i think the answer is C3H8O
<span>If the concentration of H⁺ ions will decrease then the concentration of OH⁺ ions will increase.</span>
a. mol O₂=0.5
b. volume O₂ = 25 cm³
c. i. the total volume of the two reactants = 75 cm³
c. ii. the volume of nitrogen dioxide formed = 50 cm³
<h3>Further explanation</h3>
Reaction
2NO(gas) + O₂(gas) ⇒ 2NO₂ (gas)
a.
mol NO = 1
From the equation, mol ratio NO : O₂ = 2 : 1, so mol O₂ :

b.
From Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles
Because mol ratio NO : O₂ = 2 : 1, so volume O₂ :

c.
i. total volume of reactants : 25 cm³+ 50 cm³=75 cm³
ii. the volume of nitrogen dioxide formed :
mol ratio NO : NO₂ = 2 : 2, so volume NO₂ = volume NO = 50 cm³
Moles = 15.5 g / 40 g/mol = 0.3875 mol
M = 0.3875 mol / 0.250 L = 1.55M
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!