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SIZIF [17.4K]
3 years ago
6

Steven had a sample of ethanol and wanted to see if it would boil at the temperature found in his textbook. His experiment yield

ed a temperature of 143.6oF but the literature value was 173.1oF. What is the percent error in this experiment?
Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
4 0

Answer:

17.04%

Explanation:

Actual Value = 173.1

Measured Value = 143.6

Percent error is obtained using the equation;

Percent error = (Measured - Actual) / Actual ]* 100

Percent error = [ (143.6 - 173.1) / 173.1 ] * 100

The absolute value of (Measured - Actual) is taken,

Percent Error = [29.5 / 173.1 ] * 100

Percent Error = 0.1704 * 100 = 17.04%

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3 years ago
In an acid-base titration, a student uses 21.35 mL of 0.150 M NaOH to neutralize 25.00 mL of H2SO4. How many moles of acid are i
GalinKa [24]

Answer: There are 0.006 moles of acid in the flask.

Explanation:

Given: V_{1} = 21.35 mL,        M_{1} = 0.150 M

V_{2} = 25.0 mL,           M_{2} = ?

Formula used to calculate molarity of H_{2}SO_{4} is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.15 M \times 21.35 mL = M_{2} \times 25.0 mL\\M_{2} = 0.1281 M

As molarity is the number of moles of a substance present in a liter of solution.

Total volume of solution = V_{1} + V_{2}

= 21.35 mL + 25.0 mL

= 46.36 mL  (1 mL = 0.001 L)

= 0.04636 L

Therefore, moles of acid required are calculated as follows.

Molarity = \frac{no. of moles}{Volume (in L)}\\0.1281 M = \frac{no. of moles}{0.04635 L}\\no. of moles = 0.006 mol

Thus, we can conclude that there are 0.006 moles of acid in the flask.

3 0
2 years ago
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