Nitrogen ion (Nitride) is ... Cation/Anion/Neither? # of protons? # of electrons? Charge (1-, 2-, 3-, 1+, 2+, 3+, or 0) Number o
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:
Mass, m = 1.51 grams
Explanation:
It is given that,
The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm
Length of the cylinder, h = 4.2 cm
We know that the density of the Aluminium is 2.7 g/cm³
Circumference, C = 2πr
Density is equal to mass per unit volume.
m is mass of the cylinder
V is the volume of the cylinder
So,
So, the mass of the cylinder is 1.51 grams.