Answer:
True
Explanation: Imagine the Electrons is by the nucleus which give more energy.
Answer: SO₂ + H₂O → HSO₃ ⁻ + H⁺
Justification:
1) Ionization means formation of ions.
2) Ions are species that are not neutral, they are charged, in virtue of having less or more electrons than protons.
3) Ionization may happen in different environments.
4) Ionic compunds, like Mg(OH)₂ dissociate into ions (ionize) in water. That is the example shown in the fourth option:
Mg(OH)₂ → Mg ²⁺ + 2OH⁻
5) How much a ionic compound dissociates in water (ionize) depends on the Ksp (product solubility constant) which measures the concentrations of the ions that can be in the solution.
6) The Ksp for Mg(OH)₂ is very low, meaning that it will slightly ionize.
7) SO₂ + H₂O forms H₂SO₄, which is a strong acid, meaning that it will ionize fully in water, into the ions HSO₃ ⁻ and H⁺, so the third option is a good example of ionization.
Answer:
a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) 0.0035 mole
c) 0.166 M
Explanation:
Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.
The equation of the reaction is expressed as:

1 mole 3 mole
The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

10 ml 17.50 ml
(x) M 0.200 M
Molarity = 
= 0.0035 mole
c) What was the molar concentration of phosphoric acid in the original stock solution?
By stoichiometry, converting moles of NaOH to H₃PO₄; we have
= 
= 0.00166 mole of H₃PO₄
Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:
Molar Concentration = 
Molar Concentration = 
Molar Concentration = 0.166 M
∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M
Answer:
The amount of water converted from liquid to gas with 6,768 joules is approximately 3.035 g
Explanation:
The amount of heat required to convert a given amount of liquid to gas at its boiling point is known as the latent heat of evaporation of the liquid
The latent heat of evaporation of water, Δ
≈ 2,230 J/g
The relationship between the heat supplied, 'Q', and the amount of water in grams, 'm', evaporated is given as follows
Q = m × Δ
Therefore, the amount of water, 'm', converted from liquid to gas at the boiling point temperature (100°C), when Q = 6,768 Joules, is given as follows;
6,768 J = m × 2,230 J/g
∴ m = 6,768 J /(2,230 J/g) ≈ 3.035 g
The amount of water converted from liquid to gas with 6,768 joules = m ≈ 3.035 g.