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Konuşmacı, futbol penaltı atışı örneği üzerinden beynin hareket kontrol mekanizmasını açıklamaktadır.","Video, bazal ganglionun yapısını ve işlevlerini anlatarak başlıyor, ardından indirekt yolun detaylı açıklamasına geçiyor. Bazal ganglionun beyindeki konumu, striatum, globus pallidus, substansiya nigra ve subtalamik çekirdek gibi yapıları gösteriliyor. Ayrıca talamus'un motor korteksle ilişkisi, kasların kontrol edilmesi ve duruma göre hareket seviyesinin ayarlanması gibi konular örneklerle açıklanıyor. Video, beynin hareket kontrol mekanizmasını anlamak isteyenler için faydalı bir kaynak niteliğindedir."]},"endTime":488,"title":"Bazal Ganglion ve İndirekt Yol Hakkında Eğitim Videosu","beginTime":0}],"fullResult":[{"index":0,"title":"Bazal Ganglion ve Hareket Kontrolü","list":{"type":"unordered","items":["Bir penaltı atışı gibi önemli bir anda, vücudumuzun ve kaslarımızın muntazam çalışması gerekir, aksi takdirde istenen hareketi yapamayız.","Bazal ganglion, nöronlardan oluşan bir çekirdek grubu olarak tanımlanır ve yapmak istediğimiz hareketleri yapabilmemiz için birlikte çalışır.","Bazal ganglion, yapmak istediğimiz hareketlerin gerçekleşmesini sağlayan \"direkt yol\" ve kaslarımızın hareket etmemesini sağlayan \"indirekt yol\" adı verilen iki farklı sistemden 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fazlası için: http://www.khanacademy.org.tr Matematikten sanat tarihine, ekonomiden fen bilimlerine, basit toplamadan diferansiyel denklemlere, ilkokul seviyesinden üniversite seviyesine...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/402860/07227558d7498d9d6f98972302200bb7/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/TmJJIAAAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"2","reqid":"1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=videoid:9019452702685266679","teaser":[{"list":{"type":"unordered","items":["Bu video, sembolik etkileşimcilik teorisini açıklayan eğitici bir içeriktir. George Herbert Mead ve Herbert Bloom'un teorileri üzerinden konu anlatılmaktadır.","Video, sembolik etkileşimcilik teorisinin temel prensiplerini açıklamaktadır. Bu teori, toplumun küçük ölçekli etkileşimlerine odaklanarak, bireylerin nesnelere ve olaylara yükledikleri anlamların nasıl değiştiğini incelemektedir. Bloom'un üç temel ilkesi (anlam çerçevesinde hareket etmek, sosyal etkileşimlerle anlam yüklemek ve anlamın değişebilir olması) örneklerle açıklanmakta ve teorinin sosyolojiye katkıları ele alınmaktadır."]},"endTime":254,"title":"Sembolik Etkileşimcilik Teorisi","beginTime":0}],"fullResult":[{"index":0,"title":"Sembolik Etkileşimcilik Nedir?","list":{"type":"unordered","items":["Sembolik etkileşimcilik, toplumun küçük ölçekli bir görüntüsünü alarak bireyler arasındaki etkileşimlere odaklanır.","Bu teori, küçük ölçeğe bakarak toplum içindeki bireyi ve birbirleriyle etkileşimlerini açıklar ve toplumsal düzen ve değişimi açıklayabilir.","Sembolik etkileşimcilik, 20. yüzyılın başlarında George Herbert Mead'in öğretilerinden derlenerek oluşturulmuştur."]},"beginTime":2,"endTime":39,"href":"/video/preview/9019452702685266679?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=2&ask_summarization=1"},{"index":1,"title":"Mead'in 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fazlası için: http://www.khanacademy.org.tr Matematikten sanat tarihine, ekonomiden fen bilimlerine, basit toplamadan diferansiyel denklemlere, ilkokul seviyesinden üniversite seviyesine her...