Answer:
0.0983 M
Explanation:
First, we need to find the formulas of the reactants. Potassium forms the ion K⁺, and iodide is the ion I⁻, thus potassium iodide is KI. Silver forms the ion Ag⁺, and nitrate is the ion NO₃⁻, thus silver nitrate is AgNO₃. In the reaction, the cations will be replaced:
KI(aq) + AgNO₃(aq) → KNO₃(aq) + AgI(s)
AgI is an insoluble salt, so it will precipitate, and all nitrates are soluble, thus KNO₃ will be in the ionic form: K⁺ and NO₃⁻. 1 mol of KNO₃ = 1 mol of K⁺.
The molar mass of KI is 166 g/mol, thus the number of moles that is added is:
nKI = mass/molar mass
nKI = 5.71/166 = 0.0344 mol
And the number of moles of AgNO₃ is given as 64mM = 0.064 mol. Because the stoichiometry is 1:1, AgNO₃ is in excess, thus, all the KI will react and form 0.0344 mol of KNO₃. So, nK⁺ = 0.0344 mol. The molarity is the number of moles divided by the volume (350 mL = 0.350 L):
0.0344/0.350 = 0.0983 M
Answer:
The options a and b are correct
Explanation:
This options provided to the question are the answers to the question. But for clarification. When the fat melts, the change of state that occurs here is from solid to liquid (which is called melting) while the change of state that occurs when the water evaporated is from liquid to gas (which is called evaporation).
The two samples don’t contain different atoms so it will be false
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!!
