2023年高中化學(xué)會考試卷含答案

20232023年高中化學(xué)會考試卷含答案1．考生要認(rèn)真填寫考場號和座位序號。1．考生要認(rèn)真填寫考場號和座位序號。2825〔共50分；考其次局部為非選擇題，分為必答題和選答題兩道大題〔共50。生須3．試題全部答案必需填涂或書寫在答題卡上，在試卷上作答無效。第一局部知2B4．考試完畢后，考生應(yīng)將試卷和答題卡放在桌面上，待監(jiān)考員收回。第一局部選擇題〔50〕〔250〕AA[CH3OH]C[CO(NH2)2]B[CaCl2]D[NaNO2]以下各組元素中，屬于同一周期的是A．H、Na B．Al、P C．O、Cl D．Mg、F3．電解質(zhì)是一類在水溶液里或熔融狀態(tài)下能夠?qū)щ姷幕衔?。以下物質(zhì)屬于電解質(zhì)的是2 A．Fe B．NaCl C．SiO D．KNO2 合金在生產(chǎn)及科研中具有廣泛的應(yīng)用。以下物質(zhì)屬于合金的是A．水銀 B．石墨 C．硅晶體 D．不銹鋼5．在有機(jī)化工中，氯氣是合成塑料、橡膠、農(nóng)藥和染料等的重要原料。以下關(guān)于氯氣性質(zhì)的描述的是A．無氣味 B．能溶于水 C．能與水反響 D．能與銅反響6．以下同周期元素中，原子半徑最小的是A．Mg B．Al C．Si D．S7．以下為人體供給能量的養(yǎng)分物質(zhì)中，屬于高分子化合物的是A．蔗糖 B．油脂 C．蛋白質(zhì) D．葡萄糖8．當(dāng)光束通過以下分散系時，能觀看到丁達(dá)爾效應(yīng)的是A．白酒 B．稀豆?jié){ C．食鹽水 D．白糖水9．以下物質(zhì)中，既含共價鍵又含離子鍵的是2A．KCl B．CO2

C．MgCl

D．NaOH2化學(xué)試卷第1頁〔8頁〕2化學(xué)反響的是打磨磁石制指南針B．燒結(jié)粘土制陶瓷 C．濕法煉銅 D．鐵的冶煉2 2 羰基硫〔COS〕可作為一種糧食熏蒸劑，能防止某些昆蟲的危害。在恒容密閉容器中，CO和HS混合加熱生成羰基硫的反響為CO(g)＋HS(g) COS(g)＋H(g)2 2 2正、逆反響速率都等于零 B．CO、H2

S、COS、H

的濃度相等2 2 2 2C．CO、HS、COS、H在容器中共存 D．CO、HS、COS、H2 2 2 2圖所示原電池的說法正確的選項是將電能轉(zhuǎn)化為化學(xué)能的裝置B．電子由銅片經(jīng)導(dǎo)線流向鋅片C．鋅片為負(fù)極，發(fā)生氧化反響D．銅片上發(fā)生的反響為Cu2+＋2e－ Cu鈷〔〕是一種穿透力很強(qiáng)的核輻射元素，可用來治療癌癥。以下關(guān)于的說的是A．鈷的一種核素 B．質(zhì)子數(shù)為27C．核外電子數(shù)為27 D．中子數(shù)為6014．以下反響屬于取代反響的是A．甲烷與氯氣反響生成一氯甲烷 B．苯與氫氣反響生成環(huán)己烷C．乙烯與溴反響生成1,2-二溴乙烷 D．乙醇與氧氣反響生成乙醛某粒子的構(gòu)造示意圖為 ，關(guān)于該粒子的說法正確的選項是 2核電荷數(shù)為18 B．核外有3個電子層C．屬于非金屬元素的原子 D．在化學(xué)反響中易得2個電子以下物質(zhì)中，能使酸性高錳酸鉀溶液褪色的是甲烷 B．乙烯 C．苯 D．乙酸17．以下物質(zhì)的電子式書寫正確的選項是 H.. .. ..l:l . :.:::

..C．a(chǎn)[:]－

..D．H:N:HNaCONaHCO

的說法中，不正確的選項是2 3

．．．A．均溶于水 B．均為白色固體C．受熱均易分解 D．其水溶液均能使酚酞溶液變紅19．以下有關(guān)物質(zhì)性質(zhì)的比較，的是A．金屬性：Al＞Mg B．穩(wěn)定性：HF＞HClC．酸性：HClO ＞HSO

D．堿性：NaOH＞Mg(OH)4 2 4

2化學(xué)試卷第2頁〔8頁〕對于鐵與鹽酸生成氫氣的反響，以下條件下反響速率最快的是選項 鐵的狀態(tài)

