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1d:I[3866,["51","static/chunks/795d4814-710d199cb1f51304.js","183","static/chunks/183-141f39fb5cb93742.js","23","static/chunks/app/advice/%5Bid%5D/page-26686d89cc5a4cf8.js"],"AdOnAdviceList2"] 1f:I[3866,["51","static/chunks/795d4814-710d199cb1f51304.js","183","static/chunks/183-141f39fb5cb93742.js","23","static/chunks/app/advice/%5Bid%5D/page-26686d89cc5a4cf8.js"],"AdOnAdviceList3"] 1b:Tc62,演習の際、用意している教材は問題集のみでしょうか? もしそうなら、必ず教科書や図録を見ながら取り組みましょう。 教科書や図録を使わずに学習をする人は、丸暗記に陥りがちです。 解説が的を絞ったことしか書いていないからです。 以下詳細な回答です。 *丸暗記では勝てない *暗記の心得3カ条 *人に教えて理解を深める *まとめ ------------ 【丸暗記では勝てない】 京大の化学は丸暗記で挑む人を叩きのめす構造になっています。 基本的な事柄に対しても「なぜか?」を理解して説明できる力が必要です。 大問1…酸化還元や物質の構造に関する問題 大問2…溶液や気体の化学平衡に関する問題 大問3…有機化学に関する問題 大問4…高分子化合物に関する問題 が例年のパターンです。 計算問題や有機化学のパズルは一見論証と関わりなさそうですが、立式導出過程を問題文から理解するということが必要になってくるので、丸暗記が習慣になっている人は苦戦を強いられます。 ----------- 【暗記の心得3カ条】 駿台化学科の石川正明先生は、 「論理性」 「意味性」 「感動性」 を大切に理解暗記していきましょう、と指導されています。 「論理性」 何故そうなるのかを説明できると、自信をもって理解暗記できます。 (例) 過マンガン酸イオンが、酸性溶液中で酸化剤としてはたらく半反応式 MnO4(-) 8•H( ) 5•e(-) → Mn(2 ) 4•H2O これは何故このような式になるのか、自分なりにわかりやすく説明できるようになっている人の方が自信を持って暗記できます。 教科書と化学図録を用いて詳しく調べながら学習を進めると良いです。 「意味性」 それを理解すると何が得られるのかを知ると、理解暗記する意欲が増します。 例) ステアリン酸×3とグリセリンから成る油脂は分子量890で、これを覚えておけば高分子化合物の分子量絡みの計算が速くなる。 「感動性」 面白い、すごい、ひどい、と心動くことで暗記しやすくなります。 例) 「水兵リーベ僕の船…」に代表されるような語呂合わせ。 ------------ 【人に教えて理解を深める】 重要問題集で自分が出来た問題でも、誰かにわかりやすく解説することは出来ますか? 「論理性」「意味性」「感動性」を意識して、人に教えてみてください。 これは僕が実践してきた化学勉強法です。 友達同士で質問し合いっこしましょう。 特に模試の復習でこれをやると効果的です。 ------------ 【まとめ】 問題集 教科書 図録で演習。 人と教えあいっこで理解を深める。 京大相手に丸暗記では太刀打ちできない。1c:T17da,回答させていただきます。ただ、私は受験生時代にサイトで紹介されていた暗記量を減らす方法を使っていたため、その方法を解説するサイトのリンクを貼らせていただくという形にさせていただきます。ご了承ください。 ・酸化還元、電池の暗記  酸化還元については、酸化剤と還元剤にはどんなものがあって、反応でそれらが何になるかは丸暗記する必要があります。(例:濃硫酸→二酸化硫黄)ただ、半反応式を全て覚える必要はなく、元の物質が何になるかさえ知っていれば半反応式を組み立てることができます。私は受験生時代、次のサイトを見てやり方を理解しました。 (https://kimika.net/r5hanhannousiki.html 化学のグルメ-半反応式・酸化還元反応式(作り方・覚え方・問題演習など))  また、電池についても上のサイトで全て事足ります。既に授業動画で予習されているということなので、事前知識込みで見ていただくと非常に分かりやすいです。 ・構造決定の学習法 構造決定はやればやるほど点が伸びますし、できるようになるとパズルみたいで楽しいのでとにかく量をこなしてください。ただし、構造式は問題の指示に従って書くように意識してください。これを習慣づけると理不尽な減点を防げます。  既に新演習に手を出されているということで、ある程度骨のある練習問題として東北大学の有機をおすすめします。東北大学の第3問は伝統的に構造決定が出題されますが、非常に問題の質が良いです。難易度で見てもちょうど良く大問の中で勾配があるため、基礎知識の定着と思考力の養成にうってつけです。 ・無機の暗記  イオン化傾向列を使うと覚えやすい部分があります。例えば、イオン化列で横並びの亜鉛鉄ニッケルは濃硝酸と反応して不動態を形成するというように、イオン化列で近い位置にあるものは近い性質を持ちます。また、色の暗記では既にされているように資料集を読んで写真で覚えるのがよいです。 ・理論の計算  溶解度については、出した質量が溶液のものなのか、溶質のものなのか、溶媒のものなのかを考えて解いてください。  例えば、食塩の溶解度が20g/水100gであれば、食塩:水の質量比は1:5ですが、食塩:溶液なら1:6になります。これを混同すると正解できなくなるので気をつけましょう。  気液平衡については、気体の量を体積でなく物質量で捉えると混乱しにくいです。ヘンリーの法則では標準状態基準の体積を考えることがありますが、分かりにくいので全て物質量で統一してしまいましょう。