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19:Tfc2,こんにちは。東京科学大工学院のプロトンです。無機化学は反応式だったり、製法が多すぎて、覚えるのが大変ですよね。今回は少しでも覚えやすくなるようなコツを話したいと思います。


まず、無機化学を勉強していく上で暗記に100%頼っていると途中で詰みます。理解とともに覚えるように心がけましょう。

私が受験生時代に使っていた参考書は
福間の無機化学の講義(Doシリーズ)です。ほんっとうに分かりやすいので一度手に取ってみてください。

無機化学で出てくる反応式は、限られています。

・中和反応
・酸化還元反応
・酸塩基の反応
・揮発性酸遊離反応
・熱分解による反応
・沈殿生成反応

こんなところかと思います。あとは、気体の製法で少し覚えるべき反応式があります。

反応式を作るときに、覚える必要があるのは、反応前後の物質と係数です(酸化還元反応では、酸化数も数える)。
反応前後の物質と係数を覚えていれば、慣れてくると自分で作れるようになります。

例えば、水酸化ナトリウムと塩酸の中和を考えます。
反応前の物質はNaOHとHClで、係数はどちらも1です。反応後は中和されて、NaClとH₂Oができます。
これが分かれば、反応式が書けます。

NaOH＋HCl → NaCl＋H₂O

より複雑な反応式になっても同じ方法でやりましょう。特に、酸化還元反応の半反応式を覚える際には、反応前後の状態を覚えている必要がありますので、必ず覚えましょう。
また、先程述べましたが、酸化還元反応では酸化数を数えます。酸化数とは、化合物やイオンの状態で、中性の原子と比べて電子をどれだけ失っているかを示しています。

例えば、H^+であれば、水素原子と比べて電子を１個失っているので酸化数は+1になります。
また、CO₂ の C に着目すると、電子配置が、::O::C::O::(これだと電子配置が上手く表記できないのでご了承ください)となります。

C原子は、電子数が4ですが、二酸化炭素だと、Cは8個の電子を周りに抱えている状態です。なので、4個多く電子をもらっているので、酸化数は＋4になります。

このように、酸化数を数えられるようになれば、半反応式は無双できます。

沈殿生成反応はご存知の通り、イオン結晶同士をとかした時に別の組み合わせで、溶けにくい物質が沈殿する反応です。

例えば、Ba(OH)₂と K₂SO₄の反応があります。ともにイオン結晶であり、水に溶かすとイオンとなります。
しかし、バリウムイオンと硫酸イオンは、反応して水に溶けにくい硫酸バリウムとなって沈殿します。これが沈殿生成反応です。

無機の反応の中でもこれは比較的覚えやすいかと思います。

沈殿が生成されるかは、何と何が反応するかによって変わってきますが、これは私が使っていたゴロ合わせを伝授します。

〈Cl¯との反応〉
銀(Ag+)のなま(Pb²+)※はげ(Hg+)狂ってる(Cl¯)

銀のなまはげ狂ってる

〈SO₄²¯、CO₃²¯との反応〉
そ(SO₄²¯)こ(CO₃²¯)まで、
バ(Ba²+)カ(Ca²+)にする(Sr²+)な(Pb²+)※

※鉛(なまり)はPb

そこまでバカにするな

〈CrO₄²¯との反応〉
苦労(CrO₄²¯)して買った銀(Ag+)のバ(Ba²+)ナナ(Pb²+)

苦労して買った銀のバナナ

これらの語呂合わせを参考に覚えちゃって下さい。
以上、無機化学の反応式を覚えるコツ＆語呂合わせ(一部)を紹介しました。如何でしたか。受験は大変ですが、楽しく勉強していきましょう。応援しています！1b: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。1c:Tbb0,こんにちは！現在は経済学部に在籍していますが、元理系だったのでお答えしますね。

