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英語 高校生

Reading Advantages3 です。 穴埋めが分からないので教えてください。

ble? a. president b. cleaner B. Complete the paragraph with items from the box. Two items are extra. actually commented expected made the headlines media neighboring potentially riddle significant spectacularly visible worship shapes (3) in the local (4) "This stone is for people who celebrate with fire." Archaeologists in England thought they had made an amazing discovery in July 2003, when tourists on a beach found ancient carvings on a large block of stone. The archaeologists believed that the discovery of the stone, which had been imported from Norway in the 1980s and used to make a wall, was (1). The carvings of two snakes, a dragon, and other Experts translated the stone to say, very (2) However, two months later, the archaeologists were surprised when the (5) of the carvings was solved by a fifty-year-old local builder, Barry Luxton. The man, who had seen a photograph in a newspaper, told them that he was (6) the one who had made the shapes - in 1995! Luxton said that over a period of three days he had made the carvings for a celebration on a (7) beach that was going to be held by a group of druids people who nature. However, the block did not end up being moved to the other beach and (8) was eventually covered by sand. Recent bad weather blew the sand away, making the carvings (9) again. Luxton was surprised; he really never (10) that his work would become so famous. Review 1 - 5 25

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化学 高校生

386Cまでは求められたのですがなぜ386C×2をしないのでしょうか? 私の考えだと386Cは銅の電気量で式で銅に×1/2しているので電子は2molなので×2しないといけないのかなと考えました。語彙力なくてすみません。その後900-386をしているので並列回路なので引くのは... 続きを読む

発展例題22 並列回路による電気分解 2421 硫酸銅(ⅡI)水溶液の入った電解槽Aと,希硫酸の入った電 解槽Bに, それぞれ白金電極を浸し, 図のように並列につ ないで500mAの電流を30分間流した。 このとき,電解槽 Aの陰極の質量が0.127g増加した。 電解槽Bの両極で発 生した気体は, 0℃, 1.013×105Paで何mL か。 ただし, LINE 電気分解によって発生する気体の水への溶解は無視してよ い。 考え方 並列回路では,電解槽AとB を流れた電気量の和が回路全 体を流れた電気量である。 電 解槽Aの陰極では銅が析出 する。 Cu2+ +2e- - Cu また,電解槽Bでは, 酸素と 水素が発生する。 2H2O → O2+4H + + 4e- 2H++2e¯ H2 172 1 2 解答 回路全体を流れた電気量は 0.500A×(60×30)s=900C・・・① 一方、電解槽Aの陰極では, 1mol の電子に相当する電気量 (9.65×10C) で 1/2molの銅が析出するので、流れた電気量を x [C] とすると, 63.5g/mol× -=0.127g x=386C・・・ ② 電解槽B では, 9.65×104C で 1/4molの酸素と1/2mol の水素 が発生し, 流れた電気量は①-②なので, R 22.4×103mL/mol× × 問題 305-306 電流計 可変抵抗 A q x [C] 9.65×10C/mol B H₂SO4 CENT POS <(1/1/+/)×1 900-3860. 9.65×104. mol = mol=89.5mL

解決済み 回答数: 1