Answer:
ΔG <0 , ΔH > 0 , ΔS > 0 .
Explanation:
From the data given in question , the reaction is a spontaneous process , hence , the value of change in gibbs free energy would be negative , ΔG <0
And , on dissolution process , the temperature of the water decreases , i.e. , it is an endothermic process , i.e. , the change in enthalphy value is positive , ΔH > 0
And , during the process of dissolution , the ammonia salt break does to ions , i.e. , the randomness increases , hence the ΔS > 0
Answer:
Phase changes typically occur when the temperature or pressure of a system is altered. When temperature or pressure increases, molecules interact more with each other. When pressure increases or temperature decreases, it's easier for atoms and molecules to settle into a more rigid structure.
Explanation:
Hope it helps UvU
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Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
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