Hey there!
There are 6.022 x 10²³ particles in 1 mole.
We have 3.55 moles.
6.022 x 10²³ x 3.55
2.14 x 10²⁴
This is how many particles we have.
NaCl has 2 atoms for every unit.
So we multiply this by 2.
2 x 2.14 x 10²⁴ = 4.28 x 10²⁴
There are 4.28 x 10²⁴ atoms in 3.55 moles of sodium chloride.
Hope this helps!
Answer:
3grams
Explanation:
The reaction for the production of Magnesium dioxide will be
Mg + O2 → MgO
we have 5g of MgO (molar mass 40g)
no of moles of MgO = 5/40 = 0.125
Using unitary method we have
1 mole of Mg require 1 mole of MgO
0.125 Mole of MgO = 0.125mole of Mg
n = given mass /molar mass
0.125 = mass / molar mass
mass = 0.125* 24 = 3grams
Answer:
20%
Explanation:
mass by mass percentage of a solution =(mass of solute)/(mass of solution)
mass of solute=550g
therefore 110×100/550=20%
hope u will understand .:") credit to the owner
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:

Explanation:
Given that:-
Moles of K = 1.10 moles
Moles of Te = 0.55 moles
Moles of O = 1.65 moles
Taking the simplest ratio of the moles of the elements as:-
K : Te : O
1.10 : 0.55 : 1.65
2 : 1 : 3
Empirical formulas is the simplest or reduced ratio of the elements in the compound. So, the empirical formula is:- 