Answer:
1. Caffeine, C₈H₁₀N₄O₂
Amount = 1.00/194 = 0.00515 moles
2. Ethanol, C₂H₅OH
Amount = 0.0217 moles
3. Dry Ice, CO₂
amount = 0.0227 moles
<em>Note: The question is incomplete. The compound are as follows:</em>
<em> 1. Caffeine, C₈H₁₀N₄O₂;</em>
<em>2. Ethanol, C₂H₅OH;</em>
<em>3. Dry Ice, CO₂</em>
Explanation:
Amount (moles) = mass in grams /molar mass in grams per mole
1. Caffeine, C₈H₁₀N₄O₂
molar mass of caffeine = 194 g/mol
Amount = 1.00 g/194 g/mol = 0.00515 moles
2. Ethanol, C₂H₅OH
molar mass of ethanol = 46 g/mol
Amount = 1.00 g/46 g/mol = 0.0217 moles
3. Dry Ice, CO₂
molar mass of dry ice = 44 g/mol
amount = 1.00 g/44 g/mol = 0.0227 moles
Answer: (1 Kilogram = 2.20462 pounds) . There are 2.2046226218 lb in 1 kilogram. To convert kilograms to pounds, multiply your figure by 2.205 for an approximate result. 1 kilogram is also equal to 2 lb and 3.27396195 oz. Working out a rough estimate in your head for converting to pounds and ounces may be tricky - remember that there are 16 ounces in a pound.
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
