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1a:Tb2c,こんにちは！方程式の解き方を説明していきますね。
まず、最も重要なのは問題の状況把握です。私は塾講師もやっているのですが、ここでつまづいてしまうことがほとんどです。ここで、毎回生徒には「図を使って問題文を表してください」と伝えています。愛ちゃんさんは問題文を図で表すのは得意ですかね？例えば、全体を一つの棒グラフで表し、その比率を棒グラフの中に書き込む、池の周りを走るような問題であれば、池を書いてその周りに小さい人を走らせる、など自分がわかりやすいように図を描いてみてください！そうすれば、式に直しやしやすくなりますよ。

次に注目する箇所は何を文字で置くかです。これは、基本的には問題で聞かれているものを文字で置けば大丈夫です！
例題を参考にしてみましょう。
(例題)
同じ値段のボールペンを20本買おうとしたところ、持っているお金では400円足りなかったので、15本買おうとしたが、それでも50円足りなかった。この時ボールペン1本の値段を求めなさい。

①何を文字で置くか考える。
この問題では、ボールペン1本の値段を聞かれているので、これをXと置くことにする。

②状況把握する。
ここで、何円自分が持っているのかわからないことに気づく。そのため、分からないものがボールペンの値段、所持金の二つとなり、式をどのように立てようか迷う。そこで、考えてほしい式の作り方として、「分からないもの = 分からないもの」という作り方を試してほしい。
ボールペンの値段は文字で置いているので、所持金に関する式を立てることにする。

（ボールペン20本）
このとき、400円足りなかったので、所持金は
所持金 = 20 × X － 400　
　　　　　　　　　　となる。
（ボールペン15本）
このとき、50円足りなかったので、所持金は
所持金 = 15 × X －50
　　　　　　　　　　となる。
③式を解く
これまでに求めた式を使って、Xの値を求めていく。
15 × X －50 = 20 × X － 400
を解いて、X = 70円となる。

この中の②が一番のポイントです。まあ、そこが一番難しいんですけど、、、「わからないもの=わからないもの」の式の立て方は自分のレパートリーの中にあると、多くの問題を解くことができると思います。なので、何を文字で置いて、何を基にした式を立てるのかを意識しながら、問題に取り組んでみてください！検討を祈っています。1b:Tf80,こんにちは。
計算問題はある程度できるけど暗記が絡んでくるとできなくなっちゃうということですね。
特に暗記の割合が大きい有機と無機がきついという感じですかね
僕も暗記苦手だったのですが、これから紹介するやり方で勉強したら本番で90点近く取れたので参考にしてみてください。
その方法は
「問題でわからなかったところを覚える」
です。

〈具体的なやり方〉
①参考書の問題または過去問を解く
（参考書は自分のレベルにあってれば大丈夫です）
②解説を見る
③わからなかった知識を資料集で確認する
④①〜③を繰り返す
⑤定期的に過去問を解いて自分ができるようになってるか確認する

〈おすすめポイント〉
○必要な知識だけを覚えられる
→この方法のポイントは「問題を解いてわからなかったものを確認する」というところにあります。つまり、これで使った資料集には、たくさん確認したところとほとんど見てないところができてきます。たくさん確認したところは、問題でたくさん出てくる（問題を解くのに必要な）知識なのに、自分がわかってないところです。逆にほとんど見てないところは、問題を解くにあたってそこまで重要じゃないところか、自分が理解•覚えられてるところです。そしてたくさん見てるところほど覚えやすいです。こうすることで、問題を解くときに必要な知識を効率よく身につけることができます。同時に問題の解き方も理解できるのでおすすめです。

○闇雲に覚えるより速い
→これは上に書いたことと被りますが、必要なところだけを覚えることに注力している方法なので、時間的にもあらぽんさんの状況にあっているのではないかと思います。

○資料集で色を色で覚える
→多くの資料集はカラーになっていて、色を色で覚えることができます。文字で書いてあるだけだと覚えずらかったものが、カラーのものを見ると少し覚えやすくなります。
なので資料集、特にカラーのものを使うことをお勧めします。

○過去問で自分の伸びを確認
→本番までの少しの時間で仕上げたいということなので、伸びてないのに同じやり方をやり続けるのはもったいないです。やり始めてからすぐには結果は出ませんが、過去問で伸びを確認するのはおすすめです。

〈注意点〉
○そもそもの知識がほとんどない状態だと効果が低い
→ある程度問題と戦える状態を想定しているので、どの問題も全く歯が立たないようでしたら効果が低くなってしまいます。この場合は問題集の難易度を下げるか軽く全体の知識の確認をした方がいいです。

○勉強のベースを問題を解くことにおく
→知識に問題があると感じるとインプット中心になってしまいがちですが、知識は問題を解くために身につけるものです。なので問題を解くことが勉強の中心においた方がいいです

