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Dimas [21]
3 years ago
9

The dissolution of barium hydroxide in water is an exothermic process. Which of the following statements is correct?A) The entha

lpy of solid barium hydroxide plus pure water is less than that of the solution, at the same temperature.B) The enthalpy of solid barium hydroxide plus pure water is greater than that of the solution, at the same temperature.C) The enthalpy of solid barium hydroxide plus pure water is the same as that of the solution, at the same temperature.D) The temperature of the solution is lower than of the barium hydroxide and water before mixing.E) When barium hydroxide dissolves in water, the system does work on the surroundings.
Chemistry
2 answers:
madreJ [45]3 years ago
8 0

Answer:

B

Explanation:

An exothermic reaction is one in which heat is liberated to the surrounding by a reaction.In this case,the enthalpy of the product is less than that of the reactant and such  a reaction has a negative enthalpy (ΔH).

Marta_Voda [28]3 years ago
5 0

Answer:

B

Explanation:

In an exothermic process the enthalpy of the products is smaller than the enthalpy of the reactives, in other words, the difference of enthalpy of the process (which is defined as the enthalpy at the final state of the process minus the enthalpy of the initial state) is negative. In this case, this means that the enthalpy of solid barium hydroxide plus pure water is greater than that of the solution, at the same temperature .

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The isotope that is more abundant, given the data is isotope Li7

<h3>Assumption</h3>
  • Let Li6 be isotope A
  • Let Li7 be isotope B

<h3>How to determine whiche isotope is more abundant</h3>
  • Molar mass of isotope A (Li6) = 6.02 u
  • Molar mass of isotope B (Li7) = 7.02 u
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Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]

6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]

6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]

6.94 = [6.02A% + 702 - 7.02A%] / 100

Cross multiply

6.02A% + 702 - 7.02A% = 6.94 × 100

6.02A% + 702 - 7.02A% = 694

Collect like terms

6.02A% - 7.02A% = 694 - 702

-A% = -8

A% = 8%

Thus,

Abundance of B = (100 - A)%

Abundance of B = (100 - 8)%

Abundance of B = 92%

SUMMARY

  • Abundance of A (Li6) = 8%
  • Abundance of B (Li7) = 92%

From the above, isotope Li7 is more abundant.

Learn more about isotope:

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