鹽酸的濃度

反響的溫度A． 片狀1mol/L20℃B． 片狀3mol/L40℃C． 粉末1mol/L20℃D． 粉末3mol/L40℃．的是金屬鈉著火時馬上用沙子蓋滅B．將過期的化學(xué)藥品直接倒入下水道C．少量濃硫酸濺到皮膚上馬上用大量水沖洗D．用肥皂液涂在自然氣管道接口處檢驗是否漏氣以下說法中，正確的選項是2A．44gCO22.4L23B．NH17g/mol3C．100mL1mol/LNaCl溶液中含有1molCl－D．標(biāo)準(zhǔn)狀況下，11.2LH2O3.01×1023個水分子以下反響的離子方程式正確的選項是鐵與硫酸銅溶液反響：Fe＋Cu2＋ Cu＋Fe3＋銅與稀硝酸反響：Cu＋4H＋＋2NO－ Cu2＋＋2NO↑＋2HO23 2 222鋁與氫氧化鈉溶液反響：2Al＋2OH－＋2H2O 2AlO－2

＋3H↑氫氧化鋇溶液與稀硫酸反響：Ba2+＋SO2－＋H＋＋OH－ BaSO↓＋HO4 4 2以下物質(zhì)性質(zhì)與用途的對應(yīng)關(guān)系的是選項 性質(zhì)氧化鐵呈紅棕色氧化鋁熔點(diǎn)很高濃硫酸具有脫水性液氨汽化時要吸取大量的熱

用途紅色涂料耐火材料枯燥劑制冷劑以下試驗裝置及操作到達(dá)試驗?zāi)康牡氖?Ca(OH)和2NHClNH

從碘水中提取C從食鹽水中獲碘單質(zhì) 取氯化鈉

D．用自來水制取蒸餾水4 3棉花化學(xué)試卷第3頁〔8頁〕其次局部非選擇題〔50分〕一、必答題〔30分〕2O213分〕氮在自然界的循環(huán)過程中可發(fā)生反響N ＋2O2

放電2NO。該反響中，被復(fù)原的物2質(zhì)是 〔填化學(xué)式作為復(fù)原劑的物質(zhì)是 〔填化學(xué)式假設(shè)反響中消耗1N，則生成NO mol。 223分〕煤干餾的產(chǎn)品經(jīng)加工可得到以下有機(jī)物：①甲烷、②乙烯、③苯。其中，可作清潔燃料的是〔填序號，下同劑的是。34分〕現(xiàn)有以下4種物質(zhì)：①

、②FeCl、③Al(OH)HNO。其中，既能與2 3 3 3酸反響又能與堿反響生成鹽和水的是 〔填序號下同與水反響產(chǎn)生無色氣體的是 ；遇KSCN溶液顯紅色的是 ；常溫下能使鐵外表生成致密氧化膜的是 。44分〕某小組同學(xué)利用以以下圖所示裝置探究二氧化硫氣體的性質(zhì)。SO2品紅溶液①請答復(fù)：

KMnO4溶液②

NaOH溶液③試驗室用銅和濃硫酸共熱制取二氧化硫，反響的化學(xué)方程式是 。當(dāng)觀看到②中溶液顏色變淺時，①中的現(xiàn)象是 。裝置③的作用是 。①、②中的現(xiàn)象說明二氧化硫具有的性質(zhì)是 ?；瘜W(xué)試卷第4頁〔8頁〕58分〕綠礬OH〕可用作除草劑、凈水劑或抗貧血藥等。某爭論小組同學(xué)承受4 2硫鐵礦焙燒后的燒渣〔FeO

