Answer:
2Fe⁺³ + Sn₂ → 2Fe⁺² + 2Sn⁺²
Explanation:
A redox reaction occurs between a compound that loses electrons and others that gain an electron. The first is being oxidized, and the other is being reduced.
In this situation, in the compound Fe₂O₃, the iron, has an oxidation number equal to +3, so it's Fe⁺³, and it will gain 1 electron to become Fe⁺². Because it was first dissolved in HCl, we must use the ion at the equation. The other compound Sn₂ will be oxidized to Sn⁺², so it will need to lose 2 electrons.
So, it will be necessary 2 Fe⁺³ for this reaction happen:
2Fe⁺³ + Sn₂ → 2Fe⁺² + 2Sn⁺²
Answer;
-(2) An atom is mostly empty space.
Experiment
-Rutherford conducted the "gold foil" experiment where he shot alpha particles at a thin sheet of gold. The conclusion that can be drawn from these experiment is that an atom is mostly empty space.
-Rutherford found that a small percentage of the particles were deflected, while a majority passed through the sheet. This caused Rutherford to conclude that the mass of an atom was concentrated at its center, as the tiny, dense nucleus was causing the deflections.
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
