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
I would say that the answer has to be C
Since there is no change in mols on both sides of the equation the mass is constant
4 protons the number of proton has the same number of electron [which is the same as atomic number]
Such an object makes a larger dent in the fabric of space-time than an object with little mass. (It has a greater gravitational attraction than less massive objects)
A greater force is required to accelerate such an object than a less massive object
Answer:
27.60 g urea
Explanation:
The <em>freezing-point depression</em> is expressed by the formula:
In this case,
- ΔT = 5.6 - (-0.9) = 6.5 °C
m is the molality of the urea solution in X (mol urea/kg of X)
First we<u> calculate the molality</u>:
- 6.5 °C = 7.78 °C kg·mol⁻¹ * m
Now we<u> calculate the moles of ure</u>a that were dissolved:
550 g X ⇒ 550 / 1000 = 0.550 kg X
- 0.84 m = mol Urea / 0.550 kg X
Finally we <u>calculate the mass of urea</u>, using its molecular weight:
- 0.46 mol * 60.06 g/mol = 27.60 g urea
