A regulation that should be followed to prevent accidents is called safety blank
1 answer:
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Answer : Option (A) Accelerator 2 model has the lowest percentage of energy lost as waste.
Solution : Given,
For Accelerator 1 model,
Input energy = 2078.3 J
Wasted energy = 663.1 J
Output energy = 1415.2 J
For Accelerator 2 model,
Input energy = 7690.0 J
Wasted energy = 2337.5 J
Output energy = 5353.5 J
For Accelerator 3 model,
Input energy = 4061.9 J
Wasted energy = 2259.6 J
Output energy = 1802.3 J
Formula used for lowest percentage of energy lost as waste is:
% energy lost as waste = (Total energy wasted / Total input energy ) × 100
For Accelerator 1 model,
% energy lost as waste = = 31.90%
For Accelerator 2 model,
% energy lost as waste = = 30.39%
For Accelerator 3 model,
% energy lost as waste = = 55.62%
So, we conclude that the Accelerator 2 model has the lowest percentage of energy lost as waste.
Answer:
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.
Explanation:
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In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:
Thus, by combining them, we obtain:
Which is related to the general line equation:
Whereas:
It means that we answer to the blanks as follows:
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.
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