〈最後に〉
　これも言ってしまっては元も子もありませんが、一人一人自分に合った方法は違います。あくまで参考にする程度にするのがおすすめです。これが絶対だとは思わないで欲しいですが、冒頭でも言いましたが、これは僕が実際にやっていた方法で、それなりにおすすめできます。
　長々とまとまりのない文章を失礼しました。参考になれば幸いです。本番まであと少し、頑張ってください！応援してます！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。2:["$","main",null,{"className":"px-4 pt-4 pb-4","children":["$","div",null,{"className":"max-w-3xl mx-auto w-full","children":[["$","div",null,{"className":"mb-8","children":["$","$L7",null,{"href":"https://unilink-app.onelink.me/isbO/h6xeh63x?advice=m5Jecb8kiDG8kZoJwiJZ","target":"_blank","children":["$","$L8",null,{"src":"/images/web_to_app_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink WebToAppバナー画像","className":"mb-4 rounded"}]}]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"質量パーセント濃度が簡単にわかる方法"}],["$","div",null,{"className":"flex justify-between mb-4","children":[["$","div",null,{"className":"text-left text-xs text-caption","children":["クリップ(",1,") コメント(",0,")"]}],["$","div",null,{"className":"text-right text-xs text-caption","children":"8/16 22:42"}]]}],["$","div",null,{"className":"coach-mark 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mb-2","children":"回答"}],["$","div",null,{"className":"mb-4","children":["$","$L15",null,{"adviserImageUrl":"https://firebasestorage.googleapis.com/v0/b/unilink-48e75.appspot.com/o/images%2Fs_mBLd8WqR2pOVKp8vWtIyDjP2qA03.jpg?alt=media&token=58f8baf6-452b-4be3-b749-a23ab3e01e88","adviserName":"たけなわ","adviserDepartment":"北海道大学法学部","adviceId":"m5Jecb8kiDG8kZoJwiJZ"}]}],["$","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":[["$","div","advice-part-0",{"children":[null,"質量パーセント濃度30%の砂糖水（砂糖水A）と15%の砂糖水（砂糖水B）がそれぞれ何gずつあるのかによって、混ぜ合わせた砂糖水（砂糖水C）の質量パーセント濃度も変わってきますが、考え方は変わらないので、ここではどちらも100gだと仮定することにします。質量パーセント濃度（%）とは、溶質の質量（g）を溶液の質量（g）で割った値に100をかけたものです。つまり、砂糖水Aの質量が100gで、質量パーセント濃度が30%ということは、溶質である砂糖の質量は30gです。砂糖水Bも同様に考えて、溶質である砂糖の質量は15gです。これらを混ぜ合わせた砂糖水Cは、溶液全体の質量が、100g（砂糖水Aの質量）と100g（砂糖水Bの質量）の和、すなわち200gであり、そこに含まれる溶質の質量が、30g（砂糖水Aの溶質の質量）と15g（砂糖水Bの溶質の質量）の和、すなわち45gである水溶液となります。したがって、砂糖水Cの質量パーセント濃度（%）は、45g（溶質の質量）を200g（溶液の質量）で割った値に100をかけたもの、すなわち22.5%となります。砂糖水Aと砂糖水Bをそれぞれ何gずつ混ぜるにせよ、以上のように、溶液の質量と溶質の質量に分けて考えれば、混ぜ合わせた溶液の質量パーセント濃度も簡単にわかるようになると思います。"]}]]}],["$","div",null,{"className":"mb-4","children":["$","$L16",null,{"adviserImageUrl":"https://firebasestorage.googleapis.com/v0/b/unilink-48e75.appspot.com/o/images%2Fs_mBLd8WqR2pOVKp8vWtIyDjP2qA03.jpg?alt=media&token=58f8baf6-452b-4be3-b749-a23ab3e01e88","adviserName":"たけなわ","adviserDepartment":"北海道大学法学部","adviceId":"m5Jecb8kiDG8kZoJwiJZ","numberOfFan":42,"clipsAvg":7.734104046242774,"adviceRateAvg":4.76844262295082,"profile":"気が向いたときに、気が向いたご質問に回答しています。"}]}],["$","div",null,{"children":["$","$L7",null,{"href":"https://ck.jp.ap.valuecommerce.com/servlet/referral?sid=3364577&pid=884970531&vc_url=http%3A%2F%2Fshingakunet.com%2F%3Fvos%3Dnrmnvccp0000100","rel":"nofollow","target":"_blank","children":["$","$L8",null,{"src":"/images/document_request_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink 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line-clamp-2 mb-1","children":"初めまして\n\n質問の意図は、ビーカーで最初1モル溶かしてその後メスフラスコに移動させるなら、完全にビーカーから移動させるのは不可能でどうしても水滴が少しビーカーに残ってしまう、その中にシュウ酸が含まれているから、結局メスフラスコに入っているシュウ酸は1モルより若干少ないのでは？