An equation with the equal amount and proportion of atoms of each element on both sides of the reaction is commonly referred to as a balanced chemical equation.

The law of conservation of matter asserts that no observable and empirical change in the amount of matter occurs within a conventional chemical process. As a result, each element in the product would have the same equal amount or numbers of atoms as the reactants.

5 0

3 years ago

If 2.0 mol of Zn is mixed with 3.0 mol<br> of HCl, which reactant will be limiting?

Soloha48 [4]

The following Balanced Reaction will take place:

Zn + 2HCl → ZnCl₂ + H₂

In the question, we have 2 moles of Zinc and 3 moles of HCl for this reaction

<u>Amount of HCl required to completely react with 2 moles of Zn:</u>

Since we need 2 moles of HCl for every mole of Zn, we will need 2(2) = 4 moles of HCl for every 2 moles of Zn

<u>Identifying the Limiting Reagent:</u>

But we are only given 3 moles of HCl where we need 4 moles to completely react.

So, since HCl is in less amount, it is the Limiting Reagent

8 0

2 years ago

A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Cu(s) and Cu2+(aq). W

Agata [3.3K]

Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

We are taking the value of standard reduction potential form the standard table.

$E^0_{[Ag^{+}/Ag]}=+0.80V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.

The balanced two-half reactions will be,

Oxidation half reaction (Anode) : $Cu(s)\rightarrow Cu^{2+}(aq)+2e^-$

Reduction half reaction (Cathode) : $Ag^{+}(aq)+e^-\rightarrow Ag(s)$

Thus the overall reaction will be,

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

7 0

3 years ago

Solve

stiv31 [10]

The volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.

<h3>How to calculate volume?</h3>

The volume of a given gas can be calculated using the following Charle's law equation:

V1/T1 = V2/T2

Where;

T1 = initial temperature
T2 = final temperature
V1 = initial volume
V2 = final volume

V1 = 9L
V2 = ?
T1 = 325K
T2 = 450K

9/325 = V2/450

325V2 = 4050

V2 = 4050/325

V2 = 12.46L

Therefore, the volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.

Learn more about volume at: brainly.com/question/2817451

7 0

2 years ago