N oxidation number in N₂O = 1
S oxidation number in SO₂ = 4
S oxidation number in SO₃ = 6
P oxidation number in P₄O₆ = 3
Therefore, the Sulfur in SO₃ will not react with molecular oxygen as Sulfur is already using all of its valence electrons in bonding.
Answer: option D) energy was absorbed and entropy increased.
Explanation:
1) Given balanced equation:
2H₂O (l) + 571.6 kJ → 2 H₂ (g) + O₂(g).
2) Being the energy placed on the side of the reactants means that the energy is used (consumed or absorbed). This is an endothermic reaction.
So, the first part is that energy was absorbed.
3) As for the entropy, it is a measure of the disorder or radomness of the system.
Since, two molecules of liquid water were transformed into three molecules of gas, i.e. more molecules and more kinetic energy, therefore the new state has a greater degree of radomness, is more disordered, and you conclude that the entropy increased.
With that, you have shown that the right option is D) energy was absorbed and increased.
Answer:
The answer to your question is below
Explanation:
Trial 1 Trial 2
mass of Mg 0.255 g 0.353 g
mass of MgO 0.418 g 0.576 g
Chemical reaction
2Mg(s) + O₂(g) ⇒ 2MgO(s)
Question 1.
Atomic mass of Mg = 24.31 x 2 = 48.62 g
Molecular mass of MgO = 2(24.31 + 16) = 80.62 g
Trial 1
48.62 g of Mg ----------------- 80.62 g of MgO
0.255 g ---------------- x
x = (0.255 x 80.62)/48.62
x = 0.422 g of MgO
Trial 2 48.62 g of Mg ----------------- 80.62 g of MgO
0.353 g ---------------- x
x = (0.353 x 80.62)/48.62
x = 0.585 g of MgO
Question 2
Trial 1
Percent yield = 0.418/0.422 x 100 = 99%
Trial 2
Percent yield = 0.576/0.585 x 100 = 98.5%
Question 3
Average = (99 + 98.5)/2
= 98.75%
