A.
Cr⁺¹ + Sn⁺⁴ ⇒ Cr⁺³ + Sn⁺²
Cr⁺¹ ⇒ Cr⁺³ + 2e⁻
The half-reaction of oxidation: chrome is an electron donor, losses 2 electrons ie it is oxidized
2e⁻ + Sn⁺⁴ ⇒ Sn⁺²
The half-reaction of reduction: tin is an electron acceptor, receives 2 electrons, ie it is reduced
b.
3Hg⁺² + 2Fe ⇒ 3Hg + 2Fe⁺³
2Fe ⇒ 2Fe⁺³ + 6e⁻
The half-reaction of oxidation: iron is an electron donor, losses 3 electrons ie it is oxidized
6e⁻ + 3Hg⁺² + ⇒ 3Hg
The half-reaction of reduction: mercury is an electron acceptor, receives 2 electrons, ie it is reduced
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c.
2As + 3Cl</span>₂ ⇒ 2AsCl₃
2As ⇒ 2As⁺³ + 6e⁻
The half-reaction of oxidation: arsenic is an electron donor, losses 3 electrons ie it is oxidized
6e⁻ + 3Cl₂ ⇒ 6Cl⁻
The half-reaction of reduction: chlorine is an electron acceptor, receives 1 electron, ie it is reduced
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d.
NaBr + Cl</span>₂ ⇒ NaCl + Br₂<span>
</span>2Br⁻ ⇒ Br₂ + 2e⁻
The half-reaction of oxidation: bromine is an electron donor, losses 1 electron ie it is oxidized
2e⁻ + Cl₂ ⇒ 2Cl⁻
The half-reaction of reduction: chlorine is an electron acceptor, receives 1 electron, ie it is reduced
e.
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
3C⁺² ⇒ 3C⁺⁴ + 6e⁻
The half-reaction of oxidation: carbon is an electron donor, losses 2 electrons ie it is oxidized
6e⁻ + 2Fe⁺³ ⇒ 2Fe
The half-reaction of reduction: iron is an electron acceptor, receives 3 electrons, ie it is reduced
Answer:oxygen is the central atom in water,it has two lone pairs of electrons, the bond angle is 104.27° The ideal H-O-H bond angle ought to have been 109.28°
Explanation:
The presence of lone pairs causes slight distortion in the bond angles of molecules. This distortion increases with the number of lone pairs present. Water has two lone pairs hence a greater distortion compared to ammonia.
Mg = 24.3 g/mol
23.5 g / 24.3 g/mol = 0.97 mol Mg
There are 2.2 moles in 35.7 g of CH4! To find this, you have to divide 35.7 by the molar mass of CH4, which is 16.0!