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/1780926/57625dc1676e44ca1fea8156aa8be1fa/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/pCrZOAAAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"6","reqid":"1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=videoid:7341504805365130971","teaser":[{"list":{"type":"unordered","items":["Bu video, bir eğitmen tarafından sunulan matematik dersidir. Eğitmen, vektörlerin pozitif x ekseni ile oluşturduğu açıyı bulma konusunu anlatmaktadır.","Videoda, standart formda çizilmiş vektörlerin pozitif x ekseni ile oluşturduğu açının nasıl hesaplanacağı adım adım gösterilmektedir. Eğitmen önce birim vektör gösteriminde verilen bir vektörün bileşenlerini kullanarak açıyı bulma yöntemini açıklar, ardından hesap makinesinin verdiği sonuçların hangi çeyrekte olduğu dikkate alınarak düzeltilmesi gerektiğini vurgular. Video, iki farklı örnek üzerinden konuyu pekiştirir ve izleyicilere bu tür sorularda tam olarak ne istendiğinden emin olmaları gerektiğini hatırlatır."]},"endTime":395,"title":"Vektörlerin Pozitif X Ekseniyle Oluşturduğu Açıyı Bulma","beginTime":0}],"fullResult":[{"index":0,"title":"Vektörlerin Açılarını Bulma","list":{"type":"unordered","items":["Standart formda (orijinde başlangıç noktası olan) bir vektör ile pozitif x ekseni arasındaki açıyı bulma konusunda alıştırma yapılacak.","A vektörü birim vektör gösteriminde -2i - 4j şeklinde veya x ve y bileşenleri olarak (-2, -4) şeklinde yazılabilir.","Standart formda çizilmiş vektörler için açının tanjantı y bileşeni bölü x bileşenidir."]},"beginTime":0,"endTime":88,"href":"/video/preview/7341504805365130971?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=0&ask_summarization=1"},{"index":1,"title":"İlk Örnek Hesaplama","list":{"type":"unordered","items":["Tanjant theta = -4/(-2) = 2 olarak hesaplanır.","Tanjant 2'nin ark tanjantı (tan⁻¹(2)) yaklaşık 63,40 derecedir.","Hesap makinesi açının -90 ile +90 derece arasında olduğunu varsayar, ancak bu vektör üçüncü çeyrekte olduğundan açı 180 dereceden büyük ve 270 dereceden küçüktür."]},"beginTime":88,"endTime":222,"href":"/video/preview/7341504805365130971?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=88&ask_summarization=1"},{"index":2,"title":"Doğru Açı Hesaplama","list":{"type":"unordered","items":["Vektörün açısı 63,40 + 180 = 243,40 derece olarak hesaplanır.","İkinci örnekte dördüncü çeyrekteki bir vektörün pozitif x ekseniyle oluşturduğu açının bulunması isteniyor.","Bu vektörün bileşenleri (-6, 4) olduğundan tanjant theta = -6/4 = -1,5 olarak hesaplanır."]},"beginTime":222,"endTime":288,"href":"/video/preview/7341504805365130971?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=222&ask_summarization=1"},{"index":3,"title":"İkinci Örnek Hesaplama","list":{"type":"unordered","items":["Tanjant -1,5'in ark tanjantı yaklaşık -56,30 derecedir.","Hesap makinesi negatif açı verir çünkü pozitif x ekseninde saat yönünde ilerlersek elde edilen açı negatif olur.","Pozitif açı bulmak için -56,30 + 360 = 303,70 derece olarak 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Konuşmacı, titrasyon eğrilerini detaylıca öğrendikten sonra asit-baz indikatörlerini kullanarak titrasyonun eşdeğerlik noktasını nasıl bulacağımızı açıklamaktadır.","Video, çeşitli asit-baz çiftlerinin titrasyon eğrilerini inceleyerek hangi indikatörlerin hangi durumlarda uygun olduğunu göstermektedir. Kuvvetli asit-kuvvetli baz, zayıf asit-kuvvetli baz ve zayıf baz-kuvvetli asit titrasyonları için metil kırmızısı, bromtimol mavisi ve fenolftalein gibi indikatörlerin hangi pH aralıklarında renk değiştirdiği ve bu değişimlerin eşdeğerlik noktasını belirlemedeki önemi anlatılmaktadır. Her bir titrasyon türü için uygun indikatör seçimi ve neden bu indikatörlerin tercih edildiği detaylı olarak açıklanmaktadır."]