というものと解釈して進めます 違ったらごめんなさい\n\n基本的に化学の問題や話の中で、溶液を移動したら、などという文言があれば、それは【簡単のため】100%移動したことになります もちろんこれは現実世界とは少し異なりますが、そもそも大帝の問題で理想気体として話を進めている時点で高校化学の大半は現実世界とは少し異なる都合のいい世界での話をしているので、そこは割り切るのがいいと思います\n\nあまりスッキリしない解答だったかもしれませんが、残りの受験勉強も頑張ってください🙏"}],["$","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 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"質問 何を数え落とした"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"かきふらいさん初めまして。\n文系ですが、回答させていただきます。\n\n本回答の構成は以下です。\n①数え落とした5通りはなにか\n②再考察が合っているか\n③数え上げる際のポイント\n\n以上3点です。\n\n①数え落とした5通りはなにか\n結論としては、端が白玉になる場合が想定されていないかなと思います。\n\nex.●◯●●◯●●◯●●◯●◯●◯\n\n上記のような場合です。\n\n②再考察が合っているか\n結論、誤りです。\n本問の場合、同色の玉に区別は存在しません。\n\n区別する場合は玉に番号が振ってあったりする場合でしょう。本問のように、玉を「色」でしか判別していない場合、区別する必要がないのです。\n\n③数え上げる際のポイント\n質問が若干抽象的なので、回答者様の解法に近いもので解く場合、私なら以下のように解きます。\n\nまずは、解答のための必要条件を具体的に導きます。\n必要条件は、\n「白玉同士は隣接しない」\n「赤玉は白玉に隣接する」\n以上の2つです。\n\nしたがって、初期配置は以下のようになります。\n\n◯●◯●◯●◯●◯●◯\n\n上記のように、赤5個、白6個を配置できます。残る赤4個をどこに配置するかですが、以下のようになります。\n\n1◯●2◯●3◯●4◯●5◯●6◯7\n\nつまり7箇所、赤を置く場所があります。なお、１つの場所につき置けるのは1つまでです。一見、「◯●」の間に置く場合も考えた方が良さそうな気がしますが、これは考慮する必要がありません。なぜなら、先ほど述べた通り、同色の玉はそれを区別する必要がないからです。\n例えば、\n「1◯●2」の2に赤を置けば、「1◯●●」となります。\n次に、\n「1◯●2」の◯●の間に●を置く場合を想定しても「1◯●●」となります。\nこのように、結果が同じになる事が分かるかと思います。\n\nしたがって、7箇所のスペースに4つの赤を置くケースを想定するので、\n\n7C4＝35\n\nよって、35通りとなります。"}],["$","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":[["$","div",null,{"className":"mb-1","children":"対数の計算"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"そもそも対数：logとは何でしょうか。\n　a^p＝M（a＞0, a≠1, M＞0）･･･①\nという等式が成り立つとき、\n　log a（M）＝p･･･②\nという等式が同時に成り立ちます。aを「底」、pを「指数」、Mを「真数」といい、log a（M）を「aを底とするMの対数」といいます。式②を見ればわかるように、log a（M）とは、「aを何乗したときMになるか」と言う値、すなわち、指数を表すものです。例えば、\n　log a（XY）＝log a（X）＋log a（Y）･･･③\nが成り立つのも、このとき\n　a^s＝XY\nと言う関係が常に存在し、X＝a^t、Y＝a^uとすると、\n　XY（＝a^s）＝X × Y\n　　　　　　　＝a^t × a^u\n　　　　　　　＝a^（t＋u）\nとなり、したがって、\n　s＝t＋u･･･④\nという関係を導くことができるからです。①と②から、XY、X、Yについても同様に、\n　log a（XY）＝s＝t＋u\n　log a（X）＝t\n　log a（Y）＝u\nと表せるので、結果として④は、\n　log a（XY）＝log a（X）＋log a（Ｙ）\nという式③になります。このように、logは、「対数」という名はあれど、その実「指数」のことを表しているのだということを頭に置いておくこと、つまり、①と②の対応関係を常に意識することが対数の理解の一助になるかもしれません。「logの数の大きい問題」というのがどんな問題を指すのかわからなかったので、ご期待に沿う回答ではないかもしれませんが、ご容赦ください。また、私の理解が誤っている場合は、これも申し訳ございません。"}],["$","div",null,{"className":"flex mb-1","children":[["$","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"}],["$","div",null,{"className":"text-xs","children":["北海道大学法学部"," ","たけなわ"]}]]}],["$","div",null,{"className":"flex justify-between","children":[["$","div",null,{"className":"flex","children":[["$","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 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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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