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lora16 [44]
3 years ago
5

The accepted value for the molar volume of a gas is 22.4 L. Sarah's experimental data indicated that a mole of a gas had a volum

e of 19.6L. What was her percent error?
Chemistry
1 answer:
Amanda [17]3 years ago
3 0

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%

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7 0
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Umnica [9.8K]

Answer:

The rate of disappearance of C₂H₆O = 2.46 mol/min

Explanation:

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From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.

The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus,  the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.

Rate of disappearance of C₂H₆O = 1.64 mol/min of  Cr₂(SO₄)₃  * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃

Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O

Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min

4 0
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