Use the periodic table to help you answer the following question - Edexcel - GCSE Chemistry - Question 3 - 2017 - Paper 1
Question 3
Use the periodic table to help you answer the following question.
Which of these metals will form coloured cations?
Put a cross (✗) in the box next to your answer.
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Worked Solution & Example Answer:Use the periodic table to help you answer the following question - Edexcel - GCSE Chemistry - Question 3 - 2017 - Paper 1
Step 1
Which of these metals will form coloured cations?
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Answer
The metals that will form coloured cations are generally transition metals. Therefore, based on the given options, the correct answer is to place a cross (✗) next to B chromium, Cr.
Step 2
Explain, in terms of the structure of their atoms, why caesium is more reactive than lithium.
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Answer
Caesium is more reactive than lithium because it is located lower in group 1 of the periodic table. As we move down the group, the size of the atoms increases due to the addition of electron shells. The outer electron in caesium is further from the nucleus than the outer electron in lithium, experiencing less electrostatic attraction to the positively charged nucleus. This makes it easier for caesium to lose its outer electron, resulting in higher reactivity.
Step 3
Write the balanced equation for this reaction.
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Answer
The balanced equation for the reaction of potassium with water is:
ightarrow 2KOH + H_2$$
Step 4
Which of these is the balanced equation for this reaction?
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Answer
The correct balanced equation for the reaction when chlorine is bubbled through potassium bromide solution is:
B:
ightarrow KCl + Br_2$$
Place a cross (✗) next to option B.
Step 5
Explain why filament light bulbs were filled with argon rather than left with air inside them.
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Answer
Filament light bulbs were filled with argon instead of air to prevent oxidation of the filament. In an air-filled bulb, oxygen can react with the high-temperature filament, leading to its rapid degradation. Argon is an inert gas that does not react under these conditions, which helps prolong the life of the filament and maintain the efficiency of the bulb.
