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3、安徽省皋城中学2025-2026学年九年级阶段性目标检测(四)地理试题


3、安徽省皋城中学2025-2026学年九年级阶段性目标检测(四)生物试题


四、语法填空
阅读下面短文,在空白处填入1个适当的单词或括号内单词的正确形式。
Rich traditions connect cultures, economies, emotions, and nature. Gary Sigley, a professor from Australia at Beijing Foreign Studies University, says his mother, 56 her 90s, still enjoys several cups of English-style black tea with milk and sugar each day.
A few years ago, he made her a cup of Pu’er tea from Yunnan Province, which 57 (mark) her first encounter with Chinese tea. “I placed a piece of sun-dried Pu’er on her palm, ” he recalls. “ 58 (see) her holding a cup of fresh tea grown on an ancient tree in Yunnan, I suddenly realized that tea is a medium of friendship between different 59 (nation) and peoples. ”
“Tea is a cultural treasure 60 (share) with the world by the Chinese people. Tea culture is not just about brewing (沏) or tasting tea — it’s a life philosophy. In a single cup of tea, we share not only flavor, but also knowledge, emotions and 61 (wise) , ” Sigley says.
Sigley 62 (recent) attended a cultural event to celebrate the upcoming International Tea Day, a global occasion 63 honors the rich traditions, cultural significance, and economic impact of tea.
Since 2008, Sigley has devoted 64 (he) to the landscapes and stories of Southwest China. Over 65 years, he has spent time with scholars, tea farmers and merchants. These experiences transformed him into what he calls a “tea traveler-scholar”.


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3、安徽省皋城中学2025-2026学年九年级阶段性目标检测(四)英语答案


5、安徽省皋城中学2025-2026学年九年级阶段性目标检测(四)历史答案


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6、安徽省皋城中学2025-2026学年九年级阶段性目标检测(四)英语答案


3、安徽省皋城中学2025-2026学年九年级阶段性目标检测(四)生物学试题


3、安徽省皋城中学2025-2026学年九年级阶段性目标检测(四)数学答案


500 ℃ 96.9% 97.8%
答案 D
解析 根据表格可知,在相同压强下,升高温度,A的转化率降低,说明升高温度平衡逆向移动,所以正反应为放热反应。该反应的正反应是一个气体体积减小的反应,在相同的温度下,p1→p2,A的转化率增大,说明化学平衡正向移动,即p1v正,所以平衡逆向移动,经过一段时间反应达到新平衡,C正确;增大压强,v正、v逆都增大,平衡正向移动,所以v正>v逆,最终达到新的平衡状态,D错误。
6.对于可逆反应mA(s)+nB(g)??eC(g)+fD(g),当其他条件不变时,C的体积分数[φ(C)]在不同温度(T)和不同压强(p)下随时间(t)的变化关系如图所示。下列叙述正确的是(  )
A.达到平衡后,若使用催化剂,C的体积分数将增大
B.该反应的ΔH<0
C.化学方程式中n>e+f


故选C。
9.B 【解析】A.①MnO2(s)+C(s)=MnO(s)+CO(g)ΔH1=+24.4 kJ·mol 1,②MnO(s)+CO(g)=MnO(s)+CO2(g)ΔH2= 148.1kJ·mol 1,可知,ΔH1>ΔH2,故A错误;
B.①MnO2(s)+C(s)=MnO(s)+CO(g)ΔH1=+24.4 kJ·mol 1,②MnO(s)+CO(g)=MnO(s)+CO2(g)ΔH2= 148.1kJ·mol 1,盖斯定律计算①+②得到③2MnO2(s)+C(s)=2MnO(s)+CO2(g)ΔH3=ΔH1+ΔH2,故B正确;
C.B的计算分析可知,ΔH3=ΔH1+ΔH2,故C错误;
D.①MnO2(s)+C(s)=MnO(s)+CO(g)ΔH1=+24.4kJ·mol 1,②MnO(s)+CO(g)=MnO(s)+CO2(g)ΔH2= 148.1kJ·mol 1,盖斯定律计算①+②得到③2MnO2(s)+C(s)=2MnO(s)+CO2(g)ΔH3=ΔH1+ΔH2=(+24.4kJ·mol 1)+( 148.1kJ·mol 1)= 123.6kJ·mol 1,反应③是放热反应,故D错误;
故选B。
10.C 【解析】①缩小体积,增大压强,浓度增大,反应速率加快;


3、安徽省皋城中学2025-2026学年九年级阶段性目标检测(四)化学答案