For many solids<span> dissolved in </span>liquid<span> water, the </span>solubility <span>increases with </span>temperature<span>.</span>
Answer:
kf = 1.16 x 10¹⁸
Explanation:
Step 1: [Ni(H₂O)₆]²⁺ + 1en → [Ni(H₂O)₄(en)]²⁺ ΔG°1 = -42.9 kJmol⁻¹
Step 2: [Ni(H₂O)₄(en)]²⁺ + 1en → [Ni(H₂O)₂(en)₂]²⁺ ΔG°2 = -35.8 kJmol⁻¹
Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en → [Ni(en)₃]²⁺ ΔG°3 = -24.3 kJmol⁻¹
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Overall reaction: [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ ΔG°r
ΔG°r = ΔG°1 + ΔG°2 + ΔG°3
ΔG°r = -42.9 - 35.8 - 24.3
ΔG°r = -103.0 kJmol⁻¹
ΔG°r = -RTlnKf
-103,000 Jmol⁻¹ = - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf
kf = e ^(-103,000/-8.31x298)
kf = e ^41.59
kf = 1.16 x 10¹⁸
The molar concentration of the original HF solution : 0.342 M
Further explanation
Given
31.2 ml of 0.200 M NaOH
18.2 ml of HF
Required
The molar concentration of HF
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)
Titrant = NaOH(1)
Titrate = HF(2)
Input the value :
Answer:
the answer is d.
Explanation:
because the core is deepest then is the mantle and the lithpsphere is last .
I put a picture incase you need a visual example.
The equation is:
Ca(OH)₂(s) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂<span>O(l)
</span>
n=mass in g/M.M
15 g Ca(OH)₂ is n=15 g/ 74.1 g/mol=0.2024 mol of Ca(OH)₂
no. of mol of HCl:
n=0.5 mol/L*0.075L=0.0375 mol
This could react with 0.0375/2= 0.01875 mol of Ca(OH)₂ We have a lot more than that.
Therefore, HCl is the limiting reagent and determines how much CaCl₂ forms.
Based on the balanced reaction, 2 moles of HCl gives 1 mole of CaCl₂
no. of mol of CaCl₂= 0.0375/2= 0.01875 mol
mass in g=n*MM= 0.01875*111= 2.08 g
