<span>50.2 kJ = 333 kJ/kg * mass of water
mass of water is 0.15075075075075075075075075075075 kg
therefore mass of unfrozen water is 0.10924924924924924924924924924925 kg</span>
Answer:
HClO₃ /chloric acid /suffix -ic/ ClO₃⁻ (chlorate)
HClO₂/ chlorous acid/ suffix -ous/ ClO₂⁻ (chlorite)
HNO₃ /nitric acid /suffix -ic/ NO₃⁻ (nitrate)
HNO₂/ nitrous acid/ suffix -ous/ NO₂⁻ (nitrite)
Explanation:
Chlorine has 4 positive oxidation numbers to form oxyacids: +1, +3, +5 and +7.
- When it uses the oxidation number +5, it forms HClO₃, which is named chloric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion ClO₃⁻ (chlorate).
- When it uses the oxidation number +3, it forms HClO₂, which is named chlorous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion ClO₂⁻ (chlorite).
Nitrogen has 2 positive oxidation numbers to form oxyacids: +3 and +5.
- When it uses the oxidation number +5, it forms HNO₃, which is named nitric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion NO₃⁻ (nitrate).
- When it uses the oxidation number +3, it forms HNO₂, which is named nitrous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion NO₂⁻ (nitrite).
Answer: the pH of the solution is 4.52
Explanation:
Consider the weak acid as Ha, it is dissociated as expressed below
HA H⁺ + A⁻
the Henderson -Haselbach equation can be expressed as;
pH = pKa + log( [A⁻] / [HA])
the weak acid is dissociated into H⁺ and A⁻ ions in the solution.
now the conjugate base of the weak acid HA is
HA(aq) {weak acid} H⁺(aq) + A⁻(aq) {conjugate base}
so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;
pKₐ = -logKₐ
pKₐ = -log ( 7.4×10⁻⁵ )
pKₐ = 4.13
now thw pH is
pH = pKₐ + log( [A⁻] / [HA])
pH = 4.13 + log( [0.540] / [0.220])
pH = 4.13 + 0.3899
pH = 4.5199 = 4.52
Therefore the pH of the solution is 4.52
Water is formed when this happens
Answer:
an electron in the outer energy level of an atom
