Answer:
0.558mole of SO₃
Explanation:
Given parameters:
Molar mass of SO₃ = 80.0632g/mol
Mass of S = 17.9g
Molar mass of S = 32.065g/mol
Number of moles of O₂ = 0.157mole
Molar mass of O₂ = 31.9988g/mol
Unknown:
Maximum amount of SO₃
Solution
We need to write the proper reaction equation.
2S + 3O₂ → 2SO₃
We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.
So we simply compare the molar relationship between sulfur and product formed to solve the problem:
First, find the number of moles of Sulfur, S:
Number of moles of S = 
Number of moles of S = 
= 0.558mole
Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:
2 mole of Sulfur produced 2 mole of SO₃
Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃
Answer;
The above statement is true
upon heating a copper sample will expand, leading to a lower density
Explanation;
-The density of solids decreased with increase in temperature and vice versa. The increase in temperature causes the volume of the solid to increase which as a result decreases the density as Density=Mass/Volume. The temperature of a body is the average kinetic energy of the molecules present in it.
In other words; The temperature of a body is the average kinetic energy of the molecules present in it. Therefore; when heat is supplied ( or temperature is increased) the average kinetic energy increases which increases the volume and thus density decreases.
Answer:
No, compound A and B are not the same compound
Explanation:
According to the law of definite proportion "every chemical compound contains fixed and constant proportions (by mass) of its constituent elements." (Encyclopedia Britannica)
We can see in the question that the ratio of flourine to sulphur in compound A is 1.18 while the ratio of flourine to sulphur in compound B is 2.37.
The two chemical compounds do not contain a fixed proportion by mass of their constituent elements therefore, they can not be same compound according to the law of definite proportions.
to neutralize 1 mole of H2 S o4 we need one mole of any if we are having 50 grams of H2 S o4 it means the mole of H2 S o4 in 50 gram will be 50×40)/98 hence
utilising 50 grams of H2 S o4 we need approximately 20. 5 gm of Naoh