(物質量に直すとき、分数のままにしておくと多少計算が楽) ・合成高分子の暗記  そもそも京大には合成高分子は大問としてはほとんど出題歴がありません。しかし、理論の方でイオン交換膜が出題されたりはするので、そこは暗記必須です。ただ、結局共通テストには合成高分子は出題されるので、その対策として繊維やゴムについて覚える必要はあります。  合成高分子はそれまでの化学よりも更に生活に身近になります。身の回りのものと結びつけて覚えてみましょう。例えば、アラミド繊維は強度が非常に強いため防弾チョッキに使われているといった感じで、物質の種類、性質、使用例を一気に暗記してしまいましょう。  また、反応機構に触れると、ほぼ例外のないパターンがいくつかあるだけなので無機に比べて楽に覚えられるはずです。 ・各分野の学習比、京大化学の傾向  まず、京大化学は大問4つ(2科目で180分、1課目なら90分)で、大問1は理論と無機の複合か純粋な理論、大問2は純粋な理論、大問3は有機、大問4は高分子(ほとんど天然高分子)が出題されています。これは10年間変わっていません。  全体を通して、知識を前提に思考力を問われています。記述問題だけでなく、グラフを選択したり、範囲外の内容(マルコフニコフ則やラウールの法則など)を誘導付きで出題してきます。また、数値計算の煩雑さもありますが、物理の問題のように文字のままで計算する問題が多くあり、慣れが必要です。  京大化学では、まず有機と高分子で得点を稼ぐ必要があります。比較的難易度が低く、そうでなくても誘導が丁寧なことが多いです。有機は構造決定と立体化学、高分子は糖ペプチドアミノ酸が多く出題されます。  理論は、結晶構造、電離平衡、気体が多く出題され、中でも結晶は出題頻度が高いです。  無機は全く出題されない年もありますが直近は3年連続で出題されています。対策を怠ることはできません。  以上を踏まえると、有機>理論>無機の順で重視したほうがいいと考えられます。比率に関してはなんとも言えないですが、高2から化学を勉強されているということで、受験前には全分野完成する(化学で7〜8割が目指せる)と思われます。なので比率というよりは不安な分野を都度穴埋めするぐらいでよいでしょう。  最後になりますが、暗記分野は演習をこなしていくうちにいずれ覚えていくと思ってください。演習の比重を大きくしていって、最終的に過去問にまで到達できるようにしましょう。健闘を祈ります。1e:Tc18,化学の分野別の学習量の多さは、理論:無機:有機=3:1:1くらいです。圧倒的に理論化学が難しく、点数が取りにくい。 よって、理論分野と無機・有機分野では、勉強法は全くといっていいほど違います。 理論分野は、覚えることは物理並みに少ない(ただし酸化還元の酸化数だけは暗記必須ですが)と同時に、計算量が非常に重い問題が多く、特に蒸気圧などを用いた気体の問題や、酸と塩基の平衡が絡んだ問題は、ただひたすら重い問題を解いて慣れていくしかありません。 それと真逆なのが無機分野です。 各元素の特徴など、特に色なんかは、語呂合わせなどで楽しく覚えていきましょう。例えば、斜方硫黄と単斜硫黄、どちらが安定か覚えるなら、二つで違う感じを用いているところに注目して、「方」と「単」、これらをそれぞれ床に置いた時、どちらが倒れにくそうか考えれば、「方」がより安定しそうなのは自明です笑。また、クロム酸イオンと沈殿を形成する元素は「バリウム 淡黄 鉛 黄 銀 赤褐」をまとめて、「バター王亡き銀世界」と覚えたり。 暗記が多いといっても、英語に比べれば可愛いものですから、そういった「ごり押し」で十分です。 あとは、一問一答形式の問題集で確認して、最終チェックとして入試で出た問題を扱えば、それで十分なレベルに達します。 有機分野は無機分野よりはとっつきにくいです。まずは、アルカンから順番にじっくり理解していってください。 理解したら、参考書を見ながらでいいので、問題をどんどん解いていきましょう。有機は、問題を解くときに踏む思考プロセスが非常に多く、分からないときに解答を読んでもさっぱり分からないことが多々あります。その分、解ければ、基本わかっているということなので、最終的に自分の知識以外何にも頼らず解けるように進んでいってください。 最終盤の、高分子化合物だけは、特殊かつ厄介なところがあります。ここも、無機分野と同じように、暗記から攻めていきます。有機前半と同じように問題演習していってもちっとも身につかないです。 まとめノートは、無機分野に限り作ることを強く推奨します。大した負担にはならず、大事なところだけまとめれば半日で終わ理ますしね。 理論化学と無機有機はかなり独立な分野です。理論が全くできないが、無機有機は無敵、という人もいます。復習はいずれ完了させなければなりませんが、まずはどんどん進んでいきましょう。 ただ、化学によって英数がおろそかになるのは非常にまずいので避けたほうがいいですね。2:["$","main",null,{"className":"px-4 pt-4 pb-4","children":["$","div",null,{"className":"max-w-3xl mx-auto w-full","children":[["$","div",null,{"className":"mb-8","children":["$","$L7",null,{"href":"https://unilink-app.onelink.me/isbO/h6xeh63x?advice=HVG9p30BTqPwDZPuV21c","target":"_blank","children":["$","$L8",null,{"src":"/images/web_to_app_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink WebToAppバナー画像","className":"mb-4 rounded"}]}]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"化学基礎の問題が分かりません!解き方を教えてください。"