Answer:
(3) 5.36
Explanation:
Since this is a titration of a weak acid before reaching equivalence point, we will have effectively a buffer solution. Then we can use the Henderson-Hasselbalch equation to answer this question.
The reaction is:
HAc + NaOH ⇒ NaAc + H₂O
V NaOH = 40 mL x 1 L/1000 mL = 0.040 L
mol NaOH reacted with HAc = 0.040 L x 0.05 mol/L = 0.002 mol
mol HAC originally present = 0.050 L x 0.05 mol/L = 0.0025 mol
mol HAc left after reaction = 0.0025 - 0.002 = 0.0005
Now that we have calculated the quantities of the weak acid and its conjugate base in the buffer, we just plug the values into the equation
pH = pKa + log ((Ac⁻)/(HAc))
(Notice we do not have to calculate the molarities of Ac⁻ and HAc because the volumes cancel in the quotient)
pH = -log (1.75 x 10⁻⁵) + log (0.002/0.0005) = 5.36
THe answer is 5.36
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

The number of sigma and pi bonds are,
Sigma Bonds =
16 Pi Bonds =
3Explanation: Every first bond formed between two atoms is sigma. Pi bond is formed when already a sigma bond is there. While in case of Alkyne (triple Bond) there is one sigma and one pi bond already present, so the third bond is formed by second side-to-side overlap of orbitals, hence, a second pi bond is formed.
Below all black bonds are sigma bonds, while in alkene there is one pi bond and in alkyne there are two pi bonds.
Answer:
Methane is a chemical compound with the molecular formula CH4. It is the main component in natural gas. Methane is considered the simplest of alkanes, compounds that consist only of hydrogen (H) and carbon (C) elements. Methane is an odorless, colorless, tasteless gas that is lighter than air.
Explanation:
Answer:
1.67g/cm3
Explanation:
The formula for density is
. The m variable stands for mass and the v variable stands for volume.
The mass of the brown sugar is 10.0g and the volume is 6.0cm3, so we can plug those values into the equation.



Rounded to 3 significant figures, the density of the block of brown sugar is 1.67 g/cm3. If the mass is in grams and the volume is in cm3, the unit for the final answer is
(grams per centimetres cubed).