Answer:
0.0745 mole of hydrogen gas
Explanation:
Given parameters:
Number of H₂SO₄ = 0.0745 moles
Number of moles of Li = 1.5107 moles
Unknown:
Number of moles of H₂ produced = ?
Solution:
To solve this problem, we have to work from the known specie to the unknown one.
The known specie in this expression is the sulfuric acid, H₂SO₄. We can compare its number of moles with that of the unknown using a balanced chemical equation.
Balanced chemical equation:
2Li + H₂SO₄ → Li₂SO₄ + H₂
From the balanced equation;
Before proceeding, we need to obtain the limiting reagent. This is the reagent whose given proportion is in short supply. It determines the extent of the reaction.
2 mole of Li reacted with 1 mole of H₂SO₄
1.5107 mole of lithium will react with
= 0.7554mole of H₂SO₄
But we were given 0.0745 moles,
This suggests that the limiting reagent is the sulfuric acid because it is in short supply;
since 1 mole of sulfuric acid produced 1 mole of hydrogen gas;
0.0745 mole of sulfuric acid will produce 0.0745 mole of hydrogen gas
I love these type of questions :)
Final Answer: Neon
The melting and boiling point depend on the strength of the ihydrogen bonds. Hydrogen bonding will cause the higher the melting and boiling points because more energy is needed to break bonds between molecules.
Hydrogen bonds affect solubility in water, molecules with hydrogen bonds dissolve better in water.
Answer:
n = Initial volume/22.4L
Explanation:
The molar concept is simply one that is used to find the Number of moles and explain the relationship it has with avogadro's number, molecular mass, molar mass e.t.c.
Now, in terms of molar mass, number of moles is given by the formula;
n = mass of the sample/molar mass
In terms of avogadro's number, number of moles is;
1 mole = avogadro's number = 6.02 × 10^(23)
Now, when dealing with ideal gases, the molar volume of an ideal gas is 22.4 L.
Now the relationship between this volume and the mole concept is that the number of moles is gotten by dividing the initial volume by this molar volume.
Thus;
n = Initial volume/22.4L