125 Kelvin turns into -146 C
Answer:
Explanation:
The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.
1 .Calculate the hydronium ion concentration
We can use an ICE table to organize the calculations.
HF + H₂O ⇌ H₃O⁺ + F⁻
I/mol·L⁻¹: 2.7 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 2.7 - x x x
![K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Ba%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BH%7D_%7B3%7D%5Ctext%7BO%7D%5E%7B%2B%7D%5D%20%5Ctext%7BF%7D%5E%7B-%7D%5D%7D%20%7B%5Ctext%7B%5BHF%5D%7D%7D%20%3D%207.2%20%5Ctimes%2010%5E%7B-4%7D%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B2.7%20-%20x%7D%20%3D%207.2%20%5Ctimes%2010%5E%7B-4%7D%5C%5C%5C%5C%5Ctext%7BCheck%20for%20negligibility%20of%20%7Dx%5C%5C%5C%5C%5Cdfrac%7B2.7%7D%7B7.2%20%5Ctimes%2010%5E%7B-4%7D%7D%20%3D%204000%20%3E%20400%5C%5C%5C%5C%5Ctherefore%20x%20%5Cll%202.7%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B2.7%7D%20%3D%207.2%20%5Ctimes%2010%5E%7B-4%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%202.7%20%5Ctimes%207.2%20%5Ctimes%2010%5E%7B-4%7D%20%3D%201.94%20%5Ctimes%2010%5E%7B-3%7D%5C%5Cx%20%3D%200.0441%5C%5C%5Ctext%7B%5BH%24_%7B3%7D%24O%24%5E%7B%2B%7D%24%5D%7D%3D%20%5Ctext%7Bx%20mol%24%5Ccdot%24L%24%5E%7B-1%7D%24%7D%20%3D%20%5Ctext%7B0.0441%20mol%24%5Ccdot%24L%24%5E%7B-1%7D%24%7D)
2. Calculate the pH
![\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20-%5Clog%7B%5Crm%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%20%3D%20-%5Clog%7B0.0441%7D%20%3D%20%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.36%7D%7D)
3. Calculate [C₆H₅O⁻]
C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺
2.7 x 0.0441
Answer:
The first measurement is 116.84 cm, and the second measurement is 116.85 cm.
Explanation:
The problem provides both the first and second measurements so all you have to do is fill in the blank with them.
298 g of calcium carbonate CaCO₃
Explanation:
We have the following chemical reaction:
CaCN₂ (s) + 3 H₂O (l) → CaCO₃ (s)+ 2 NH₃ (g)
number of moles = mass / molar weight
number of moles of H₂O = 161 / 18 = 8.94 moles
Knowing the chemical reaction we devise the following reasoning:
if 3 moles of H₂O produces 1 mole of CaCO₃
then 8.94 moles of H₂O produces X moles of CaCO₃
X = (8.94 × 1) / 3 = 2.98 moles of CaCO₃
mass = number of moles × molar weight
mass of CaCO₃ = 2.98 × 100 = 298 g
Learn more about:
number of moles
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shared to form an ionic bond.
I hope this helps!
