Hey there!:
Given % of Mn=59.1% means 59.1 g of Mn present in 100 g of manganese fluoride.
Molar mass of Mn= 54.938 g/mol
Moles of Mn = mass / molar mass
59.1 /54.938 => 1.07 ≈ 1 mol.
and % of F=40.9% means 40.9 g of of F present in 100 g of manganese fluoride.
Molar mass of F=18.998 g/mol
Moles of F :
40.9 / 18.999 => 2.15 mol ≈ 2 mol.
The mole ratio between Mn:F= 1 : 2
Therefore the empirical formula of manganese fluoride:
=> MnF2=Mn1F2
Hope that helps!
Number of neutrons is 121.
Answer:
ΔH = -20kJ
Explanation:
The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:
H₂(g) + S(g) → H₂S(g)
Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:
<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
The sum of -(1) + (2) + (3) gives:
<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>
<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>
<em />
I hope it helps!
