You must know the concentration of the acetic acid. Suppose the concentration is 0.1 M. The solution is as follows:
CH₃COOH → CH₃COO⁻ + H⁺
I 0.1 0 0
C -x +x +x
E 0.1 - x x x
Ka = (x)(x)/(0.1 - x)
1.8×10⁻⁵ = x²/(0.1 - x)
Solving for x,
x = 1.333×10⁻³ = H⁺
pH = -log[H⁺] = -log(1.333×10⁻³)
pH = 2.88
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Your answer is C. its level of dissolved minerals ~A
Answer:
4.7 kJ/kmol-K
Explanation:
Using the Debye model the specific heat capacity in kJ/kmol-K
c = 12π⁴Nk(T/θ)³/5
where N = avogadro's number = 6.02 × 10²³ mol⁻¹, k = 1.38 × 10⁻²³ JK⁻¹, T = room temperature = 298 K and θ = Debye temperature = 2219 K
Substituting these values into c we have
c = 12π⁴Nk(T/θ)³/5
= 12π⁴(6.02 × 10²³ mol⁻¹)(1.38 × 10⁻²³ JK⁻¹)(298 K/2219 K)³/5
= 9710.83(298 K/2219 K)³/5
= 1942.17(0.1343)³
= 4.704 J/mol-K
= 4.704 × 10⁻³ kJ/10⁻³ kmol-K
= 4.704 kJ/kmol-K
≅ 4.7 kJ/kmol-K
So, the specific heat of diamond in kJ/kmol-K is 4.7 kJ/kmol-K
11.0 kg = (11.0 kg)(1000 g/kg) = 11000 g
(11000 g)/(1400 cm3) = 7.857 g/cm3
Simplified = 7.86 g/cm3
