Answer:
0.583 kilojoules
Explanation:
The amount of heat required to pop a single kernel can be calculated using the formula as follows:
Q = m × c × ∆T
Where;
Q = amount of heat (J)
m = mass of water (g)
c = specific heat capacity of water (4.184 J/g°C)
∆T = change in temperature
From the given information, m = 0.905 g, initial temperature (room temperature) = 21°C , final temperature = 175°C, Q = ?
Q = m × c × ∆T
Q = 0.905 × 4.184 × (175°C - 21°C)
Q = 3.786 × 154
Q = 583.044 Joules
In kilojoules i.e. we divide by 1000, the amount of heat is:
= 583.04/1000
= 0.583 kilojoules
Polarity, cohesion, adhesion, surface tension, high specific heat, and evaporating cooling
Hydrophilic;
Polar compounds
water
ions
glucose
Hydrophobic;
Triglycerides
cholesterol
Lipids
Oil
Carbohydrates
<h3>What are hydrophilic substances?</h3>
The term hydrophilic substances is used to describe the substances that could dissolve in water. They are those substances that are water soluble. Somehow, the substances must be polar because in chemistry like dissolves like.
The hydrophobic substances are those substances that do not dissolve in water. They are nonpolar and interact with other nonpolar substances.
The classification of the substances took place below;
Hydrophilic;
Polar compounds
water
ions
glucose
Hydrophobic;
Triglycerides
cholesterol
Lipids
Oil
Carbohydrates
Learn more about hydrophilic substances:brainly.com/question/4692308
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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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