<span>50.2 kJ = 333 kJ/kg * mass of water
mass of water is 0.15075075075075075075075075075075 kg
therefore mass of unfrozen water is 0.10924924924924924924924924924925 kg</span>
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Answer:
Molecularity of the rate determining step = 2
Explanation:
Step 1 (slow): H₂O₂ + I⁻ -----> H₂O + OI⁻
Step 2 (fast): H₂O₂ + OI⁻ -----> H₂O + O₂ + I⁻
The rate determining step in a reaction mechanism is also considered as slowest step.
Slowest step is also considered its highest activation energy in energy profile diagram.
In this case intermediate (IO⁻) is formed.
Step 1 considered as a slowest step.
So, Rate = K [H₂O₂][I⁻]
Molecularity = 2
Answer:
Please, see attached two figures:
- The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.
- The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.
Explanation:
The red arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the <em>solubility</em> of <em>NH₄Cl.</em>
From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.
Assuming density 1.0 g/mol for water, 10 mL of water is:
Thus, the solutibily is:
Answer:
To increase the yield of H₂ we would use a low temperature.
For an exothermic reaction such as this, decreasing temperature increases the value of K and the amount of products at equilibrium. Low temperature increases the value of K and the amount of products at equilibrium.
Explanation:
Let´s consider the following reaction:
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
When a system at equilibrium is disturbed, the response of the system is explained by Le Chatelier's Principle: <em>If a system at equilibrium suffers a perturbation (in temperature, pressure, concentration), the system will shift its equilibrium position to counteract such perturbation</em>.
In this case, we have an exothermic reaction (ΔH° < 0). We can imagine heat as one of the products. If we decrease the temperature, the system will try to raise it favoring the forward reaction to release heat and, at the same time, increasing the yield of H₂. By having more products, the value of the equilibrium constant K increases.
