Answer:
pKb = 10.96
Explanation:
Tartaric acid is a dyprotic acid. It reacts to water like this:
H₂Tart + H₂O ⇄ H₃O⁺ + HTart⁻ Ka1
HTart⁻ + H₂O ⇄ H₃O⁺ + Tart⁻² Ka2
When we anaylse the base, we have
Tart⁻² + H₂O ⇄ OH⁻ + HTart⁻ Kb1
HTart⁻ + H₂O ⇄ OH⁻ + H₂Tart Kb2
Remember that Ka1 . Kb2 = Kw, plus pKa1 + pKb2 = 14
Kb2 = Kw / Ka1 → 1×10⁻¹⁴ / 9.20×10⁻⁴ = 1.08×10⁻¹¹
so pKb = - log Kb2 → - log 1.08×10⁻¹¹ = 10.96
Answer: 63.88 atm
Explanation:
To answer this, we use the formula PV = nRT since the asumption is that the gas has an ideal behavior
where number of mole = 2.60 mol, R(gas constant) = 0.08205746 L atm/K mol,
T = 251 ∘C = (251 + 273) K = 524 K, Volume = 1.75 L
Making Pressure the subject of the formula, we have
P = nRT/V = 2.6 * 0.08205746 * 524/2.75 = 63.88 atm
Answer:
No it is not because it is not in group 3 or 12.
Explanation:
Answer:
sodium + chlorine --> sodium chloride
