Answer:
1341541345134514355143
Explanation:
THIS DUMB B~!TCH GAVE ME THIS ANSWER
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
There should be 4.5 moles
Active metals are those metals in the group 1 of the periodic table.
Electronegativity is the trend to atract electrons.
Active metals have few valence electron, because their last shell is of the kind ns^1 or ns^2
Then, these atoms do not trend to attract electrons. The most electronegative atomos are those who have 7 valence elecfrons; this is their last shell is of the kind ns^7, because when they attract one electron to its valence shell they will complete 8 electrons which is the most stable configuration.