有機化学では、炭素と炭素の繋がりによって作られる物質が無数にあり、それらが色々な化学反応を起こして複雑に変化したりと、覚えることが多いと思うかもしれません。ですが、これも無秩序に反応を起こしてるわけではなく、実はちゃんとれっきとした規則性があるんです！
化学の世界は、電子に着目するとある程度統一的に説明することができます。これを電子論といいます。これは有機化学だけでなく無機化学でももちろん有効な考え方ですが、とりわけ有機化学で用いられます。

電子論は一般に大学でやることであって高校範囲を逸脱してるとされており、高校では教えてくれないでしょう。したがって教科書も実験ベースの記述で、〇〇と△△とで反応させたら□□になる、という話しか載ってないわけです。しかし、本格的に学ぼうとすればこそ量子力学が必要になってきますが、大雑把に電子がどう動くのか？等の話であれば高校生でも十分理解できる内容です。電子論を用いると今まで見えなかった反応のメカニズム等が視覚的に見えてきます。現在暗記で苦しんでいるのであれば、電子論を用いて学習することである程度暗記から解放されることでしょう。おすすめする参考書は「原点からの化学」シリーズと映像授業になりますがスタディサプリのトップ・ハイレベル化学がとてもわかりやすくておすすめです。

次に、構造についてですが、やはりこれも無数にパターンがあるのに一つ一つ別々に頭に入れていこうとするとキリがありません(笑)。
まずは最低限として各アルカンの名称、各官能基、命名法、そして教科書に出てくる物質の慣用名だけ覚えましょう。その際英語(ラテン語？)を意識すると意味が理解できて覚えやすいかなと思います。(例: 炭素数6のアルカン: ヘキサン(hexa: 6を意味する接頭辞)、カルボキシ基: -COOH(carbon + oxygen))

