Answer:
-21 kJ·mol⁻¹
Explanation:
Data:
H₃O⁺ + OH⁻ ⟶ 2H₂O
V/mL: 50 50
c/mol·dm⁻³: 1.0 1.0
ΔT = 4.5 °C
C = 4.184 J·°C⁻¹g⁻¹
C_cal = 50 J·°C⁻¹
Calculations:
(a) Moles of acid
So, we have 0.050 mol of reaction
(b) Volume of solution
V = 50 dm³ + 50 dm³ = 100 dm³
(c) Mass of solution
(d) Calorimetry
There are three energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
q₃ = heat to warm the calorimeter
q₁ + q₂ + q₃ = 0
nΔH + mCΔT + C_calΔT = 0
0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0
0.050ΔH + 1883 + 225 = 0
0.050ΔH + 2108 = 0
0.050ΔH = -2108
ΔH = -2108/0.0500
= -42 000 J/mol
= -42 kJ/mol
This is the heat of reaction for the formation of 2 mol of water
The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.
Answer:
The De Broglie wavelength decreases when the momentum increases
Explanation:
The De Broglie wavelength of a particle (or any object) is given by
where
h is the Planck constant
p is the momentum of the object
As we can see, the wavelength is inversely proportional to the momentum of the object: therefore we can say that, if the momentum increases, the De Broglie wavelength will decrease.
- The change in color from blue to pink of the cobalt complexes here has been the basis of cobalt chloride indicator papers for the detection of the presence of water. It is also used in self-indicating silica gel desiccant granules.
- Pink cobalt species + chloride ions ⇌ Blue cobalt species + water molecules
<u>Explanation</u>:
- The adjustment in color from blue to the pink of the cobalt complexes here has been the premise of cobalt chloride indicator papers for the detection of the presence of water. It is likewise utilized in self-demonstrating silica gel desiccant granules.
Pink cobalt species + chloride particles ⇌ Blue cobalt species + water molecules
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The response of [Co(H2O)6]2+(aq) + 4Cl–(aq) → [CoCl4]2–(aq) + 6H2O(l) is endothermic. In this manner, as per Le Chatelier's rule, when the temperature is raised, the situation of the balance will move to one side, shaping a greater amount of the blue complex particle at the expense of the pink species.
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Including concentrated hydrochloric raises the chloride particle fixation, making the equilibrium move to one side, as per Le Chatelier. Including water brings down the chloride particle fixation, moving the equilibrium the other way.
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As an extension, it is conceivable to show that it is the Cl–particles in the hydrochloric acid that move the balance by including a spatula of sodium chloride rather than the pink arrangement. This delivers a bluer color, however, this may take some time because the salt is delayed to dissolve.
H₂O₂ + 2FeSO₄ + H₂SO₄ → Fe₂(SO₄)₃ + 2H₂O
H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O k=1
Fe²⁺ → Fe³⁺ + e⁻ k=2
H₂O₂ + 2H⁺ + 2Fe²⁺ → 2H₂O + 2Fe³⁺
