紵3.鍦ㄦ爣鍑嗙姸鍐典笅锛屾阿姘斿拰姘ф皵娣峰悎姘斾綋AL銆傚紩鐖...
绛旓細楂樹竴鍖栧娴嬭瘯棰 涓銆 閫夋嫨棰 1锛庡鍥炬墍绀虹殑瀹為獙鎿嶄綔涓紝姝g‘鐨勬槸()2锛庝笅鍒楁湁鍏冲疄楠岀敤鍝佺殑浣跨敤鎴栨搷浣滅殑鍙欒堪涓紝姝g‘鐨勬槸()A锛庣敤绾㈣壊鐭宠晩璇曠焊妫楠屾憾娑茬殑閰告 B锛庣敤瀹归噺鐡堕厤鍒跺噯纭祿搴︾殑婧舵恫锛屼絾涓嶈兘浣滀负鍙嶅簲瀹瑰櫒浣跨敤 C锛庨厤鍒剁█纭吀鏃讹紝鍏堝悜鐑ф澂涓敞鍏ユ祿纭吀锛屽啀鍔犳按绋閲 D锛庣О閲廚aOH鍥轰綋鏃讹紝灏哊aOH...
绛旓細BCDDAC 7.2Cu+O2=鍔犵儹2CuO CuO+H2SO4=H2O+CuSO4 CuSO4+2NaOH=Cu(OH)2+Na2SO4 Cu(OH)2=鍔犵儹CuO+H2O CuSO4+Fe=FeSO4+Cu 8.CO2+2OH-=CO3 2-+H2O 2Al+2OH-+2H2O=2AlO2-+3H2姘斾綋 9.CaCO3=楂樻俯CaO+CO2姘斾綋 CaO+H2O=Ca(OH)2 Na2CO3+Ca(OH)2=CaCO3娌夋穩+2NaOH 10.KAl(SO4)2 ...
绛旓細璁惧師NaNO3婧舵恫鐨勮川閲忎负x锛屽垯22%x=(100+x)14%锛岃В寰梮=175g 鎵浠ワ紝m(NaNO3) = 175g脳22% = 38.5g c(NaNO3) = n(NaNO3)/V(aq)= 3.02mol/L 鍙傝冭祫鏂欙細http://gzhx.cooco.net.cn/testdetail/26576/
绛旓細绗崄浜岄锛氾紙1锛変簩 锛2锛塂 (3) NaHCO3 (4锛 Ca2+ + OH- + HCO3- =H2O +CaCO3鈫 锛5锛塏a2CO3 鍔犲叆婢勬竻鐭崇伆姘达紝鍙樻祽娴婏紝鍦ㄥ姞鍏ョ█鐩愰吀锛屼骇鐢熸棤鑹叉棤鍛虫皵浣 绗崄涓夐 锛1锛塏aCl + CO2 + H2O + NH3= NaHCO3 鈫 + NH4Cl 锛2锛塁O2 NH3 (3) NH4CI CaCl...