Answer:C.people with perfect understanding of science.
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
8.03 solutions report is described below.
Explanation:
8.03 Solutions Lab Report
In this laboratory activity, you will investigate how temperature, agitation, particle size, and dilution affect the taste of a drink. Fill in each section of this lab report and submit it and your pre-lab answers to your instructor for grading.
Pre-lab Questions:
In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.
Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).
Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.
Answer: Larger molecules have stronger London forces.
Explanation:
Pentane therefore has a stronger force than butane.
Answer:
3.
D. Philippine Atmospheric, Geophysical and Astronomical Services
Administration (PAGASA)
Explanation:
4.C. 62.0 kph
