The answer is most likely C
Answer:
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
Explanation:
Volume of NaOH = 1.7 ml = 0.0017 L
Molarity of NaOH = 0.0811 M
Moles of NaOH = n
n = 0.0001378 mol
According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.
Then 0.0001378 mol of NaOH will neutralize:
of sulfuric acid.
Concentration of sulfuric acid in the acid rain sample: x
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
Answer:
When a molecule is excited to a higher state it often ends up in its lowest excited state S1 and then emits radiation.
Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
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Answer:
A - Chemical potential energy being converted to heat energy
Explanation:
