a. Organic: C₁₀H₁₆KNO₉S₂; (CH₃)₄As₂; C₆H₁₂O₆
b. Inorganic: NaAsO₂; HSiCl₃; (BiO)₂CO₃; H₂P₂O₇; H₂O; CO₂
Compounds containing <em>both C and H</em> are organic.
Compounds that are <em>not organic</em> are inorganic.
The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.
<h3>How do we calculate pH of weak base?</h3>
pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:
pH = pKb + log([HB⁺]/[B])
pKb = -log(1.8×10⁻⁶) = 5.7
Chemical reaction for C₆H₅NH₂ is:
C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻
Initial: 0.25 0 0
Change: -x x x
Equilibrium: 0.25-x x x
Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]
Kb = x² / 0.25 - x
x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:
1.8×10⁻⁶ = x² / 0.25
x² = (1.8×10⁻⁶)(0.25)
x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]
On putting all these values on the above equation of pH, we get
pH = 5.7 + log(0.67×10⁻³/0.25)
pH = 3.13
Hence pH of the solution is 3.13.
To know more about Henderson Hasselbalch equation, visit the below link:
brainly.com/question/13651361
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Because each element has an exactly defined line emission spectrum, scientists are able to identify them by the color of flame they produce.
High tide or Neap tides which are normally really high.
PH + pOH = 14
12.52 + pOH = 14
pOH = 14 - 12.52
pOH = 1.48
[OH⁻] = 10^ -pOH
[OH⁻] = 10 ^- 1.48
[OH⁻] = 0.033 M
