Answer:
For each system listed in the first column of the table below, decide (If possible) whether the change described in the second column will Increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the not enough information" button in the last column.
System As A few grams of water vapor (H,0). The water condenses to a liquid at a not enough The carbon dioxide is heated from -5.0 °C to 13.0°c and is also compressed from a volume of 9.0 1 to a volume of 5.0L A few moles of carbon dioxide (CO2) gas. not enough information -15.0 °C to 86.0 °C while the volume is held constant at 2.0 L A few moles of carbon dioxide (CO,) gas. S > 0 not enough 1 Don't Know
Number of moles :
n = mass solute / molar mass
n = 20 / 18
n = 1.111 moles
Therefore:
Molarity = moles solute / volume
Molarity = 1.111 / 30 => 0.37 M
Answer : The enthalpy change for the reaction is 1043 kJ/mol.
Explanation :
The given chemical reaction is:
As we know that:
The enthalpy change of reaction = E(bonds broken) - E(bonds formed)
![\Delta H=[(2\times B.E_{C\equiv O})+(1\times B.E_{O\equiv O})]-[2\times B.E_{C=O}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%282%5Ctimes%20B.E_%7BC%5Cequiv%20O%7D%29%2B%281%5Ctimes%20B.E_%7BO%5Cequiv%20O%7D%29%5D-%5B2%5Ctimes%20B.E_%7BC%3DO%7D%5D)
Given:
= 1074 kJ/mol
= 499 kJ/mol
= 802 kJ/mol
Now put all the given values in the above expression, we get:
![\Delta H=[(2\times 1074kJ/mol)+(1\times 499kJ/mol)]-[2\times 802kJ/mol]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%282%5Ctimes%201074kJ%2Fmol%29%2B%281%5Ctimes%20499kJ%2Fmol%29%5D-%5B2%5Ctimes%20802kJ%2Fmol%5D)
Therefore, the enthalpy change for the reaction is 1043 kJ/mol.
Nonmetals is most likely the answer
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