Its leading to contaminated drinking water.
In normal conditions, warm water does "pile up" in the" Western Pacific Ocean.
The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
<h3>What is Enthalpy of Vaporization ?</h3>
The amount of enthalpy or energy that must be added to a liquid substance into gas substance is called Enthalpy of Vaporization. It is also known as Latent heat of vaporization.
<h3>How to find the energy change from enthalpy of vaporization ?</h3>
To calculate the energy use this expression:

where,
Q = Energy change
n = number of moles
= Molar enthalpy of vaporization
Now find the number of moles
Number of moles (n) = 
= 
= 0.5 mol
Now put the values in above formula we get
[Negative sign is used because Br₂ condensed here]
= - (0.5 mol × 15.4 kJ/mol)
= - 7.7 kJ
Thus from the above conclusion we can say that The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
Learn more about the Enthalpy of Vaporization here: brainly.com/question/13776849
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Answer:
7.04 g
Explanation:
Let's consider the reaction in the last step of the Ostwald process.
3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)
The molar mass of HNO₃ is 63.01 g/mol. The moles corresponding to 6.40 g are:
6.40 g × (1 mol/63.01 g) = 0.102 mol
The molar ratio of NO₂ to HNO₃ is 3:2. The reacting moles of NO₂ are:
0.102 mol HNO₃ × (3 mol NO₂/2 mol HNO₃) = 0.153 mol NO₂
The molar mass of NO₂ is 46.01 g/mol. The mass corresponding to 0.153 moles is:
0.153 mol × (46.01 g/mol) = 7.04 g
Given , Speed of light = 3 × 10⁸ m/s
Distance = 40 × 1.61× 1000 m
= 264400 m
We know that, speed = distance/time
so, time = distance/speed
t = 264400 / 3 × 10⁸
t = 0.00881333 sec Approx
The answer is 0.00881333 sec Approx