Answer:
1.82 L
Explanation:
We are given the following information;
- Initial volume as 2.0 L
- Initial temperature as 60.0°C
- New volume as 30.0 °C
We are required to determine the new volume;
From Charles's law;

Where,
are initial and new volume respectively, while
are initial and new temperatures respectively;



Rearranging the formula;


Therefore, the new volume that would be occupied by the gas is 1.82 L
Answer:
The object at 50°C will have a higher kinetic energy.
Explanation:
Temperature is a measure of the average kinetic energy of the particles in an object. As you introduce more energy into the system (e.g. heat the object), the particles on average move faster because they have more kinetic energy.
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.
Empirical formula is the simplest ratio of components making up a compound.
The percentage composition of each element has been given
therefore the mass present of each element in 100 g of compound is
B N H
mass 40.28 g 52.20 g 7.53 g
number of moles
40.28 g / 11 g/mol 52.20 g / 14 g/mol 7.53 g / 1 g/mol
= 3.662 mol = 3.729 mol = 7.53 mol
divide the number of moles by the least number of moles, that is 3.662
3.662 / 3.662 3.729 / 3.662 7.53 / 3.662
= 1.000 = 1.018 = 2.056
the ratio of the elements after rounding off to the nearest whole number is
B : N : H = 1 : 1 : 2
therefore empirical formula for the compound is B₁N₁H₂
that can be written as BNH₂