最後に、問題集の進め方ですが、それで問題ないと思います。個人的には、新演習はかなり重いので余裕があればやるのが良いですが、そうでなければ重要問題集までだけでも過去問に十分スムーズに入っていけるのではと考えています。
東大化学は物理との兼ね合い上、問題の難易度以上に時間的にかなり厳しくなる戦いになってくると思います。ですので、過去問やるときはそのこともしっかり意識して頑張ってほしいなと思います。応援しています。2:["$","main",null,{"className":"px-4 pt-4 pb-4","children":["$","div",null,{"className":"max-w-3xl mx-auto w-full","children":[["$","div",null,{"className":"mb-8","children":["$","$L7",null,{"href":"https://unilink-app.onelink.me/isbO/h6xeh63x?advice=FcNz8DtHt9cGVLkMfoYA","target":"_blank","children":["$","$L8",null,{"src":"/images/web_to_app_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink WebToAppバナー画像","className":"mb-4 rounded"}]}]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"式のたて方がよく分かりません"}],["$","div",null,{"className":"flex justify-between mb-4","children":[["$","div",null,{"className":"text-left text-xs text-caption","children":["クリップ(",1,") コメント(",0,")"]}],["$","div",null,{"className":"text-right text-xs text-caption","children":"8/7 0:21"}]]}],["$","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。"}],["$","div",null,{"className":"mb-4","children":["$","$L13",null,{"clientImageUrl":"https://firebasestorage.googleapis.com/v0/b/unilink-48e75.appspot.com/o/images%2Fs_uvBpvuNhooQaVZD9q6W6Z38i4ZK2.jpg?alt=media&token=0017d380-4ec6-4fa8-9a94-740d83e73981","clientUserName":"なーちゃん","infoString":"中学 埼玉県 愛国学園大学人間文化学部(35)志望","adviceId":"FcNz8DtHt9cGVLkMfoYA"}]}],["$","div",null,{"className":"mb-8","children":[["$","div",null,{"className":"leading-loose whitespace-pre-wrap","children":[["$","div","consultation-part-0",{"children":[null,"15.9gの酸化銅と、炭素1.2gが過不足なく反応するとする。\n酸化銅の粉末1.2gと炭素0.6gの場合、15.9×1.2分の0.6＝7.95gという式で7.95gの酸化銅と炭素0.6gが過不足なく反応することが分かるそうなのですが、どうしてこのような式になるのかが分かりません。なぜ15.9に1.2分の0.6をかけるのですか？どうしたらこの式を導けるのかを教えて欲しいです。"]}]]}],["$","div",null,{"className":"pt-4","children":["$","$L14",null,{}]}],null]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"回答"}],["$","div",null,{"className":"mb-4","children":["$","$L15",null,{"adviserImageUrl":null,"adviserName":"有為の罪","adviserDepartment":"京都大学理学部","adviceId":"FcNz8DtHt9cGVLkMfoYA"}]}],["$","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 mb-4","children":[["$","div","advice-part-0",{"children":[null,"条件より反応比率は、酸化銅 : 炭素=15.9 : 1.2です。この比率は、定比例の法則より問題の場合でも保存されます。\n反応する酸化銅をx(g)とすると、x : 0.6=15.9 : 1.2となり、xが求められます。また、xが与えられた酸化銅以下なので、過不足なく反応が進行したとわかります。（酸化銅を未知数とするとき、炭素は全て消費されたと仮定することになります。この問題ではその仮定は正しいですが、もし炭素を未知数とした場合、与えられた量以上の炭素が必要な計算になり、不適となります。）"]}]]}],["$","div",null,{"className":"mb-4","children":["$","$L16",null,{"adviserImageUrl":null,"adviserName":"有為の罪","adviserDepartment":"京都大学理学部","adviceId":"FcNz8DtHt9cGVLkMfoYA","numberOfFan":3,"clipsAvg":3.6,"adviceRateAvg":5,"profile":""}]}],["$","div",null,{"children":["$","$L7",null,{"href":"https://ck.jp.ap.valuecommerce.com/servlet/referral?sid=3364577&pid=884970531&vc_url=http%3A%2F%2Fshingakunet.com%2F%3Fvos%3Dnrmnvccp0000100","rel":"nofollow","target":"_blank","children":["$","$L8",null,{"src":"/images/document_request_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink パンフレットバナー画像","className":"mt-4 rounded"}]}]}],["$","div",null,{"className":"pt-4","children":["$","$L17",null,{"id":"adsbygoogle-init-under-advice"}]}]]}],["$","div",null,{"className":"flex 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"イオン反応式と係数"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"こんにちは！勉強お疲れ様です。