}],["$","div",null,{"className":"flex justify-between mb-4","children":[["$","div",null,{"className":"text-left text-xs text-caption","children":["クリップ(",5,") コメント(",2,")"]}],["$","div",null,{"className":"text-right text-xs text-caption","children":"12/11 13:25"}]]}],["$","div",null,{"className":"coach-mark mb-4","children":"UniLink利用者の80%以上は、難関大学を志望する受験生です。これまでのデータから、偏差値の高いユーザーほど毎日UniLinkアプリを起動することが分かっています。"}],["$","div",null,{"className":"mb-4","children":["$","$L13",null,{"clientImageUrl":null,"clientUserName":"aka","infoString":"中学 広島県 大阪大学経済学部(65)志望","adviceId":"HVG9p30BTqPwDZPuV21c"}]}],["$","div",null,{"className":"mb-8","children":[["$","div",null,{"className":"leading-loose whitespace-pre-wrap","children":[["$","div","consultation-part-0",{"children":[null,"10%の希硫酸(分子量98)19.6gを中和するのに1.0 mol/Lの水酸化カリウム水溶液が何ml必要か。"]}]]}],["$","div",null,{"className":"pt-4","children":["$","$L14",null,{}]}]]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"回答"}],["$","div",null,{"className":"mb-4","children":["$","$L15",null,{"adviserImageUrl":"https://firebasestorage.googleapis.com/v0/b/unilink-48e75.appspot.com/o/images%2Fs_Oou0tJz4wTbzaJ6HpZzQAgIh9GZ2.jpg?alt=media&token=690b2ef6-c85b-45da-9920-7d340e4fc684","adviserName":"yuya","adviserDepartment":"東京工業大学物質理工学院","adviceId":"HVG9p30BTqPwDZPuV21c"}]}],["$","div",null,{"className":"coach-mark mb-4","children":"すべての回答者は、学生証などを使用してUniLinkによって審査された東大・京大・慶應・早稲田・一橋・東工大・旧帝大のいずれかに所属する現役難関大生です。加えて、実際の回答をUniLinkが確認して一定の水準をクリアした合格者だけが登録できる仕組みとなっています。"}],["$","div",null,{"className":"mb-8","children":[["$","div",null,{"className":"leading-loose whitespace-pre-wrap","children":[["$","div","advice-part-0",{"children":[null,"H₂SO₄ + 2KOH→ K₂SO₄ + 2H₂O\n\r\n希硫酸の水素イオン[H+]のmol数と水酸化カリウムの水酸化物イオン[OH-]のmol数が同じになった時に中和されるよ!\n\nmol数ってどうやって出すか覚えてるかな?\n\n\n*希硫酸について\n\n分子量っていうのはもともと1molあたりの重さだから、希硫酸19.6gのmol数は19.6÷98=0.20molだね!\n\nでも、これ10%の希硫酸なんだよね。だから19.6gのうち1.96gが希硫酸であとは水だね。\n\nだから正しくは1.96÷98=0.020molだね!\n\n\n\n*水酸化カリウムについて\n\n希硫酸の[H+]0.020molと同じ量の水酸化カリウムの[OH-]を用意しよう!\n\n2KOHだから2倍、つまり0.040molの[OH-]を用意するにはどうすれば良いかな?\n\n1Lつまり1000mlで1molだから、何mlで0.040molになるかは自分で考えてみよう!"]}]]}],["$","div",null,{"children":["$","$L7",null,{"href":"https://ck.jp.ap.valuecommerce.com/servlet/referral?sid=3364577&pid=884970531&vc_url=http%3A%2F%2Fshingakunet.com%2F%3Fvos%3Dnrmnvccp0000100","rel":"nofollow","target":"_blank","children":["$","$L8",null,{"src":"/images/document_request_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink パンフレットバナー画像","className":"mt-4 rounded"}]}]}],["$","div",null,{"className":"pt-4","children":["$","$L16",null,{"id":"adsbygoogle-init-under-advice"}]}]]}],["$","div",null,{"className":"flex