423 psi
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Answer is: n (number of fluorine atoms) is 6, formula of the compound is XeF₆.
m(F) = 0.8682 g; mass of fluorine.
n(F) = m(F) ÷ M(F).
n(F) = 0.8682 g ÷ 19 g/mol.
n(F) = 0.0457 mol; amount of substance.
m(Xe) = 1.00 g.
n(Xe) = 1.00 g ÷ 131.293 g/mol.
n(Xe) = 0.00761 mol.
n(Xe) : n(F) = 0.00761 mol : 0.0457 mol.
n(Xe) : n(F) = 1 mol : 6 mol.
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!!
To figure this out, you need to figure out how many molecules of mgso4 you have.
If you have 1.04 moles of MgSO4, you have that many moles times avogadro's number (6.022*10^23 molecules/mole).
1.04 moles MgSO4 * (6.022*10^23 molecules MgSO4)/1 mole MgSO4 =
6.263 *10^23 molecules MgSO4
Next, there are 4 oxygen atoms per molecule of MgSO4, so we multiply the number of molecules of MgSO4 by 4 to get the number of oxygen atoms.
6.263 *10^23 molecules MgSO4 * (4 oxygen atoms/1 MgSO4 molecule) = 2.51 * 10^24 oxygen atoms in 1.04 moles of MgSO4