},"endTime":531,"title":"Asit-Baz Titrasyonlarında İndikatör Seçimi","beginTime":0}],"fullResult":[{"index":0,"title":"Asit-Baz İndikatörleri ve Renk Değişimleri","list":{"type":"unordered","items":["İndikatörler belirli pH aralıklarında renk değiştirmektedir.","Metil kırmızısı 4,40-6,20 pH aralığında (yaklaşık 4-6) kırmızıdan sarıya döner.","Bromtimol mavisi 6-7,60 pH aralığında (yaklaşık 6-8) sarıdan maviye döner.","Fenolftalein 8,20-10 pH aralığında (yaklaşık 8-10) renksizden pembe/mora döner."]},"beginTime":1,"endTime":71,"href":"/video/preview/10356845823538992556?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=1&ask_summarization=1"},{"index":1,"title":"Kuvvetli Asit-Kuvvetli Baz Titrasyonu","list":{"type":"unordered","items":["Kuvvetli asit-kuvvetli baz titrasyonunda eşdeğerlik noktasındaki pH 7'dir.","Bu titrasyon için bromtimol mavisi (6-8 pH aralığı) uygun bir indikatördür.","Dik bir titrasyon eğrisi olduğundan, metil kırmızısı (4-6 pH) ve fenolftalein (8-10 pH) de kullanılabilir."]},"beginTime":71,"endTime":259,"href":"/video/preview/10356845823538992556?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=71&ask_summarization=1"},{"index":2,"title":"Zayıf Asit-Kuvvetli Baz Titrasyonu","list":{"type":"unordered","items":["Zayıf asit-kuvvetli baz titrasyonunda eşdeğerlik noktasındaki pH 7'den büyüktür (örnekte 9'a yakın).","Bu titrasyon için fenolftalein (8-10 pH aralığı) uygun bir indikatördür.","Metil kırmızısı (4-6 pH aralığı) bu titrasyon için uygun değildir çünkü renk değişimi eşdeğerlik noktasını 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Videoda Napolyon, Prusya Kralı Frederick Wilhelm III, Rus Çarı Alexander I ve diğer tarihi karakterler ele alınmaktadır.","Video, üçüncü koalisyon savaşı sonrası Napolyon'un durumunu tekrar ederek başlayıp, 1806-1807 yılları arasında gerçekleşen dördüncü koalisyon savaşı hakkında detaylı bilgi vermektedir. Jena-Auerstedt, Austerlitz ve Friedland muharebeleri, Tilsit Antlaşması ve sonrasında Napolyon'un Avrupa'daki siyasi ve askeri gücünün artışı anlatılmaktadır.","Videoda ayrıca Napolyon'un kardeşini Vestfalya Krallığı'nın başına getirmesi, kıta ablukası sisteminin oluşturulması ve Rusya ile olan ilişkiler de detaylandırılmaktadır. Harita gösterimleri kullanılarak savaşların coğrafi konumları açıklanmakta ve tarihçilerin dördüncü koalisondan sonra Napolyon'un durumuna dair görüşleri paylaşılmaktadır."]},"endTime":891,"title":"Napolyon Dönemi Dördüncü Koalisyon Savaşı","beginTime":0}],"fullResult":[{"index":0,"title":"Napolyon'un Gücü ve Kutsal Roma-Cermen İmparatorluğu'nun Sonu","list":{"type":"unordered","items":["Napolyon 1804 yılında kendisini imparator ilan etmiş, 1805 yılında gerçekleşen Austerlitz Muharebesi'nde Rusya ve Avusturya'nın çoğunluğunu oluşturduğu üçüncü koalisyonu devirmiş.","Austerlitz zaferi sonucunda Kutsal Roma-Cermen İmparatorluğu'nun da sonu gelmiş, bu devlet sadece Almanca konuşan federal hanedanlıkların oluşturduğu bir topluluktu.","Napolyon, Austerlitz zaferini kutlamak için Paris'te \"Zafer Takı\" adında bir anıt inşa ettirmiş, bu anıt şu an Fransa için ölmüş olan bütün askerleri temsil ediyor."]},"beginTime":1,"endTime":89,"href":"/video/preview/4078124855193924428?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=1&ask_summarization=1"},{"index":1,"title":"Prusya'nın Savaş Açması","list":{"type":"unordered","items":["Napolyon için işler iyi gidiyorken, diğer Avrupa devletleri kendilerini tehdit altında hissediyor ve özellikle Prusya Kralı III. 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tarafından bozguna uğrayarak yoldan çekiliyor.","Napolyon, Prusya'yı yenikten sonra Rusya'yı Polonya sınırlarına dahil olan topraklarda takip ediyor."]},"beginTime":308,"endTime":436,"href":"/video/preview/4078124855193924428?