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:
Radiation is being released from the reactor.
Explanation:
( A P E X )
Answer:me being a tanned ginger
Explanation: i cant date u
Explanation:
The balanced chemical equation of the reaction is:
From the balanced chemical equation,
1 mole of propane forms ------ 3 mol. of 
gas.
The molar mass of propane is 44.1 g/mol.
One mole of any gas at STP occupies --- 22.4 L.
Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.
Answer:
Thus, 67.2 L of CO2 is formed at STP.
Answer:
[SO2Cl2] = = 0.015 M
[SO2] = = 0.0027 M
[Cl2] = = 0.0027 M
Q = = = 4.8 × 10−4
No. Q < Kc, so reaction will shift to the right.
Explanation:
