Which substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes? a)CH4 b)H2S c)CO2 d)NaCl
1 answer:
Answer:
Explanation:
Given the amount of heat absorbed and the amount of substance in moles, we may calculate the heat of vaporization. Heat of vaporization is defined as the amount of heat per 1 mole of substance required to evaporate that specific substance.
Based on the value of heat of vaporization, we will identify the substance. Firstly, let's calculate the heat of vaporization:
Secondly, let's use any table for heat of vaporization values for substances. We identify that the heat of vaporization of is 18.7 kJ/mol
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First, let's write down the balanced chemical reaction between the given reactants:
NO₂ + NO → N₂O + O₂
The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:
Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)
For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1
NO₂ + NO → N₂O + O₂
The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:
Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)
For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1