answer:To identify the salt, we need to first understand how each salt will affect the pH of the solution. 1. NaF: When dissolved in water, NaF will dissociate into Na+ and F- ions. The F- ion can react with water to form HF and OH- ions, causing an increase in pH (alkaline solution). F- + H2O -> HF + OH- 2. NaCl: When dissolved in water, NaCl will dissociate into Na+ and Cl- ions. Neither of these ions will react significantly with water, so the pH of the solution will be neutral, around 7. 3. NaOCl: When dissolved in water, NaOCl will dissociate into Na+ and OCl- ions. The OCl- ion can react with water to form HOCl and OH- ions, causing an increase in pH (alkaline solution). OCl- + H2O -> HOCl + OH- Now, we need to determine the pH of the solution when 0.050 mol of each salt is dissolved in 0.500 L of water. For both NaF and NaOCl, we can use the Kb (base-dissociation constant) expression to determine the alkalinity of the solution, and therefore, the pH. Kb = [OH-][HF]/[F-] for NaF Kb = [OH-][HOCl]/[OCl-] for NaOCl First, we need to find the Kb values for F- and OCl- ions. For F-, the Ka (acid-dissociation constant) of HF is known (Ka = 6.76 × 10^(-4)). We can find the Kb using the relationship: Kw = Ka × Kb, where Kw is the ion product of water (1.0 × 10^(-14)). Kb for F- = Kw/Ka = (1.0 × 10^(-14))/(6.76 × 10^(-4)) = 1.48 × 10^(-11) For OCl-, the Ka of HOCl is known (Ka = 3.5 × 10^(-8)). We can find the Kb using the same relationship. Kb for OCl- = Kw/Ka = (1.0 × 10^(-14))/(3.5 × 10^(-8)) = 2.86 × 10^(-7) Next, we can use the Kb expression and the given molar concentrations of the salts to find the OH- concentration, and ultimately, the pH of the solution. For NaF: 0.050 mol of NaF in 0.500 L of water gives a [F-] = 0.1 M. Let x be the [OH-] produced. Then Kb = 1.48 × 10^(-11) = (x)(x)/(0.1 - x). If we assume x is small, we can approximate 0.1 - x ≈ 0.1. 1.48 × 10^(-11) = x^2/0.1. Solving for x yields [OH-] ≈ 1.29 × 10^(-6) M. pOH = -log10(1.29 × 10^(-6)) ≈ 5.89. Since pH + pOH = 14, pH ≈ 8.11. For NaOCl: 0.050 mol of NaOCl in 0.500 L of water gives a [OCl-] = 0.1 M. Let x be the [OH-] produced. Then Kb = 2.86 × 10^(-7) = (x)(x)/(0.1 - x). If we assume x is small, we can approximate 0.1 - x ≈ 0.1. 2.86 × 10^(-7) = x^2/0.1. Solving for x yields [OH-] ≈ 5.35 × 10^(-4) M. pOH = -log10(5.35 × 10^(-4)) ≈ 3.27. Since pH + pOH = 14, pH ≈ 10.73. The experimental pH value given is 8.08, which is closest to the calculated pH value for NaF at 8.11. Therefore, the identity of the unknown salt is NaF.

answer:The unknown salt can be either NaF, NaCl, or NaOCl. When 0.050 mol of salt is dissolved in 0.500 L of water, the pH is 8.08. 1. NaF forms F- ions, causing an increase in pH (alkaline solution). 2. NaCl doesn't affect pH significantly, so the pH remains neutral (~7). 3. NaOCl forms OCl- ions, causing an increase in pH (alkaline solution). Using base-dissociation constants, we find the pH for each salt: 1. pH for NaF ≈ 8.11 2. pH for NaOCl ≈ 10.73 Comparing pH values, NaF is closest to the experimental pH of 8.08. Therefore, the unknown salt is NaF.

answer:1) Реакция aминоуксусной кислоты с KOH: CH3COOH + KOH -> CH3COOK + H2O В результате реакции образуется соль калия аминоуксусной кислоты и вода. 2) Реакция aминоуксусной кислоты с HCl: CH3COOH + HCl -> CH3COOHCl + H2O В результате реакции образуется соль аминоуксусной кислоты и воды. 3) Реакция aминоуксусной кислоты с CH3OH: CH3COOH + CH3OH -> CH3COOCH3 + H2O В результате реакции образуется метиловый эфир аминоуксусной кислоты и вода.

answer:Массовая доля азота в аминоуксусной кислоте H2N-CH2-COOH равна: молярная масса азота = 14 г/моль молярная масса аминоуксусной кислоты = 75 г/моль Массовая доля азота = (масса азота / масса аминоуксусной кислоты) x 100% Масса азота = (2 x 14 г/моль) = 28 г/моль Массовая доля азота = (28 г/моль / 75 г/моль) x 100% = 37.3% Формула гомолога аминоуксусной кислоты: H2N-CH2-(CH2)n-COOH, где n - целое число, обозначающее количество метиловых групп в углеводородной цепи.