\n\n早速、質問にお答えしていくんですが、\n\nイオン反応式である\nCa²+＋CO₃²-→CaCO₃\nの係数は実はこれでOKなんです。\n\nカルシウムイオン、炭酸イオンともに係数1で(1:1の比)炭酸カルシウムが沈殿するという反応です。\n勿論、もっと難しい係数決定の問題が出てくると思うので、例題を通じて、理解を深めましょう。\n\n(例題)\n水素イオンと硫化物イオンから硫化水素ができるイオン反応式の係数を揃えよ\n\n(  )H⁺＋(  )S²⁻ → (   )H₂S\n\n(ポイント)\n①化学反応式と同様に、原子数を揃える\n②イオン反応式では、加えて電子を揃える\n\n①化学反応式の係数決定と同様に、原子数が左右でそろっているか確認します。今回は右辺に、Hが2個、Sが1個ありますので、左辺であわせてあげると、H⁺が2個、S²⁻が1個というようになります。\n\n②イオン反応式ではさらに、左右で電子数を揃えなければいけません。ちなみに、+の電荷と-の電荷が同じ数の場合、打ち消して0になることに注意してください。\nそうすると、上の式の右辺は、電気的に中性(つまり電子数0)なので、左辺が打ち消してゼロになってくれればいいはずです。\n先程、H⁺の係数が2、S²⁻の係数が1であるとしていたので、当てはめると、H⁺の左上の+が2個あって、S²⁻の右上に2-があるので打ち消し合って0になるはずです。\n\n(解答)\n\n(2)H⁺＋(1)S²⁻ → 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","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","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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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","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 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 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"質量パーセント濃度が簡単にわかる方法"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"質量パーセント濃度30%の砂糖水（砂糖水A）と15%の砂糖水（砂糖水B）がそれぞれ何gずつあるのかによって、混ぜ合わせた砂糖水（砂糖水C）の質量パーセント濃度も変わってきますが、考え方は変わらないので、ここではどちらも100gだと仮定することにします。質量パーセント濃度（%）とは、溶質の質量（g）を溶液の質量（g）で割った値に100をかけたものです。つまり、砂糖水Aの質量が100gで、質量パーセント濃度が30%ということは、溶質である砂糖の質量は30gです。砂糖水Bも同様に考えて、溶質である砂糖の質量は15gです。これらを混ぜ合わせた砂糖水Cは、溶液全体の質量が、100g（砂糖水Aの質量）と100g（砂糖水Bの質量）の和、すなわち200gであり、そこに含まれる溶質の質量が、30g（砂糖水Aの溶質の質量）と15g（砂糖水Bの溶質の質量）の和、すなわち45gである水溶液となります。したがって、砂糖水Cの質量パーセント濃度（%）は、45g（溶質の質量）を200g（溶液の質量）で割った値に100をかけたもの、すなわち22.5%となります。砂糖水Aと砂糖水Bをそれぞれ何gずつ混ぜるにせよ、以上のように、溶液の質量と溶質の質量に分けて考えれば、混ぜ合わせた溶液の質量パーセント濃度も簡単にわかるようになると思います。"}],["$","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","children":[["$","div",null,{"className":"mb-1","children":"化学反応式やその計算を得意にするには"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"こんにちは。\n\n化学反応式の係数を合わせるのが苦手、ということでしょうか？\n具体的に書かれていないのであまりわかりませんが、そういうことだと思って回答させていただきます。\n\n化学反応式の係数合わせに関しては、左辺と右辺でそれぞれの元素種の数が等しくなるようにすればいいです。\n文字を使って連立方程式を解くだけで簡単に導くことができます。\n\nまた、物質量や濃度の計算に関しては「単位」を意識して考えてみましょう。\n例えば、密度[g/m^3]に体積[m^3]をかければ質量[g]が出てきます。\nこれは覚えることではなく、単位に注目すれば一目でわかります。\nその考え方が身につけば、そういった計算は簡単にできるようになると思いますよ。"}],["$","div",null,{"className":"flex mb-1","children":[["$","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":[]}],["$","path","1",{"d":"M12 6c1.1 0 2 .9 2 2s-.9 2-2 2-2-.9-2-2 .9-2 2-2m0 10c2.7 0 5.8 1.29 6 2H6c.23-.72 3.31-2 6-2m0-12C9.79 4 8 5.79 8 8s1.79 4 4 4 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"大学入試化学の問題です計算過程を含めて解答を教えてください！"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"こんにちは！お疲れ様です！\n端末だと見にくいと思うので、数式や文字を紙に全て書き写してから読むことをおすすめします。\n\n\nでは解説していきます。\n\n\n最初にあるH2SO4の物質量を計算しておきます。\n\n2.0(mol/L)×0.25(L)＝0.5(mol)\n\n発生したNH3の物質量を、x(mol)とおきます。\n\n2NH3+H2SO4→(NH4)2SO4\n\nこの反応式より、反応したH2SO4は0.5x(mol)と分かります。\n\nよって反応せずに残ったH2SO4は0.5-0.5x(mol)･･･①となります。\n\n滴定で用いたH2SO4の物質量は1000(mL)のうち、5(mL)を取りだして使ったので\n\n5/1000×(0.5-0.5x)molとなります。\n\n滴定で用いたNaOHの物質量を計算しておきます。\n\n0.40(mol/L)×5.0×10^-3(L)＝2.0×10^-3(mol)\n\n2NaOH+H2SO4→Na2SO4+2H2O\n\nこの反応式より、\n2.0×10^-3＝2×5/1000×(0.5-0.5x)\n                  2.0＝5(1.0-1.0x)\n                   2/5＝1-x\n                       x＝3/5＝0.60\n\nよって(2)が、0.60molとなります。\n\n(１)は①より残った硫酸は0.5-0.5×0.60＝0.5-0.3\n                                                                             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