The mass in grams of butane at standard room temperature is 53.21 grams.
<h3>How can we determine the mass of an organic substance at room temperature?</h3>
The gram of an organic substance at room temperature can be determined by using the ideal gas equation which can be expressed as:
PV = nRT
- Pressure = 1.00 atm
- Volume = 22.4 L
- Rate = 0.0821 atm*L/mol*K
- Temperature = 25° C = 298 k
1 × 22.4 L = n × (0.0821 atm*L/mol*K× 298 K)
n = 22.4/24.4658 moles
n = 0.91556 moles
Recall that:
- number of moles = mass(in grams)/molar mass
mass of butane = 0.91556 moles × 58.12 g/mole
mass of butane = 53.21 grams
Learn more about calculating the mass of an organic substance here:
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Answer:
6 oxygen atoms
Explanation:
From the equation,
2Fe(OH)₃ → Fe₂O₃ + 3H₂O
From the reactant (left hand side) we have 2 moles of Fe(OH)₃ having (2 * 3 = 6) atoms of oxygen and decomposed to give Fe₂O₃ which contains 3 atoms of oxygen and 3 moles of water that also contains 3 atoms of oxygen.
Since the number of oxygen participating in the reaction is independent on the product (not a reversible reaction) then the total number of oxygen atoms participating in the reaction is 6
#1. The condition of the atmosphere at a certain time and certain place
#2. The rate of evaporation is EQUAL to the rate of condensation
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