First, consider the steps to heat the sample from 209 K to 367K.
1) Heating in liquid state from 209 K to 239.82 K
2) Vaporaizing at 239.82 K
3) Heating in gaseous state from 239.82 K to 367 K.
Second, calculate the amount of heat required for each step.
1) Liquid heating
Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol
=> number of moles = 12.62 g / 17 g/mol = 0.742 mol
Heat1 = #moles * heat capacity * ΔT
Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J
2) Vaporization
Heat2 = # moles * H vap
Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J
3) Vapor heating
Heat3 = #moles * heat capacity * ΔT
Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J
Third, add up the heats for every steps:
Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J
Fourth, divide the total heat by the heat rate:
Time = 22,466.3 J / (6000.0 J/min) = 3.7 min
Answer: 3.7 min
Answer:
When chlorine and bromine atoms come into contact with ozone in the stratosphere, they destroy ozone molecules.
Explanation:
Answer:- 537 kJ of heat is released.
Solution:- For the given equation, 
is -657 kJ and the coefficient of 
in the balanced equation is 2. It means 657 kJ of heat is released when 2 moles of chlorine are used. We need to calculate the heat released when 116 g of are used.
Grams of chlorine are converted to moles and then multiplied by the value and divided by the coefficient of chlorine and the set could be shown using dimensional analysis as:
= 537.46 kJ
If we use the correct sig figs then it needs to be round off to three sig figs as the given grams of chlorine has only three sig figs. So, 537 kJ of heat is released.
An element is made up of only one type of atom.
