Answer:
7. A) I, II
; 8. D) 2.34e9 kJ
Step-by-step explanation:
7. Combustion of ethanol
I. The negative sign for ΔH shows that the reaction is exothermic.
II. The enthalpy change would be different if gaseous water were produced.
That's because it takes energy to convert liquid water to gaseous water, and this energy is included in the value of ΔH.
III. The reaction is a redox reaction, because
- Oxygen is reacting with a compound
- The oxidation number of C increases
- The oxidation number of O decreases.
IV. The products of the reaction occupy a smaller volume than the reactants, because 3 mol of gaseous reactant are forming 2 mol of gaseous product.
Therefore, only I and II are correct.
7. Hindenburg
Data:
V = 2.00 × 10⁸ L
p = 1.00 atm
T = 25.1 °C
ΔH = -286 kJ·mol⁻¹
Calculations:
(a) Convert temperature to kelvins
T = (25.1 + 273.15) K = 298.25 K
(b) Moles of hydrogen
Use the <em>Ideal Gas Law</em>:
pV = nRT
n = (pV)/(RT)
n = (1.00 × 2.00 × 10⁸)/(0.082 06 × 298.25) = 8.172 × 10⁶ mol
(c) Heat evolved
q = nΔH = 8.172 × 10⁶ × (-286) = -2.34 × 10⁹ kJ
The hydrogen in the Hindenburg released 2.34e9 kJ
.
That is a true statement but you have to ask your teachet
Full Question;
What volume of a 0.150 M solution of KOH must be added to 450.0 mL of the acidic solution of 300ml of 0.450M HCL to completely neutralize all of the acid? Express the volume in liters to three significant figures.
Answer:
0.9l
Explanation:
First thing's first, we have to write out the balanced chemical equation.
KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)
Potassium hydroxide, KOH, and hydrochloric acid, HCl, react in a 1:1 mole ratio to produce aqueous potassium chloride, KCl, and water.
From the reaction;
Na = Nb
Where Na = Number of moles of acid
Na = Ca * Va = 0.450 * 0.300 = 0.135
Nb = Cb * Vb = Cb * 0.150
Na = Cb * 0.150
0.135 = Cb * 0.150
Cb = 0.135 / 0.150 = 0.9L
<h3>
Answer:</h3>
13 g CO₂
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] 6.7 L O₂
[Solve] g O₂
<u>Step 2: Identify Conversions</u>
[STP] 22.4 L = 1 mol
[PT] Molar Mass of O: 16.00 g/mol
[PT] Molar Mass of C: 12.01 g/mol
Molar Mass of CO₂: 12.01 + 2(16.00) = 44.01 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide/Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
13.1637 g CO₂ ≈ 13 g CO₂
