Lesson of generalizing repetition on the topic "Metals" - SUBSTANCES AND their PROPERTIES - LESSON plans FOR CHEMISTRY 11 class-lesson plans - lesson plans-author's lessons-plan-lesson summary - chemistry
The purpose of the lesson: to teach you to apply the knowledge of theory in solving exercises and calculation problems of the topic.
Equipment: two-option task cards for paired work; responses to task cards, if self-checking is provided.
Lesson progress
I. Organizational moment
Setting goals and tasks for the lesson. Monitoring the completion of homework. Organizing students to work based on task cards.
Recommendation to the teacher
During these lessons, you can self-check completed tasks. In this case, students should be offered answers to the task questions at the end of the lesson. This will give students the opportunity not only to perform tasks together, but also to compare their answers with the reference answer, закрепляяthus fixing the key questions of the topic and working on mistakes made.
II. monitoring the completion of homework
III. Conducting verification work
Contents of task cards
Option I |
Option II |
1. Determine the Maximum number of elements: |
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№ 3, № 23 |
№ 20, № 43 |
2. Why are the elements located in the same group of Mendeleev's PSC III? |
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# 16 and # 24 |
# 17 and # 25 |
3. How does metallicity change? |
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in periods |
in major subgroups |
What is the reason? |
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Arrange the elements in descending order of metallicity |
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№ 23, 26, 20, 30, 19 |
№ 49, 13, 81, 31 |
4. Is it possible to completely dissolve the alloy in dilute sulfuric acid? |
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Ni, Si, Al, Mn |
Cu, Fe, Zn, Cr |
Give a reasonable answer. Draw up equations of chemical reactions. |
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5. Draw up equations of reactions of obtaining copper metals |
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from copper sulfide (II) |
manganese from manganese oxide (IV) |
The equation to figure out how IAB. |
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6. Is it safe to iron the design, if it is reinforced with bolts from |
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a) the lead-acid batterie. b) manganese |
a) zinc b) Nickel |
Give a reasonable answer. |
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7. Solve the transformation scheme. Write the equations of reactions |
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8. The solution to computational problems. |
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7 g of iron filings were added to a solution of copper (II) chloride weighing 135.0 g with WCuCl2= 20%. Determine the mass of copper released as a result of the reaction. |
Determine the mass of the formed salt at the interaction of 5.4 g of aluminum with sulfuric acid weighing 490.0 g, WH2SO4= 10 % |
Answers to questions in task cards
Option I |
Option II |
They exhibit the same maximum CP +6, oxides have an acidic character, form strong acids: SO3— H2SO4; CGO3-H2CGO4 3. In periods with increasing nuclear charge, the radius of the atom decreases — metallicity weakens: K, CA, V, Fe, Zn. 4. Si is present in the composition of the alloy Si, it does not dissolve in sulfuric acid. Metals Ni, Al, Mn in the stress series are located up to hydrogen. Interact with dilute H2SO4and displace hydrogen.
6. The iron structure will not be protected by lead, because iron is more active than lead and will be destroyed first, in the case of bolts made of manganese-the manganese will first be destroyed. However, in any case, the stability of the structure will be violated. y = 0.25 mol CuCl2 is required, and given 0.2 mol / CuCl СuСl2is a disadvantage. The calculation is based on a lack. |
They exhibit the same maximum CP +7, oxides have an acidic character, form strong acids: SL2O7- Nclo4; MP2O7– Nmpo4 3. In groups, the main subgroups, the radius of the atom increases with increasing nuclear charge — metallicity increases: Those In, Ja, Al. 4. The alloy contains Cu, which does not react with dilute sulfuric acid. In the series of stresses is located behind the hydrogen. Metals Fe, Cr, Zn are located in the stress series up to hydrogen and displace hydrogen from the dilute H2SO4. 6. The iron structure will be protected by zinc, because zinc is more active than iron and will be destroyed first, in the case of Nickel, iron will first be destroyed, because it is more active than Nickel. But the stability of the structure will be violated in both cases. y = 0.667 mol Andl is required, and given 0.2 mol ? Al is a drawback. The calculation is based on a lack. |
IV. Homework assignment
Calculation issues § 18 № 28, 29, 30, 31, 32.
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