Answer is: 4)Mass
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Answer:
ΔS° = 180.5 J/mol.K
Explanation:
Let's consider the following reaction.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)
The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.
ΔS° = ∑np × S°p - ∑nr × S°r
where,
ni are the moles of reactants and products
S°i are the standard molar entropies of reactants and products
ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))
ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol
ΔS° = 180.5 J/K
This is the change in the entropy per mole of reaction.
Answer:
Thermo means heat/related to heat and haline means salty/of salt
Answer:
Mass = 13.23 g
Explanation:
Given data:
Mass of oxygen = 48.0 g
Mass of propane burn = ?
Solution:
Chemical equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 48.0 g/ 32 g/mol
Number of moles = 1.5 mol
now we will compare the moles of propane and oxygen.
O₂ : C₃H₈
5 : 1
1.5 : 1/5×1.5 = 0.3 mol
Mass of propane burn:
Mass = number of moles × molar mass
Mass = 0.3 mol × 44.1 g/mol
Mass = 13.23 g
