The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
Answer:175⋅mL of the given sulfuric acid
Explanation:
First, since l = n-1,
5,4,-5,1/2 and 2,1,0,1/2 are the only answer choices left.
Next, since ml = -l to l,
2,1,0,1/2
is the answer because in 5,4,-5,1/2, the ml value of -5 is not in the range of -4 to 4, as notes by the value 4 for l.
Q = ?
Cp = 0.397 J/ºC
Δt = 40.3 - 21.0<span> => 19.3</span><span> ºC</span>
m = 15.2 g
Q = m x Cp x Δt
Q = 15.2 x 0.397 x 19.3
Q ≈ 116.46 J
<span>hope this helps! </span>