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=308&ask_summarization=1"},{"index":3,"title":"Friedland Savaşı ve Tilsit Antlaşması","list":{"type":"unordered","items":["Auersted'de kanlı bir çıkmaz gerçekleşiyor ve her iki tarafta da 15 ile 25 arasında ölü sayısı var, Napolyon savaşı kazanıyor ancak Rusya karşısında aldıkları galibiyet o kadar da büyük değil.","1807 Haziran ayında Friedland Savaşı gerçekleşiyor ve Napolyon dördüncü koalisyonu devirmeyi başarıyor.","Temmuz ayında Napolyon, Rusya ile ayrı, Prusya ile ayrı antlaşmalar imzalıyor; Rusya ile yapılan antlaşma daha saygıdeğerken, Prusya parçalara 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girmiştir."]},"beginTime":599,"endTime":664,"href":"/video/preview/4078124855193924428?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=599&ask_summarization=1"},{"index":5,"title":"Napolyon'un Britanya ile Mücadelesi","list":{"type":"unordered","items":["Endüstri devrimi başlamış ve Napolyon Britanya'yı okyanuslarda yenemeyeceğini anlayarak ekonomik bir savaş başlatmıştır.","1806 yılının Kasım ayında Napolyon, Prusyalıları Jena-Auerstedt'te yenmesinden hemen sonra \"Kıta Ablukası\" adı verilen bir sistem yaratmıştır.","Bu sisteme göre Fransız İmparatorluğu'nun parçasıysanız veya onlara bağlıysanız Büyük Britanya'ya ambargo uygulamanız gerekiyordu."]},"beginTime":664,"endTime":726,"href":"/video/preview/4078124855193924428?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=664&ask_summarization=1"},{"index":6,"title":"Rusya ve Diğer Güçlerle 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fazlası için: http://www.khanacademy.org.tr Matematikten sanat tarihine, ekonomiden fen bilimlerine, basit toplamadan diferansiyel denklemlere, ilkokul seviyesinden üniversite seviyesine...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/4078022/176af7e4f2dbe666f49e875633114cf1/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/DxbiGwAAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"17","reqid":"1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=videoid:5932028472907402418","teaser":[{"list":{"type":"unordered","items":["Bu video, bir eğitmen tarafından sunulan matematik dersidir. Eğitmen, iki boyutlu vektörlerde toplama ve çıkarma işlemlerini anlatmaktadır.","Video, iki boyutlu vektörlerin toplama ve çıkarma işlemlerini hem matematiksel hem de görsel olarak açıklamaktadır. Önce A ve B vektörlerinin bileşenleri (A: x=3, y=-1; B: x=2, y=3) kullanılarak toplama ve çıkarma işlemlerinin nasıl yapılacağı gösterilir, ardından bu işlemler grafik üzerinde vektörlerin yer değiştirilmesi yöntemiyle görsel olarak anlatılır. Eğitmen, toplama işleminde vektörlerin bileşenlerinin toplanması, çıkarma işleminde ise bileşenlerinin birbirinden çıkarılması gerektiğini vurgulamaktadır."]},"endTime":342,"title":"İki Boyutlu Vektörlerde Toplama ve Çıkarma Dersi","beginTime":0}],"fullResult":[{"index":0,"title":"İki Boyutlu Vektörlerde Toplama İşlemi","list":{"type":"unordered","items":["A vektörünün x bileşeni 3, y bileşeni -1; B vektörünün x bileşeni 2, y bileşeni 3 olarak tanımlanmıştır.","Vektörleri toplarken, toplam vektörünün x bileşeni A'nın x bileşeni ile B'nin x bileşeninin toplamı, y bileşeni de A'nın y bileşeni ile B'nin y bileşeninin toplamıdır.","A+B vektörünün x bileşeni 3+2=5, y bileşeni -1+3=2 olarak hesaplanmıştır."]},"beginTime":0,"endTime":51,"href":"/video/preview/5932028472907402418?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=0&ask_summarization=1"},{"index":1,"title":"İki Boyutlu Vektörlerde Çıkarma 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Eğitmen, öğrencilere bir vektör problemi çözerek adım adım anlatmaktadır.","Videoda, güçlü bir mıknatısın düzgün bir yüzey üzerinde bulunan metal topu çekme kuvveti problemi ele alınmaktadır. Eğitmen önce problemi okuyup, ardından öğrencilere kendi başlarına çözmelerini önererek başlar. Ardından, verilen vektör seçeneklerini inceleyerek doğru cevabı bulma sürecini gösterir. Video, vektörlerin sadece büyüklüğü ve yönüyle tanımlandığını vurgulayarak sona erer."]