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:
brainly.com/question/14686462
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Answer:
FeCl₃
Explanation:
4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂
Given => 7moles 9moles
A simple way to determine which reagent is the limiting reactant is to convert all given data to moles then divide by the respective coefficients of the balanced equation. The smaller value will be the limiting reactant.
4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂
Given => 7/4 = 1.75* 9/3 = 3
*Smaller value => FeCl₃ is limiting reactant.
NOTE: However, when working problems, one must use original mole values given.
Well you would think yeah because it’s a liquid but the answer is no
<span>1.00 atm of each gas, in what direction will the system shift to reach equilibrium</span>