、AlO

，不考慮其他雜質(zhì)〕制備綠礬，設(shè)計了如下流程。

2 3 2 3 225.0225.02gAFeSO·7HO4 210.00g燒渣足量酸①固體B足量酸③CC④FeSO4溶液足量堿A②溶液B請答復(fù)：①~③所涉及的試驗操作方法中，用到過濾的是 〔填序號。③中發(fā)生反響的離子方程式是 。以下說法正確的選項是 〔填字母。①中可選用鹽酸，③中只能用硫酸溶液BOH－、SiO2－、AlO－等陰離子3 2④中發(fā)生了氧化復(fù)原反響，固體C做復(fù)原劑原燒渣中FeO的質(zhì)量分?jǐn)?shù)為 。2 3AB確定條件CDBE68分〕AEAB確定條件CDBE請答復(fù)：假設(shè)C的焰色反響為黃色；常溫下，D為無色氣體，能參與綠色植物的光合作用。①組成A的元素在周期表中的位置是 。②以下表達(dá)正確的選項是 〔填字母。A與冷水猛烈反響C與水反響可生成BE與酸反響可生成D假設(shè)C是能使潮濕的紅色石蕊試紙變藍(lán)的氣體，D是黃綠色氣體。①C與D反響的化學(xué)方程式是 。②上述反響中轉(zhuǎn)移1.2mol電子時，生成B的體積為 〔標(biāo)準(zhǔn)狀況。5頁〔8頁〕二、選答題〔共20分。請在以下三個模塊試題中任選一個模塊試題作答，假設(shè)選答了多個模塊的試題，以所答第一模塊的試題評分〕《化學(xué)與生活》模塊試題14分“北京烤鴨”舉世著名。制作烤鴨套餐的局部原料如下：①鴨肉、②面粉、③黃瓜、④食鹽。請答復(fù)：富含淀粉的是 〔填序號，下同。富含蛋白質(zhì)的是 。富含維生素的是 。屬于調(diào)味劑的是 。28分材料之王”。它最有潛力的應(yīng)用是硅的替代品，用來生產(chǎn)將來的超級計算機(jī)。石墨烯原來就存在于自然界，將石墨烯一層層疊起來就是石墨，厚1mm300萬層石墨烯。制備石墨烯有多種方法。其中，機(jī)械剝離法是利用物體與石墨烯之間的摩擦得到石墨烯薄層。氧化復(fù)原法是先將石墨氧化，增大石墨層之間的間距，再通過物理方法將其分別，最終用復(fù)原劑復(fù)原得到石墨烯。中國在石墨烯應(yīng)用領(lǐng)域探究中已獲得重大發(fā)現(xiàn)，必將為世界的科技進(jìn)步作出重大奉獻(xiàn)。請依據(jù)以上短文，推斷以下說法是否正確〔填“對”或“錯。自然界中不存在石墨烯。石墨烯與石墨的組成元素一樣。工業(yè)上，可以通過機(jī)械剝離法或氧化復(fù)原法得到石墨烯。石墨烯是型納米材料，具有優(yōu)良的導(dǎo)電性能和機(jī)械性能。38分〕含汞廢水主要來源于氯堿工業(yè)和塑料工業(yè)，未經(jīng)處理不得排放。某含汞酸性廢水〔Hg2+、H+、Na+、Cl－等〕的處理工藝流程如以以下圖所示。溶液A NaS、FeSO含汞酸性廢水

pH＝9

2 4沉淀 超濾 活性炭吸附 出水請答復(fù)：含汞廢水未經(jīng)處理直接排放的危害是 〔填字母。造成水體污染 b.引起人體中毒調(diào)整廢水pH時參與的溶液A可能是 〔填字母。鹽酸 b. NaOH溶液NaS與Hg2+發(fā)生反響的離子方程式是 。2該廢水經(jīng)處理后，測得有關(guān)數(shù)據(jù)如下〔其他離子無視不計：離子濃度〔mol/L〕

Na+3.0×10-4

Fe2+ Cl－2.0×10-4

SO2－42.5×10-44mol/L?！队袡C(jī)化學(xué)根底》模塊試題14分〕現(xiàn)有以下6種與生產(chǎn)生活相關(guān)的有機(jī)物：

化學(xué)試卷第6頁〔8頁〕①CH3CHCH2OH

〔增塑劑〕 ②CH C—CH CH 〔合成橡膠原料〕2 CH3 CH2 3 2 2 2 2 ③CHCHCHCHOH〔萃取劑〕 ④CH CH—CH CH3 2 2 2 2 ⑤ CHO

〔香料〕

—CHCOOH2⑥2

〔植物生長刺激素〕請答復(fù)：屬于羧酸的是 〔填序號。與②互為同系物的是 〔填序號。與③互為同分異構(gòu)體的是 〔填序號。用系統(tǒng)命名法給③命名，其名稱為 。2〔8分〕現(xiàn)有以下4種重要的有機(jī)物：3 ①CHCH3 請答復(fù)：

②CHCHCH ③3 3

OH ④

CH3能與銀氨溶液反響生成丙酸銨的是 〔填序號。能與濃硝酸反響生成TNT的是 〔填序號。能與甲醛在酸催化下相互縮合成高分子化合物的是 〔填序號。寫出②與NaOH的乙醇溶液共熱生成丙烯的化學(xué)方程式： 。3.〔8分〕〔MMA〕〔有機(jī)玻璃的單體，也用于制造其他樹脂、塑料、涂料、粘合劑和潤滑劑等。MMA的合成路線如以以下圖所示〔反響條件和試劑略〕。O〔甲〕請答復(fù)：

