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:
2.0202 grams
Explanation:
1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.
so ?g glucose = 144.3 mL soln
Now apply the conversion factor, and you have:
?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).
so you have (144.3x1.4/100) g glucose= 2.0202 grams
The subscriot 2 means that in the formula there are two parts of K, and the subscript 1 (implicit) for S, indicates that there is one part of S.
This is, the formula gives the ratio of the elements K and S in the compound, which is:
2 atoms of K : 1 atom of S.
Answer: there are 2 atoms of K and 1 atom of S in a molecule of K2S.
Answer:
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<em>Hope That Helps!</em>
a. KCl = strong electrolyte
b. CCl4 = non-electrolyte
c. LiCl = strong electrolyte
d. Na2SO4 = strong electrolyte
