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
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C. Digesting a sandwich
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She would observe a yellowish solid precipitate which is the lead iodide and a white solid precipitate which is the potassium nitrate.
This is because the lead nitrate solution which contains particles of lead will mix with the potassium iodide solution containing particles of iodide. Upon mixing,the lead particles from the Lead nitrate solution combines with the iodide particles from Potassium iodide and form two compounds, a yellowish solid precipitate called lead iodide and a white solid precipitate called Potassium nitrate.
The formation of entirely two new compounds is known as the double displacement reaction and can be written in a chemical equation as
2KI(aq.)+Pb(NO₃)₂(aq.)------>2KNO₃(aq.)+PbI ₂(s)
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