justify-between","children":[["$","h1",null,{"className":"text-xl font-semibold","children":["コメント(",2,")"]}],["$","$L17",null,{"adviceId":"HVG9p30BTqPwDZPuV21c"}]]}],["$","div",null,{"className":"mb-8","children":["$","div",null,{"className":"divide-y","children":[["$","div",null,{"className":"flex py-4","children":[["$","div",null,{"className":"mr-2","children":["$","$L18",null,{"avatarUrl":null,"contributorName":"aka","adviceId":"HVG9p30BTqPwDZPuV21c"}]}],["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"flex justify-between","children":[["$","div",null,{"className":"mb-2","children":["$","$L19",null,{"contributorName":"aka","adviceId":"HVG9p30BTqPwDZPuV21c"}]}],["$","div",null,{"className":"text-xs text-caption","children":"12/11 14:36"}]]}],["$","div",null,{"className":"text-xs whitespace-pre-wrap","children":"お忙しい中ありがとうございました!\n理解できました。"}]]}]]}],["$","div",null,{"className":"flex py-4","children":[["$","div",null,{"className":"mr-2","children":["$","$L18",null,{"avatarUrl":"https://firebasestorage.googleapis.com/v0/b/unilink-48e75.appspot.com/o/images%2Fs_Oou0tJz4wTbzaJ6HpZzQAgIh9GZ2.jpg?alt=media&token=690b2ef6-c85b-45da-9920-7d340e4fc684","contributorName":"yuya","adviceId":"HVG9p30BTqPwDZPuV21c"}]}],["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"flex justify-between","children":[["$","div",null,{"className":"mb-2","children":["$","$L19",null,{"contributorName":"yuya","adviceId":"HVG9p30BTqPwDZPuV21c"}]}],["$","div",null,{"className":"text-xs text-caption","children":"12/11 16:03"}]]}],["$","div",null,{"className":"text-xs whitespace-pre-wrap","children":"理解できた?それは嬉しい😭\n\nまた何かあったら質問してね!"}]]}]]}]]}]}],["$","h1",null,{"className":"text-xl font-semibold","children":"よく一緒に読まれている人気の回答"}],["$","div",null,{"className":"mb-8","children":["$","div",null,{"className":"divide-y","children":[["$","div",null,{"children":["$","$L7",null,{"href":"/advice/HVG9p30BTqPwDZPuV21c","children":["$","div",null,{"className":"flex items-center py-4","children":[["$","div",null,{"className":"flex-1 mr-3","children":[["$","div",null,{"className":"mb-1","children":"化学基礎の問題が分かりません!解き方を教えてください。"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"H₂SO₄ + 2KOH→ K₂SO₄ + 2H₂O\n\r\n希硫酸の水素イオン[H+]のmol数と水酸化カリウムの水酸化物イオン[OH-]のmol数が同じになった時に中和されるよ!\n\nmol数ってどうやって出すか覚えてるかな?\n\n\n*希硫酸について\n\n分子量っていうのはもともと1molあたりの重さだから、希硫酸19.6gのmol数は19.6÷98=0.20molだね!\n\nでも、これ10%の希硫酸なんだよね。だから19.6gのうち1.96gが希硫酸であとは水だね。\n\nだから正しくは1.96÷98=0.020molだね!\n\n\n\n*水酸化カリウムについて\n\n希硫酸の[H+]0.020molと同じ量の水酸化カリウムの[OH-]を用意しよう!\n\n2KOHだから2倍、つまり0.040molの[OH-]を用意するにはどうすれば良いかな?\n\n1Lつまり1000mlで1molだから、何mlで0.040molになるかは自分で考えてみよう!"