① 乙 ② CH C C2CH32〕

OCH3甲分子中所含官能團(tuán)的名稱是 。MMA的分子式是 。乙的構(gòu)造簡式是 。MMA在確定條件下生成聚甲基丙烯酸甲酯的化學(xué)方程式是 ?；瘜W(xué)試卷第7頁〔8頁〕《化學(xué)反響原理》模塊試題14分a氣化爐是用焦炭為原料生產(chǎn)一氧化碳的裝置，它既可用來氣化無煙煤，又可用來氣化焦炭和煙煤。該氣化爐內(nèi)有以下反響發(fā)生：2 ①2C(s)＋O(g) 2CO(g) ΔH ＝－225.82 ②C(s)＋O(g) CO(g)

＝－394.4kJ/mol2 2 2請答復(fù)：反響①屬于 〔填“吸熱”或“放熱〕反響。反響①消耗24gC(s)時，?H＝ kJ/mol。2反響②生成1molCO(g)時，?H＝ kJ/mol。22反響C(s)＋CO(g) 2CO(g) ΔH＝ kJ/mol。222 28分)＋O) O) ?＜0是制造硝酸的重要反響之一。在0℃時，容積為1L的密閉容器中充入0.010molNO和0.005molO，反響過程中NO22 間變化如以以下圖所示：請答復(fù)：2〔1〕2min內(nèi)，v(O)= mol/(L·min)。2上升溫度，平衡向 〔填“正”或“逆〕反響方向移動。平衡時，NO的轉(zhuǎn)化率是 。〔4〕800℃時，該反響的化學(xué)平衡常數(shù)數(shù)值為 。38分〕侯氏制堿法曾廣泛用于工業(yè)制純堿aO，是我國科學(xué)家為人類做出的重大2 3250.10mol/LNaCO

溶液。請答復(fù)：2 3〔〕該溶液中H) c－)〔填“＞“＝”或“＜。用離子方程式解釋〔1〕的緣由： 。以下說法正確的選項是 〔填字母。a．NaCO在水中完全電離2 3b．c(CO2－)＋c(HCO－)＋c(HCO)＝0.1mol/L3 3 2 3c．c(Na+)＞c(CO2－)＞c(OH－)＞c(HCO－)＞c(H+)3 3用石墨作電極電解NaCO水溶液時，陰極反響式是 。2 3化學(xué)試卷第8頁〔8頁〕xx年北京市春季一般高中會考化學(xué)試卷答案及評分參考說明：考生答案如與本答案不同，假設(shè)答得合理，可酌情給分，但不得超過原題所規(guī)定的分?jǐn)?shù)。第一局部選擇題〔50分〕選擇題〔250分〕題號12345678910答案CBBDADCBDA題號11121314151617181920答案DCDABBCCAD題號2122232425答案BBCCB一、必答題〔30分〕

其次局部非選擇題〔50分〕213分〕O ·1分22N ················································································································1分22 ···············································································································1分23分〕①·1分② ··············································································································1分③················································································································1分34分〕③·1分①················································································································1分②················································································································1分④··················································································································1分2 4 4 2 44分〕u＋HO〔濃〕 O ＋O↑＋HO …12 4 4 2 〔2〕品紅溶液褪色〔或變淺〕 ………1分2吸取未反響的SO，防止污染環(huán)境 …………1分2漂白性、復(fù)原性 1分258分1〕①② 2分23〔2〕Fe(OH)3

＋3H+ ＋3H

O ……………2分〔3〕ac 2分〔4〕48.00%〔0.48〕················································································2分68分〕①第3周期第A族 2分②abc ····························································································2分3 2 2 〔2〕①8NH ＋3Cl N ＋6NH3 2 2 3 2 或2NH ＋3Cl N ＋6HCl 23 2 ②4.48L …………2分二、選答題〔20分〕說明：考生可任選其中一個模塊的試題作答，假設(shè)選答了多個模塊的試題，以所答第一模塊的試題評分?！痘瘜W(xué)與生活》模塊試題14分〕② ·1分〔2〕①··········································································································1分〔3〕③········································································································1分〔4〕④············································································································1分28分〕錯·2分〔2〕對·········································································································2分〔3〕對·········································································································2分〔4〕對·········································································································2分38分〕ab ·2分〔2〕b ············································································································2分〔3〕S2—＋Hg2+ HgS↓ 2分〔4〕2.0×10-4 ·····························································································2分《有機(jī)化學(xué)根底》模塊試題14分〕⑥·1分〔2〕④········································································································1分〔3〕①··········································································································1分〔4〕1-丁醇··································································································1分2〔8分〕①·2分〔2〕④········································································································2分〔3〕③········································································································2分乙醇〔4〕CHCHClCH ＋NaOH CHCH＝CH↑＋NaCl＋HO3 3 △ 3 2 2························································2分38分〕碳碳雙鍵、醛基·2分58 〔2〕CHO ································································································258 〔3〕 ………2分COOCH32〔4〕nCH2

C

確定條件

CH2 CnCH3 CH3…2《化學(xué)反響原理》模塊試題14分〕放熱 ·1