The molecular weight for Calcium Chloride Dihydrate (CaCl2•2H20) is 2H2O.
mass of 64g/mol = 256.52 g/mol
mass of 11g/mol = 26.981538 g/mol
mass of 147g/mol= 146.9149 g/mol
258g/mol = 258.09843 g/mol
Find the pH using the concentration of hydrogen ions [H+] of 0.001M (molarity) by taking the negative log of the value:
-log[H+]=pH
-log[0.001] = pH
-log[1x10^-3]=pH
3=pH
I dont get it someone please help
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!!
<span>Missing question: The first-order rate
constant for the decomposition of N2O5, 2N2O5(g)→4NO2(g) + O2(g) at 70°</span><span>C is 6.82×10−3 s−1. Suppose we start with 2.70×10−2 mol of
N2O5(g) in a volume of 1.8 L .
</span>c₀(N₂O₅) = 0,027 mol ÷ 1,8 L.<span>
c</span>₀(N₂O₅) =
0,015 mol/L.<span>
c(N</span>₂O₅) = 0,019 mol/ 1,8 L = 0,01055 mol/L.<span>
k = 6,82·10</span>⁻³ s⁻¹.<span>
ln c(N</span>₂O₅) =
ln c₀(N₂O₅) -
k·t.<span>
t = (ln c</span>₀(N₂O₅) - ln c(N₂O₅)) ÷ k.<span>
t = 0,35 ÷ 0,00682 s</span>⁻¹.<span>
t = 51 s = 0,86 min.</span>
