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英語 高校生

速読のミニテストで、文章を全部読んでから問題を確認するのと、問題を読んでから文章を読んで答えを探すのはどっちがいいですか? 時間は2分半です!

Read the text and the graph and answer the three questions. (10 thousand) Number of food vending machines in Japan 10 2 8 2013 14 15 16 17 18 19 20 21 22 23 A vending machine is a machine that sells drinks, train tickets, or other things. In 2023 there were about 3.93 million vending machines in Japan. That number has been falling year by year. Japan had the most vending machines in 2013, with about 5 million. chines/sell 5 about 250,000 of these machines in 1985, but the number/fell/after Some vending machines/sell food such as bread and frozen foods. Japan had after that. However, in 2021 the number started to rise. In 2023, there were about 81,000 food vending machines. about One of the reasons for this rise is a change in people's lifestyles. Since the 10 coronavirus pandemic/people/have been eating at home more. Also, food vending 144 83 2 1 There were about before. vending machines in 2023 than there were 10 years (10点) 01 million fewer 23.93 million more 35 million more 250,000 fewer (①) 2 There were about | ①10,000 more 2 food vending machines in 2023 than in 2021. (10点) 281,000 more ③9,000 fewer 170,000 fewer 3 Many of today's food vending machines | have food products of different sizes and weights sell food that is easy for people to eat outside the home 3 are much larger than they used to be were developed after the coronavirus pandemic (10 ①

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化学 高校生

この問題でCH3COO-+H+=ch3coohとなっているのはなぜですか?可逆反応ではないのでしょうか?

日 000 H+ +酢酸ナトリウムの緩衝液 00ml= 濃度 アンモニ CH3COO- ごくわずか 混合水溶液中に含まれる各物質の濃度を求める。 CH3COO + H+→CHCOOH HCIより生じた H+ と CH3COOからCH COOH が生じる。 [CH3COO] = 0.05mol/L-0.02mol/L=0.03mol/L [CH3COOH] = 0.10mol/L+0.02mol/L=0.12mol/L 電離定数より、混合水溶液の [H+] およびpHを求める 1.8×10mol/L= 0.03mol/L × [H+] [H+] = 7.2 x 10mol/L 0.12mol/L pH = -log10 (7.2×10-)=-logio (2°×32 × 10 ) == - 3 × 0.30 + 2 × 0.48-6)=4.14≒4.1 (2) NaOH より生じた OH と CH3COOH から CH3COO が 生じる。混合水溶液中に含まれる各物質の濃度を求める。 CHCOOH+OH — CH,COO +HẠO [CH3COO-] = 0.05mol/L +0.05mol/L=0.10mol/L [CH3COOH] = 0.10mol/L-0.05mol/L=0.05mol/L 電離定数より,混合水溶液の[H+] およびpHを求める。 0.10mol/L× [H+] 18×10mol/L= [H+] = 9.0×10mol/L 0.05mol/L pH = -log10 ( 9.0×10)=-log10 (32×10) == 3 -(2×0.48-6) = 5.04≒5.0 (3)過剰に加えたNaOH と CH3COOH の中和反応後、余った NaOHの濃度を求める。 中和前 CH3COOH + 0.10mol/L CH3COONa+ け。 夏の OK,= [CH,COO][H] [CH COOH] と ②log102=0.30.log103=0.48 混合 合 ③log103 = 0.48 この後 H2O NaOH 0.15mol/L 0.05mol/L -0.10mol/L + 0.10mol/L +0.10mol/L 0.05mol/L 0.15mol/L 変化量 0.10mol/L 中和後 0mol/L [OH] = 0.05mol/L 水のイオン積K より 混合水溶液の [H*] およびpHを求め る。 [H+] x 0.05mol/L=1.0×10 -4 (mol/L) 2 [H] = 2.0×1013mol/L DH=logio ( 2.0×10-13) = (0.30-13)=12.7 この 酸の道 答えよ。 度が水の 及ぼす。 log102=0.30 酸化物イ 数式で ol/L)と

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