八下理化1-3

Quiz
•
Science
•
8th Grade
•
Hard
房繼誠教務處教師 房繼誠教務處教師
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15 questions
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1.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
已知 X、Y 和 Z 是三種不同的純物質,其分子量分別為 28、2、17。若 X 和 Y 反應可生成 Z,則下列何者可能為其均衡的化學反應式?[95. 基測Ⅱ]
英詳解
It can be known from the law of conservation of mass: 【】 28 × 1 + 【】 × 2 = 【】 × 2, which is consistent.
X+Y → 2Z
2X+Y → 2Z
X+3Y → 2Z
2X+3Y → 4Z
2.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
葡萄糖在無氧密閉環境中經由酵母菌發酵的反應式為:C6H12O6 2C2H5OH+2CO2,已知二氧化碳、葡萄糖的分子量分別為 44 與 180。現有 6×1024 個葡萄糖分子,經由酵母菌發酵後,最多約可產生 CO2 多少公克?
220 公克
230 公克
440 公克
880 公克
3.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
化學反應式的係數,可以代表參加反應的反應物及生成物之間的何種比例?
分子數比
原子數比
質量比
重量比
4.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
已知 H、O、S 的原子量分別為 1、16、32。取 20 公克的金屬 X 氧化物(XO),在適當條件下與足量的硫酸完全反應,理論上會產生 60 公克的 XSO4 和 m 公克的 H2O,反應式為:XO+H2SO4 → XSO4+H2O,此反應式的係數已平衡,m 值應為下列何者?〔103. 會考〕
英講解
The mole numbers of XO, XSO4, and H2O are the same.
,
x = 【】, m = 【】
9
18
24
40
5.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
二氧化碳生成的化學反應式如下:C+O2 → CO2,由此反應式可推知下列何種資訊?[97. 基測Ⅱ]
英詳解
Coefficient ratio is equal to mole ratio, C + O2 → CO2
Therefore, the mass ratio C: O2: CO2 = 1 × 12: 1 × 32: 1 × 【】 =【】: 8: 11 can be obtained.
化學反應的速率
各物質反應的濃度大小
各物質反應時的質量比
反應進行所需要的溫度
6.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
已知甲和乙二種物質反應會生成丙和丁,其反應式為:3 甲+乙→ 2 丙+2 丁。附表是甲和乙反應的一組實驗數據,若改取 24 g 的甲與 24 g 的乙進行上述反應,最多可以生成多少 g 的物質丙?
英詳解
It can be seen from the table that A reacts with 96 g, and B reacts with 28 g to produce 36 g; from the equation, it can be seen that the molecular weight of 甲 is 96 3 ÷ 32, the molecular weight of 乙 is 28 ÷ 1 = 28, and the molecular weight of propyl is 36 ÷ 2 = 18. Because the coefficient ratio is equal to the mole number ratio, the reaction of A and B with 24 g will show that the complete use of nail A will produce 0.5 Mole C, so the mass of C is 【】 × 18 =【】 g
9
18
44
88
7.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
乙醇在充足的氧氣下,燃燒產生水與二氧化碳。已知氫的原子量為 1,碳的原子量為 12,氧的原子量為 16。點燃盛有 100 公克乙醇的酒精燈,在充足的氧氣下燃燒,一段時間後,還餘有 54 公克的乙醇,此段時間燃燒所排放的二氧化碳應為多少公克?〔100. 基測Ⅱ〕
英詳解
C2H5OH + 3O2 → 2CO2 + 3H2O, C2H5OH Mole number = 1, CO2 Mole number = 【】 × 2 = 【】, CO2 mass = 2 ×【】 = 【】 (g).
44
46
88
92
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