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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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Explanation:
Once solid ammonium nitrate interacts with water, the molecules of polar water intermingle with these ions and attract individual ions from the structure of the lattice, that actually will break down. E.g;-NH4NO3(s) — NH4+(aq)+ NO3-(aq) To split the ionic bonds that bind the lattice intact takes energy that is drained from the surroundings to cool the solution.
Some heat energy is produced once the ammonium and nitrate ions react with the water molecules (exothermic reaction), however this heat is far below that is needed by the H2O molecules to split the powerful ionic bonds in the solid ammonium nitrate.
Hence, we can say that the dissolution of ammonium nitrate in water is highly endothermic reaction.
The southern pacific ocean has a similar symmetric pattern to the seafloor ages of the Atlantic ocean. In the Pacific, the seafloor on one side of the youngest crust gets very old but the seafloor on the other side is much younger.