The atoms of hydrogen that are present in 7.63 g of ammonia(NH3)
find the moles of NH3 =mass/molar mass
7.63 g/ 17 g/mol = 0.449 moles
since there is 3 atoms of H in NH3 the moles of H = 0.449 x 3 = 1.347 moles
by use of 1 mole = 6.02 x10^23 atoms
what about 1.347 moles
= 1.347 moles/1 moles x 6.02 x10^23 atoms = 8.11 x10^23 atoms of Hydrogen
collect this money and use it to finance social projects
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The answer to this is Codon.
I think it’s :answer choice c
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