Answer:
D. 44.2 g O₂
General Formulas and Concepts:
<u>Gas Laws</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at <em>1 atm, 273 K</em>
<u>Stoichiometry</u>
- Dimensional Analysis
- Mole Ratio
Explanation:
<u>Step 1: Define</u>
<em>Identify given.</em>
61.9 L O₂ at STP
<u>Step 2: Convert</u>
We know that the oxygen gas is at STP. Therefore, we can set up and solve for how many <em>moles</em> of O₂ is present:
Recall the Periodic Table (Refer to attachments). Oxygen's atomic mass is roughly 16.00 grams per mole (g/mol). We can use a mole ratio to convert from <em>moles</em> to <em>grams</em>:
Now we deal with sig figs. From the original problem, we are given 3 significant figures. Round your answer to the <u>exact</u> same number of sig figs:
∴ our answer is letter choice D.
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Topic: AP Chemistry
Unit: Stoichiometry
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
Learn more about empirical formula:
brainly.com/question/9459553
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Answer:
1. d. The reaction is spontaneous in the reverse direction at all temperatures.
2. c. The reaction is spontaneous at low temperatures.
Explanation:
The spontaneity of a reaction is associated with the Gibbs free energy (ΔG). When ΔG < 0, the reaction is spontaneous. When ΔG > 0, the reaction is non-spontaneous. ΔG is related to the enthalpy (ΔH) and the entropy (ΔS) through the following expression:
ΔG = ΔH - T. ΔS [1]
where,
T is the absolute temperature (T is always positive)
<em>1. What can be said about an Endothermic reaction with a negative entropy change?</em>
If the reaction is endothermic, ΔH > 0. Let's consider ΔS < 0. According to eq. [1], ΔG is always positive. The reaction is not spontaneous in the forward direction at any temperature. This means that the reaction is spontaneous in the reverse direction at all temperatures.
<em>2. What can be said about an Exothermic reaction with a negative entropy change?</em>
If the reaction is exothermic, ΔH < 0. Let's consider ΔS < 0. According to eq. [1], ΔG will be negative when |ΔH| > |T.ΔS|, that is, at low temperatures.
The question is incomplete, the question is;
Which drawing best accounts for the polarity of methanol, CH3OH, and the bond polarities that make a major contribution to the overall molecular polarity?
A) drawing (1) B) drawing (2)
D) drawing (4) C) drawing (3)
Answer:
B) drawing (2)
Explanation:
In Chemistry, the direction of dipole is shown from positive end to negative end.
The image that contains the options in the question asked has been attached.
We can see in image 2 that the oxygen atom was correctly designated as the negative end of the dipole while the carbon and hydrogen atoms were each designated as positive ends of the dipole in accordance with the magnitude of electronegativity difference between the two atoms. The net dipole moment is now taken in the direction shown in image 2. This is the correct answer.
