Answer:
0.135 mole of H2.
Explanation:
We'll begin by calculating the number of mole in 3.24 g of Mg. This can be obtained as follow:
Mass of Mg = 3.24 g
Molar mass of Mg = 24 g/mol
Mole of Mg =?
Mole = mass /Molar mass
Mole of Mg = 3.24/24
Mole of Mg = 0.135 mole
Next, we shall write the balanced equation for the reaction. This is illustrated below:
Mg + 2HCl —> MgCl2 + H2
From the balanced equation above,
1 mole of Mg reacted to produce 1 mole of H2.
Finally, we shall determine the number of mole of H2 produced by reacting 3.24 g (i.e 0.135 mole) of Mg. This can be obtained as follow:
From the balanced equation above,
1 mole of Mg reacted to produce 1 mole of H2.
Therefore, 0.135 mole of Mg will also react to produce 0.135 mole of H2.
Thus, 0.135 mole of H2 can be obtained from the reaction.
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I guess is weak alkaline. when the substance is more acidic, there will be less alkalinity
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
Answer:
The reaction would shift toward the reactants
When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm
Explanation:
For the reaction:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Where K is defined as:
As initial pressures of all 3 gases is 1.0atm, reaction quotient, Q, is:
As Q > K, <em>the reaction will produce more NH₃ until Q = K consuming N₂ and H₂.</em>
Thus, there are true:
<h3>The reaction would shift toward the reactants</h3><h3>When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm</h3>
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