If 34.7 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be for
1 answer:
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Answer : The balanced half-reaction in a basic solution will be,
Explanation :
Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.
Rules for the balanced chemical equation in basic solution are :
First we have to write into the two half-reactions.
Now balance the main atoms in the reaction.
Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion at that side where the less number of hydrogen are present.
Now balance the charge.
- The half reaction is :
- Now balance the oxygen atoms.
- Now balance the hydrogen atoms.
- Now balance the charge.
The balanced half-reaction in a basic solution will be,
The equilibrium constant of a reaction is defined as:
"The ratio between equilibrium concentrations of products powered to their reaction quotient and equilibrium concentration of reactants powered to thier reaction quotient".
The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.
To solve this question we need this additional information:
<em>For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:</em>
<em>[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M</em>
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Thus, for the reaction:
N₂ + 3H₂ ⇄ 2NH₃
The equilibrium constant, K, of this reaction, is defined as:
And Q, is:
Where actual concentrations are:
[NH₃] = 1.0x10⁻⁴M
[N₂] = 4.0M
[H₂] = 2.5x10⁻¹M
Replacing:
As Q < K,
<h3>The chemical system will shift to the right in order to produce more NH₃</h3>Learn more about chemical equililbrium in:
Answer:
The mass of the solute and the volume of the solution.
Explanation:
Hello,
In this case, given the formula of molarity:
In such a way, since the moles could not be directly measured, we must measure the mass of the solute and by using its molar mass, one could compute its moles. Moreover, since the solution is composed by the solvent (typically water) and the solute, we consequently must measure the volume of the solution needed for the preparation of such concentration-known solution. In such a way, we can actually prepare the required solution.
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