Answer:
Explanation:
(a) Firstly, caesium abd potassium are both in Group 1 of the periodic table. Group 1 metals (also called alkali metals) are the most reactive metals of the periodic table. Caesium is more reactive than Potassium because it has a higher electropositivity than Potassium. Electropositivity is the tendency of a metal to donate electron(s) to form a cation. Electropositivity increases down the group; this is because it is easier for atoms to loose electrons on the outermost shell that are far away from the central nucleus as against atoms whose outermost electrons are closer to the central nucleus. <u>Thus, the more "bulky" an atom is, the farther it's outermost electrons (valence electrons) get from the central nucleus and the easier it is to lose the outermost electron(s). And the easier it is for the valence electron(s) to be removed, the more reactive the atom would be and vice-versa.</u>
Caesium is more reactive than potassium because it is more bulky than potassium, with an atomic number of 55, while potassium has an atomic number of 19.
NOTE: The closer an electron is to the nucleus, the more difficult it is to be removed from it's shell.
(b) i. Formula for Caesium Nitrate:
Symbol for Caesium is Cs and Nitrate is NO₃⁻.
Cs⁺ + NO₃⁻ ↔ CsNO₃
Formula for Caesium Nitrate is CsNO₃
ii. Formula for Caesium sulphate
Symbol for caesium is Cs and Sulphate is SO₄²⁻
Cs⁺ + SO₄²⁻ ↔ Cs₂SO₄
Formula for Caesium sulphate is Cs₂SO₄
NOTE: When writing the formulae, the charges would be exchanged to form the subscript as seen on the product sides above.
The smallest particle of a covalently bonded compound is an atom.
The answers that are correct are a, b, and d
AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be a subject to electrolysis. Therefore, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. The most preferred reduction reaction will be Ag+ + e- = Ag (Emf=0.7996 V) which will occur at the cathode, on the other hand, the most favorable oxidation reaction will be
2H2O = O2 +4H+ + 4e- (Emf = -1.3 V) that will occur at the anode. Thus, the product at the anode is oxygen gas and at the cathode electrode is silver metal.
Answer & Explanation:
At high temperatures or in the presence of catalysts, sulfur dioxide reacts with hydrogen sulfide to form elemental sulfur and water. This reaction is exploited in the Claus process, an important industrial method to dispose of hydrogen sulfide.
