Answer: 0.172 M
Explanation:
a) To calculate theconcentration of base, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
b) To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
The concentration of the phosphoric acid solution is 1.172 M
Mg(s) + 2 AgNO₃(aq) = Mg(NO₃)₂(aq) + 2 Ag(s)
reaction is :<span> single replacement
hope this helps!</span>
Answer:
(a) X electrode
(b) Y electrode
(c) Y electrode
(d) X electrode
(e) Y electrode
Explanation:
<em>A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y, that is, X more readily reacts to form a cation than Y does.</em>
In the X electrode occurs the oxidation whereas in the Y electrode occurs the reduction.
Oxidation: X(s) → X⁺ⁿ(aq) + n e⁻
Reduction: Y⁺ˣ(aq) + x e⁻ → Y(s)
<em>Classify the descriptions by whether they apply to the X or Y electrode.
</em>
<em>(a) anode.</em> Is where the oxidation takes place (X electrode).
<em>(b) cathode.</em> Is where the reduction takes place (Y electrode).
<em>(c) electrons in the wire flow toward.</em> Electrons in the wire flow toward the cathode (Y electrode).
<em>(d) electrons in the wire flow away.</em> Electrons in the wire flow away from the anode (X electrode).
<em>(e) cations from salt bridge flow toward.</em> Cations from the salt bridge flow toward the cathode (Y electrode) to maintain the electroneutrality.
