Answer:
Carbon-Carbon Bonds
Carbon can form single, double, or even triple bonds with other carbon atoms. In a single bond, two carbon atoms share one pair of electrons. In a double bond, they share two pairs of electrons, and in a triple bond they share three pairs of electrons
Answer:
The electrons are attracted to the nuclei of both atoms.
Explanation:
Electrons are attracted to nuclei because nuclei are positively charged and electrons are negatively charged. Opposites attract
The concentration
pertains to the molarity of the solution with an equation of mol of solution
per liter of solvent. <span> </span>
<span>M (molarity)
= moles of solute/L of solvent</span>
M (molarity)
= 0.5 mol NaOH / 0.5 L
M (molarity)
= <span>1 M NaOH</span>
Answer:
yes it is a chemical reaction
Explanation:
because the substances combined and made something new
Answer:
(a) 5.04; (b) 5.18; (c) 12.30
Explanation:
(a) pH of buffer
HA + H₂O ⇌ H₃O⁺ + A⁻
![\begin{array}{rcl}\text{pH}& = &\text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\& = &4.74 + \log\dfrac{0.2}{0.1}\\\\& = &4.74 + \log 2\\& = &4.74 + 0.30\\& = & \mathbf{5.04}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Ctext%7BpH%7D%26%20%3D%20%26%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%26%20%3D%20%264.74%20%2B%20%5Clog%5Cdfrac%7B0.2%7D%7B0.1%7D%5C%5C%5C%5C%26%20%3D%20%264.74%20%2B%20%5Clog%202%5C%5C%26%20%3D%20%264.74%20%2B%200.30%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B5.04%7D%5C%5C%5Cend%7Barray%7D)
(b) pH after addition of base
![\text{Moles of NaOH} = \text{0.002 L} \times \dfrac{\text{10 mol}}{\text{1 L }} = \text{0.02 mol}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20NaOH%7D%20%3D%20%5Ctext%7B0.002%20L%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B10%20mol%7D%7D%7B%5Ctext%7B1%20L%20%7D%7D%20%3D%20%5Ctext%7B0.02%20mol%7D)
HA + H₂O ⇌ H₃O⁺ + A⁻
I/mol: 0.1 0 0.2
C/mol: -0.02 +x +0.02
E/mol: 0.08 x 0.22
![\text{pH}& =4.74 + \log\dfrac{0.22}{0.08}\\\\& = & 4.74 + \log 2.75\\& = & 4.74 + 0.44\\& = & \mathbf{5.18}\\\end{array}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%26%20%3D4.74%20%2B%20%5Clog%5Cdfrac%7B0.22%7D%7B0.08%7D%5C%5C%5C%5C%26%20%3D%20%26%204.74%20%2B%20%5Clog%202.75%5C%5C%26%20%3D%20%26%204.74%20%2B%200.44%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B5.18%7D%5C%5C%5Cend%7Barray%7D)
(c) pH of water after addition of base
![\text{[OH$^{-}$]} = \dfrac{\text{0.02 mol}}{\text{1.002 L}} = \text{0.02 mol/L}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%24%5E%7B-%7D%24%5D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.02%20mol%7D%7D%7B%5Ctext%7B1.002%20L%7D%7D%20%3D%20%5Ctext%7B0.02%20mol%2FL%7D)
pOH} = -log0.02 = 1.70
pH = 14.00 - pOH = 14.00 - 1.70 = 12.30