Energy released from changing the phase of a substance from the gas phase to liquid phase can be calculated by using the specific latent heat of vaporization. The heat of fusion of water at 0 degrees Celsius is 40.7 kJ/mol. Calculation are as follows:<span> </span>
Energy = 27.9 g (1 mol / 18.02 g) x 40.7 kJ/mol
Energy = 63.09 kJ
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Explanation:
To solve this we assume
that the gas is an ideal gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant temperature and number of moles of the gas
the product of PV is equal to some constant. At another set of condition of
temperature, the constant is still the same. Calculations are as follows:
P1V1 =P2V2
P2 = P1V1/V2
P2 = 740mmhg x 19 mL / 30 mL
<span>P2 = 468.67 mmHg = 0.62 atm</span>
Answer:
The change in entropy is -1083.112 joules per kilogram-Kelvin.
Explanation:
If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (
), in joules per gram-Kelvin, by the following model:

(1)
Where:
- Mass, in kilograms.
- Specific heat of water, in joules per kilogram-Kelvin.
,
- Initial and final temperatures of water, in Kelvin.
If we know that
,
,
and
, then the change in entropy for the entire process is:


The change in entropy is -1083.112 joules per kilogram-Kelvin.
Answer:
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