Đốt cháy hoàn toàn 6,2 gam metylamin rồi cho sản phẩm chảy qua dung dịch Ca(OH)2 dư. Khối lượng bình đựng dung dịch Ca(OH)2 tăng là
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaamisamaaBaaameaacaaIZaaabeaaliaad6eacaWG % ibWaaSbaaWqaaiaaikdaaeqaaaWcbeaakiabg2da9iaaicdacaGGSa % GaaGOmaiabgkziUkaad6gadaWgaaWcbaGaam4qaiaad+eadaWgaaad % baGaaGOmaaqabaaaleqaaOGaeyypa0JaaGimaiaacYcacaaIYaGaai % 4oaiaad6gadaWgaaWcbaGaamisamaaBaaameaacaaIYaaabeaaliaa % d+eaaeqaaOGaeyypa0JaaGimaiaacYcacaaI1aaaaa!4FE1! {n_{C{H_3}N{H_2}}} = 0,2 \to {n_{C{O_2}}} = 0,2;{n_{{H_2}O}} = 0,5\)
m bình tăng = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH % RaWkcaWGTbWaaSbaaSqaaiaadIeadaWgaaadbaGaaGOmaaqabaWcca % WGpbaabeaakiabg2da9iaaigdacaaI3aGaaiilaiaaiIdacaWGNbGa % amyyaiaad2gacaGGUaaaaa!45B1! {m_{C{O_2}}} + {m_{{H_2}O}} = 17,8gam.\)
Đề thi thử tốt nghiệp THPT QG môn Hóa năm 2020
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