Answer:
λ = 5.68×10⁻⁷ m
Explanation:
Given data:
Energy of photon = 3.50 ×10⁻¹⁹ J
Wavelength of photon = ?
Solution:
E = hc/λ
h = planck's constant = 6.63×10⁻³⁴ Js
c = 3×10⁸ m/s
Now we will put the values in formula.
3.50 ×10⁻¹⁹ J = 6.63×10⁻³⁴ Js × 3×10⁸ m/s/ λ
λ = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 3.50 ×10⁻¹⁹ J
λ = 19.89×10⁻²⁶ J.m / 3.50 ×10⁻¹⁹ J
λ = 5.68×10⁻⁷ m
Grams ethanol = 33 ml times .789 gms/ml = 26.037 gms
<span>Moles ethanol = 26.037 gms / 46 gms/mole = .57 moles </span>
<span>Moles water = 67 ml or 67 grams/18 gms/mole = 3.22 moles </span>
<span>total moles = .57 + 3.72 = 4.29 moles </span>
<span>Mole fraction ethanol = .57 moles ethanol / 4.29 moles total = 0.13</span>
<span>Moles fraction water = 3.72 moles water / 4.29 moles total = 0.87</span>
<span>Partial pressure of ethanol = mole fraction ethanol (.13) _ times VP ethanol 43.9 torr) = 5.707 torr </span>
<span>partial pressure water = mole fraction water .87) times VP water (l7.5 torr) = 15.23 torr </span>
<span>Total vapor pressure over solution = 5.71 torr + 15.23 torr = 20.94 torr</span>
Answer:
pH = 12.7
Explanation:
First, we have to calculate the [Ca²⁺] in a solution of about 250 ppm CaCO₃.

Now, let's consider the dissolution of Ca(OH)₂ in water.
Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)
The solubility product Ksp is:
Ksp = [Ca²⁺] × [OH⁻]²
[OH⁻] = √(Ksp/[Ca²⁺]) = √(6.5 × 10⁻⁶/2.5 × 10⁻³) = 5.1 × 10⁻² M
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log (5.1 × 10⁻²) = 1.3
pH + pOH = 14 ⇒ pH = 14 - pOH = 14 - 1.3 = 12.7
Answer:

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