the two process that occur in a cell are
oxidation: this is loss of electron by electrode. the metal electrode loaes electrons and get oxidized and forms ions
the ions get migrated to solution
Reduction: here the ions present in solution gains electron and get deposited on electrodes.
so gain of electrons is by ions
electrode gains electrons is where reduction occurs, and the half cell in which the electrode loses electrons is where oxidation occurs.
Answer:
I believe it is a hope this helps
Complete Question:
A chemist prepares a solution of silver (I) perchlorate (AgCIO4) by measuring out 134.g of silver (I) perchlorate into a 50.ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the silver (I) perchlorate solution. Round your answer to 2 significant digits.
Answer:
13 mol/L
Explanation:
The concentration in mol/L is the molarity of the solution and indicates how much moles have in 1 L of it. So, the molarity (M) is the number of moles (n) divided by the volume (V) in L:
M = n/V
The number of moles is the mass (m) divided by the molar mass (MM). The molar mass of silver(I) perchlorate is 207.319 g/mol, so:
n = 134/207.319
n = 0.646 mol
So, for a volume of 50 mL (0.05 L), the concentration is:
M = 0.646/0.05
M = 12.92 mol/L
Rounded to 2 significant digits, M = 13 mol/L
Answer:
Atoms of sulfur = 9.60⋅g32.06⋅g⋅mol−1×6.022×1023⋅mol−1
Explanation:
because the units all cancel out, the answer is clearly a number, ≅2×1023 as required.
