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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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It is the last one.

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Write a balanced equation for the complete combustion of each compound. a. formaldehyde (CH2O(g)) b. heptane (C7H17(l)) c. benze
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b. heptane (C₇H₁₇(l)): 4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)

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Explanation:

In a reaction of combustion, a hydrocarbon compound (composed of C, H and O) reacts with oxygen gas (O₂). The <u>complete</u> combustion of a hydrocarbon - such as formaldehyde, heptane, and benzene - produces carbon dioxide (CO₂) and water (H₂O).  

Thus, we write the reactants and products for each combustion reaction and then we balance the atoms: C, H, and O.

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CH₂O(g) + O₂(g) → CO₂(g) + H₂O(g)

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b. heptane (C₇H₁₇(l)):

C₇H₁₇(l) + O₂(g) → CO₂(g) + H₂O(g)

Here we have to write a coefficient 28 for CO₂ to balance the C atoms in the products side, and for C₇H₁₇ we write 28/4 = 7. With similar reasoning we found the coefficients for O₂ and H₂O:

4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)

c. benzene (C6H6(l))​:

C₆H₆(l) +  O₂(g) → CO₂(g) +  H₂O(g)

First, we write a coefficient of 6 for CO₂ to balance the C atoms. Then, we have to balance H atoms: we write a coefficient 3 in H₂O. Now, we have 12 + 3 = 15 atoms of O on the reactants side. So, we write a half of these number of atoms in the coefficient for O₂: 15/2. We obtain the balanced equation:

C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3 H₂O(g)

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