The entropy of the process in which the individual ions first leave the crystal lattice is positive while the entropy of the process whereby the each ion becomes surrounded by a cluster of polar water molecules is negative.
<h3>What is entropy?</h3>
The term entropy has to do with the degree of disorder in a system. The higher the entropy of the system, the more the disorderliness of the system.
Now, the entropy of the process in which the individual ions first leave the crystal lattice is positive while the entropy of the process whereby the each ion becomes surrounded by a cluster of polar water molecules is negative.
Learn more about entropy: brainly.com/question/13146879
Answer:
(slow)xy2+z→xy2z (fast) c step1:step2:xy2+z2→xy2z2
Explanation:
Step1: xy2+z2→xy2z2 (slow)
Step2: xy2z2→xy2z+z (fast)
2XY 2 + Z 2 → 2XY 2 Z
Rate= k[xy2][z2]
When the two elementary steps are summed up, the result is equivalent to the stoichiometric equation. Hence, this mechanism is acceptable. The order of both elementary steps is 2, which is ‘≤3’; this also makes this mechanism acceptable. Furthermore, the rate equation aligns with the experimentally determined rate equation, and this also makes this mechanism acceptable. Therefore, since all the three rules have been observed, this mechanism is possible.
Answer is: adding NaCl will lower the freezing point of a solution.
A solution (in this example solution of sodium chloride) freezes at a lower temperature than does the pure solvent (deionized water).
The higher the solute concentration (sodium chloride), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
Dissociation of sodium chloride in water: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
Explanation:
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Answer:
3.15 × 10⁻⁶ mol H₂/L.s
1.05 × 10⁻⁶ mol N₂/L.s
Explanation:
Step 1: Write the balanced equation
2 NH₃ ⇒ 3 H₂ + N₂
Step 2: Calculate the rate of production of H₂
The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s
Step 3: Calculate the rate of production of N₂
The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s
