Answer: (C) conservation of matter
Solution: Law of conservation of matter or mass states that' total mass of the reactants should always be equal to the total mass of the product that is the total mass is remained conserved in a chemical reaction.
A balanced chemical equation always follow this law.
For example:
Mass of hydrogen = 1 g/mol
Mass of Oxygen = 16 g/mol
Total mass on the reactants = 2(2×1)+(2×16)= 36g/mol
Total mass on the product side = 2[(2×1) +16] = 36 g/mol
As,
Mass on reactant side = Mass on the product side
Therefore, a balanced chemical reaction follows Law of Conservation of mass.
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](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.80V)
![E^0_{[Cu^{2+}/Cu]}=+0.34V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D%2B0.34V)
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) : 
Reduction half reaction (Cathode) : 
Thus the overall reaction will be,
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.
Answer: Microscope
Explanation: This is a kid answering :)
Answer:
See explanation
Explanation:
The shorthand nuclear reaction equations have been given; the first particle in the parentheses is a reactant particle while the second particle is a product particle. These can now be rewritten as the longhand equations as follows;
238/92U + 4/2 He -------> 241/94Pu + 1/0 n
238/92U + 4/2 He ------> 241/94Pu + 1/0 n
14/7N + 4/2 He------> 17/8O + 1/1 p
56/26Fe + 2 4/2 He----> 60/29Cu + 4/2 He