},"endTime":167,"title":"Fizik Dersinde Vektör Problemi Çözümü","beginTime":0}],"fullResult":[{"index":0,"title":"Mıknatıs ve Top Problemi","list":{"type":"unordered","items":["Bir mıknatıs, düzgün bir yüzey üzerinde bulunan metal topu 15 Newtonluk bir kuvvetle çekmektedir.","Mıknatıs, topa göre doğu yönünün 20 derece güneyinde bulunmaktadır.","Soruda A ve C vektörlerinin büyüklüğünün 15 Newton, B ve D vektörlerinin büyüklüğünün 20 Newton olduğu belirtilmiştir."]},"beginTime":1,"endTime":49,"href":"/video/preview/8980014790209043999?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=1&ask_summarization=1"},{"index":1,"title":"Vektörlerin İncelenmesi","list":{"type":"unordered","items":["Mıknatısın yönü doğu yönünün 20 derece güneyinde, çekme kuvvetinin yönü de bu yönde olmalıdır.","A seçeneği doğru görünmektedir çünkü yönü doğru ve 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Anlatıcı, gümüş klorürün molar çözünürlüğünü hesaplama yöntemini adım adım göstermektedir.","Video, öncelikle gümüş klorürün saf su içindeki molar çözünürlüğünü bulmak için çözünürlük dengesi ve denge bağıntısı oluşturma sürecini açıklar, ardından amonyak çözeltisi içindeki çözünürlüğünün nasıl arttığını Louis asit-baz tepkimi ve kompleks iyon oluşumu kavramları üzerinden anlatır. İki tepkime denkleminin net tepkime denklemine dönüştürülmesi, denge sabitinin hesaplanması ve son olarak molar çözünürlüğün bulunması gösterilmektedir.","Videoda, gümüş klorürün suda az çözünürken amonyak çözeltisinde daha fazla çözünebildiği ve bu durumun çözünürlük çarpımı sabiti üzerindeki etkisi vurgulanmaktadır."]},"endTime":815,"title":"Gümüş Klorürün Çözünürlüğü ve Amonyak Çözeltisi Etkisi","beginTime":0}],"fullResult":[{"index":0,"title":"Gümüş Klorürün Suydaki Çözünürlüğü","list":{"type":"unordered","items":["Gümüş klorür az çözünür bir bileşik olup, saf su içinde çözünür ve doymuş bir çözelti oluşturur.","Gümüş klorürün molar çözünürlüğünü bulmak için çözünürlük dengesi tablosu oluşturulur.","Çözünür her bir gümüş klorür molü için bir mol gümüş +1 katyonu ve bir mol klorür anyonu 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tepkimeye girerek kompleks iyon oluşturur.","Bu tepkime için denge sabiti (Kf) 1,6 × 10⁷ olup, bu yüksek değer dengenin sağa doğru kaydığını gösterir.","Amonyak çözeltisinde gümüş katyonlarının derişimi azaldığı için Le Chatelier ilkesine göre gümüş klorürün çözünürlüğü artar."]},"beginTime":210,"endTime":352,"href":"/video/preview/2077987883459525223?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=210&ask_summarization=1"},{"index":3,"title":"Sonuç ve Hesaplama","list":{"type":"unordered","items":["Suya amonyak çözeltisi eklendiğinde daha fazla gümüş klorür çözebilir çünkü kompleks iyonun oluşmasıyla gümüş klorürün çözünürlüğü artar.","Gümüş klorürün 3 molar amonyak çözeltisi içindeki çözünürlüğünü bulmak için iki ayrı tepkimeyi birbirine eklemek gerekiyor."]},"beginTime":352,"endTime":397,"href":"/video/preview/2077987883459525223?parent-reqid=1773724240062469-2950183374082675987-balancer-l7leveler-kubr-yp-sas-161-BAL&text=KnowledgeForAll&t=352&ask_summarization=1"},{"index":4,"title":"Gümüş Klorürü ve Amonyak Çözeltisi Arasındaki Tepkime","list":{"type":"unordered","items":["İki tepkime toplanarak gümüş klorürü amonyak çözeltisine konulduğunda neler olduğunu anlayabilmek için net tepkime elde edilmiştir.","Reaktan ve ürün tarafında gümüş katyonları olduğu için denklemde çıkarılarak, reaktant olarak katı haldeki gümüş klorür ve amonyak, ürün olarak kompleks iyon ve klorür anyonu elde edilmiştir.","Net tepkimenin denge sabiti (K), iki tepkimenin denge sabitlerinin çarpımıdır ve değeri 2,9 × 10⁻³ olarak 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