Select all that are true
1 answer:
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Answer:
Explanation:
We have three equations:
1. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ
2. S(s, rhombic) + O₂(g) ⟶ SO₂(g); ∆H = -296.8 kJ
3. PbO(s) + H₂S(g) ⟶ PbS(s) + SO₂(g); ∆H = -104.3 kJ
From these, we must devise the target equation:
4. 2PbS(s) + 3O₂(g) ⟶2PbO(s) + 2SO₂(g); ΔH = ?
The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.
When you reverse an equation, you reverse the sign of its ΔH.
When you double an equation, you double its ΔH.
5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ
Equation 5 has 2H₂O on the left. That is not in the target equation.
You need an equation with 2H₂O on the right, so you copy Equation 1.
6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ
Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.
You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.
7. 2S(s, rhombic) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ
Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.
When you add equations, you add their ΔH values.
You get the target equation 4:
5. 2PbS(s) + <u>2H₂O(g</u>) ⟶ 2PbO(s) + <u>2H₂S(g</u>); ∆H = 208.6 kJ
6. <u>2H₂S(g)</u> + O₂(g) ⟶ <u>2S(s</u>) + <u>2H₂O(g)</u> ; ∆H = -442.4 kJ
<u>7</u><u>. </u><u>2S(s)</u><u> + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ </u>
4 . 2PbS(s) + 3O₂(g) ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ
Answer:
Hi.
The temperature is approximately zero degrees (0°C)
Explanation:
It is important to keep in mind that in the production of ice cream the decrease in the freezing point of the water present in the mixture is called the antifreeze power of the mixture. In ice cream, the freezing point decrease will be caused by each substance that is dissolved in the mixture: lactose, salts, sugars and any other substance. Each of these substances will contribute to the decrease in the freezing point of the mixture. The phase diagram attached in the file shows the sugar solutions in water. When a solution cools (point A), there comes a time when the freezing curve is reached (point B). At that moment ice begins to appear. As shown in the diagram this temperature is approximately zero degrees (0 ° C).
