When Cr dissolves in 1 M H₂Cr₂O₇
Cr(s) ⇄ Cr³⁺ + 3 e⁻
Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ ⇄ 2 Cr³⁺ + 7 H₂O
by multiplying the first equation by 2 and add the 2 equations;
The balanced redox reaction will be:
2 Cr(s) + Cr₂O₇²⁻(aq) + 14 H⁺(aq) → 4 Cr³⁺(aq) + 7 H₂O(l)
Answer:
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Given molecule Lithium iodide (LiI)
Heat of hydration = -793 kj/mol
Lattice energy = -730 kJ/mol
Heat of hydration = Heat of solution - Lattice energy
Heat of solution = Hydration + Lattice = -793 + (- 730) = -1523 kJ/mol
Now,
Mass of LiI = 15.0 g
molar mass of LiI = 134 g/mol
# moles of LiI = 15/134 = 0.112 moles
Heat of solution for 1 mole of LiI = -1523 KJ
Therefore, for 0.112 moles of LiI the corresponding heat is
= 0.112 *(-1532) = 171.6 kJ
The concentration of the solution is 8 M
<u><em>calculation</em></u>
<em> </em>Concentrati<em>on = moles/volume in liters</em>
<em> </em><em>step 1: find moles of HF</em>
<em>moles of HF =mass/molar mass</em>
<em>molar mass of HF = 1+ 19 )= 20 g/mol</em>
<em> moles is therefore = 32.0 g/ 20 g/mol= 1.6 moles</em>
<em>Step 2: convert ml to L</em>
<em>volume in liters = 2.0 x 10^2 / 1000 =0.2 l</em>
<em>step 3: find the concentration</em>
<em>concentration = 1.6 mol / 0.2 l = </em><em>8 M</em>
