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试题答案

本文从以下几个角度介绍。

    1、2024云师大附中高考适应性月考四答案
    2、2024云师大附中高考适应性月考四

1、云南师大附中2026届高考适应性月考卷(四)地理试题


5、云南师大附中2026届高考适应性月考卷(四)生物试题


I gladly accepted her new offer.
______
16．（2025?黄梅县校级模拟）阅读下面材料，根据其内容和所给段落开头语续写两段，使之构成一篇完整的短文。
It was during one holiday，a time when people usually reunite with loved ones or take vacations.But for me，this particular holiday became a turning point，one that would deeply influence how I view kindness and compassion（同情）.
It all began when I decided to surprise my best friend，who was studying in Leeds.I hadn't seen her in months，and we had been talking about catching up for weeks.It was already nearly 7 p.m.when I arrived at the city of Leeds.Excited but tired，I was looking forward to a pleasant evening with my friend.
However，my excitement quickly turned into disappointment when I was told that the last bus to her university had already left.Panic began to set in as I stood on the busy street unsure of what to do next.I couldn't even reach her on the phone it seemed that she had switched it off，probably to avoid interruptions for the evening.
I wandered aimlessly along the street.The feeling of loneliness started to weigh heavily on me.Suddenly，I noticed a motorbike following me，and I instinctively（本能地） felt a wave of unease.I quickened my pace，but the motorbike kept following.My anxiety intensified，and before I could think clearly，I rushed into the nearest store.


In summary, AI is set to integrate into every aspect of student life, enriching the educational experience and fostering a new era of personalised, efficient, and innovative learning.


2、云南师大附中2026届高考适应性月考卷(四)物理答案


6、云南师大附中2026届高考适应性月考卷(四)生物学答案


2、云南师大附中2026届高考适应性月考卷(四)生物学答案


1、云南师大附中2026届高考适应性月考卷(四)语文答案


1、云南师大附中2026届高考适应性月考卷(四)数学试题


2、云南师大附中2026届高考适应性月考卷(四) 答案


B. 和具有不同化学性质构成物质的分子不同，不同分子具有不同的化学性质
C. 液态水受热变成水蒸气，体积膨胀升高温度，分子体积变大
D. 红热的铁丝在空气中不易燃烧，在氧气中剧烈燃烧单位体积内氧分子的数目不同
9.借助模型可以更好地认识物质的微观构成。下列模型可以用来表示“”的是( )
A. B. C. D.
10.如图是表示气体微粒的示意图，图中黑球和白球分别表示相中不同的原子，那么其中表示纯净物的是( )
A. B.


A.Na2S4O6、H2O、Na2S2O3、Au2O、NaOH
B.Na2S2O3、NaOH、Na2S4O6、Au2O、H2O
C.Na2S2O3、H2O、Na2S4O6、Au2O、NaOH
D.当1 mol Au2O3完全反应时,转移电子8 mol
【答案】C
【解析】Au2O3是反应物,则Au2O一定是生成物,其中Au元素的化合价由+3价变成+1价,化合价降低,则必然有化合价升高的元素,即Na2S2O3(硫元素为+2价)是反应物,Na2S4O6(硫元素平均为+2.5价)是生成物。根据反应前后硫原子守恒可得如下关系:2Na2S2O3Na2S4O6,根据钠原子守恒,可知生成物中缺少钠原子,所以NaOH是生成物,再根据氢原子守恒,可知水是反应物。由关系式:Au2O3~Au2O~4e-,所以当1molAu2O3完全反应时,转移电子的物质的量为4mol。
10.高锰酸钾溶液在酸性条件下可以与硫酸亚铁反应,反应的化学方程式如下(未配平):


3、云南师大附中2026届高考适应性月考卷(四)化学答案