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
.
Answer:
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Answer:
Sodium hydroxide and water will react at room temperature. What does this indicate about its activation energy? The activation energy is at exactly 600 kJ. C. The activation energy is very low.
Explanation:
Place the solid on an electronic balance. Read the balance reading.
Answer:
The element with electron configuration 1s² 2s² 2p⁶ will most likely not react with an element having seven Valence electrons.
The electron configuration of the element in discuss is 1s² 2s² 2p⁶.
The element has enough electrons to fill it's energy level, n = 2 shell.
In essence, the element in discuss is unreactive as it has attained it's octet configuration and as such is neither in need of an electron nor ready to donate an electron.
As such, although an atom having seven Valence electrons is highly electronegative and as such is an electron attractor, the element with the full octet configuration does not react with it.
This unreactive nature of noble gases is attributed to the full octet configuration of noble gases.
P.S: The electron configuration above is the electron configuration of Neon, Ne.
