Answer:
Percent error = 12.5%
Explanation:
In a measurement you can find percent error following the formula:
Percent error = |Measured value - Accepted Value| / Acepted value * 100
Based on the data of the problem, accepted value is 22.4L and the measured Value (Value of Sara) was 19.6L.
Replacing:
Percent error = |Measured value - Accepted Value| / Acepted value * 100
Percent error = |19.6L - 22.4L| / 22.4L * 100
Percent error = |-2.8L| / 22.4L * 100
Percent error = 2.8L / 22.4L * 100
Percent error = 12.5%
Answer:
Volume = 45.62L
Explanation:
Data;
V1 = 54.9L
T1 = 64°C = (64 + 273.15)k = 337.15K
T2 = 7°C = (7 + 273.15)k = 280.15K
V2 = ?
From Charles law,
The volume of a fixed mass of gas is directly proportional to its temperature provided that pressure remains constant
V = KT, K = V / T = V1 / T1 = V2 / T2 = V3 / T3 =.........= Vn / Tn
(54.9 / 337.15) = (V2 / 280.15)
V2 = (54.9 * 280.15) / 337.15
V2 = 45.618L
V2 = 45.62L
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Considering the ideal gas law, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
- P= 2 atm
- V= ?
- n=
being 2g/mole the molar mass of H2, that is, the amount of mass that a substance contains in one mole. - R= 0.082

- T= 353 K
Replacing:
2 atm× V = 4.745 moles× 0.082
× 353 K
Solving:
V = (4.745 moles× 0.082
× 353 K)÷ 2 atm
<u><em>V= 68.67 L</em></u>
Finally, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
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