<span>128 g/mol
Using Graham's law of effusion we have the formula:
r1/r2 = sqrt(m2/m1)
where
r1 = rate of effusion for gas 1
r2 = rate of effusion for gas 2
m1 = molar mass of gas 1
m2 = molar mass of gas 2
Since the atomic weight of oxygen is 15.999, the molar mass for O2 = 2 * 15.999 = 31.998
Now let's subsitute the known values into Graham's equation and solve for m2.
r1/r2 = sqrt(m2/m1)
2/1 = sqrt(m2/31.998)
4/1 = m2/31.998
127.992 = m2
So the molar mass of the unknown gas is 127.992 g/mol.
Rounding to 3 significant figures gives 128 g/mol</span>
Answer:
K = 6.5 × 10⁻⁶
Explanation:
C₅H₆O₃ ⇄ C₂H₆ + 3CO
Use PV=nRT to find the initial pressure of C₅H₆O₃
P (2.50) = (0.0493) (0.08206) (473)
P = 0.78atm
C₅H₆O₃ ⇄ C₂H₆ + 3CO
0.78atm 0 0
0.78 - x x 3x
1.63atm = 0.78 - x + x + 3x
P(total) = 0.288atm
C₅H₆O₃ = 0.78 - 0.288
= 0.489atm
C₂H₆ = 0.288atm
CO = 0.846atm
= 0.379
= 6.5 × 10⁻⁶
The mass would also be 150g like the first block of wood. Both of their volumes are 300 cm^2, and since they are the same type of wood they have the same density. Therefore following d=m/v, they have the same mass
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