}],["$","div",null,{"className":"flex 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items-center py-4","children":[["$","div",null,{"className":"flex-1 mr-3","children":[["$","div",null,{"className":"mb-1","children":"中和滴定に電離度がなぜ関係ないのか"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"それは確かに私も不思議に思っていました。\n\n私は化学の専門家ではないので、あくまで一つの考え方として見て下さい。詳しいことは、化学の先生などにうかがったほうがいいと思います。\n\n後々習うとは思いますが、物質は電離するとき、電離平衡というバランスをとります。たとえばアンモニアでは、\nNH3 H2O ⇌NH4 OH-\nとなります。\n矢印が両方向にあることからも分かると思いますが、反応は右にも左にも進みます。NH3とNH4 ,OH-の濃度の比が一定になるように反応が起こります。\n\n塩酸を加えると、H とOH-が反応して、H2Oが生成します。すると、OH-の濃度は減少するわけです。そうなると先ほどの電離式の左右の比が崩れてしまうため、それを補うように反応が右に進みます。すると、アンモニアがさらに電離します。\nこれが続いていくと、アンモニアは続々と電離していき、最終的には全て電離することになります。\n\nこう考えると、少し納得がいく気がします。\n\nちなみに、電離度というのは、先ほどの電離式のそれぞれの濃度の比から導き出された値になります。高3になると、その算出の仕方も学ぶと思いますので、楽しみにしておいて下さい。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary 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items-center py-4","children":[["$","div",null,{"className":"flex-1 mr-3","children":[["$","div",null,{"className":"mb-1","children":"化学基礎からが全くわからない"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"こんにちは。\n\nmolの計算がよくわからない為に、その後の計算(中和やph、電池の問題など)も解けないという推測のもとに回答しますね。そうではなく、単元ごとに意味のわからない部分があると言うことであれば、個別にご質問なさると良いと思います。\n\nまず、モルを難しく考えすぎてはいないですか?\n6.02×10^23個=1 mol にしよう!と決めただけです。12個で1ダースと同じですね。\nそして、同じ原子を6.02×10^23個集めると、[その原子の原子量]g になります。\n同じ原子を6.02×10^23個集めたときの重さ、といちいち書くのは面倒なので、モル質量と呼びましょう。単位は1molで○gを分数にして g/mol ですね。\n\n炭素の原子量が12とします。いま、炭素が24 gあります。炭素の原子は何molありますか? 簡単ですね、2 molです。個数でいえば2×6.02×10^23個ですね!\n(分子の時は少し気をつけて下さい。水素の原子量を1とします。水素ガスが4 gあります。このとき水素原子は4 mol存在しますが、水素ガス(分子)は2 molです。)\n\nこのくらいは分かってる!とお思いかも知れませんが、mol計算が苦手な方は単位変換が苦手な印象です。計算の時、単位同士も掛け算、割り算していますか? もっとハイレベルな部分でわからないと言うことでしたら申し訳ありません…\n\nmolができれば、phや化学式も解けるはずです。\n電気分解などは多少慣れが必要ですが、一個の原子から何個の電子が出るのかを知っていれば後は比の計算ですから、こちらも大丈夫だと思います。\n\n最後に参考書ですが、私自身はあまり役に立ったと思えたことがありません。読んだ時は分かった気になるのですが…。結局、自分の言葉で説明できるように頭を使って理解するしかないと思います。参考書はその補助に使うものだと考えて、頑張ってください!"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary mr-1","children":["$undefined",[["$","path","0",{"fill":"none","d":"M0 0h24v24H0V0z","children":[]}],["$","path","1",{"d":"M12 6c1.1 0 2 .9 2 2s-.9 2-2 2-2-.9-2-2 .9-2 2-2m0 10c2.7 0 5.8 1.29 6 2H6c.23-.72 3.31-2 6-2m0-12C9.79 4 8 5.79 8 8s1.79 4 4 4 4-1.79 4-4-1.79-4-4-4zm0 10c-2.67 0-8 1.34-8 4v2h16v-2c0-2.66-5.33-4-8-4z","children":[]}]]],"style":{"color":"$undefined"},"height":16,"width":16,"xmlns":"http://www.w3.org/2000/svg"}],["$","div",null,{"className":"text-xs","children":["北海道大学工学部"," ","かんな"]}]]}],["$","div",null,{"className":"flex 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