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:
1.0 *10^(-4) mol
Explanation:
For gases:
n1/n2 = V1/V2
n1/3.8*10^(-4) mol = 230 mL/ 860 mL
n1 = 3.8*10^(-4)*230/860 = 1.0 *10^(-4) mol
It would be 1 or 2 because if the number is higher than 5 you need to round up , if its lower than 5 you need to round down.
Start by adding the numbers then divide
C, to form bonds between each monomers
