Answer:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
Explanation:
Let's consider the notation of a galvanic cell.
Cr(s) | Cr³⁺(aq) || Ag⁺(aq) | Ag(s)
On the left, it is represented the anode (oxidation) and on the right, it is represented the cathode (reduction).
The half-reactions are:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
To have the global reaction, we have to multiply the reduction by 3 (so the number of electrons gained and lost are the same) and add both half-reactions.
Global reaction: Cr(s) + 3 Ag⁺(aq) ⇒ Cr³⁺(aq) + 3 Ag(s)
Answer:
c yun Lang
Explanation:
<h2>#carryonlearning</h2>
Explanation:
(a) Molar mass of is 691.92 g/mol. And, there is 0.76 g of sodium monofluorophosphate () in 100 ml.
Molar mass of fluorine is 19 g/mol.
This means that in 691.92 g/mol there are 19 g/mol of fluorine is present. Hence, mass of fluorine present in 0.76 g is calculated as follows.
Mass of fluorine =
= 0.021 g
As, 1 g = 1000 mg. Hence, 0.021 g = 21 mg.
Therefore, mass of florine atoms in present was 21 mg.
(b) As we know that number of moles equal mass divided by molar mass.
Therefore, No. of moles of fluorine =
=
= 0.001 mol
Hence, according to mole concept in one mole there are atoms.
So, in 0.001 mole number of fluorine atoms will be calculated as follows.
atoms.
= atoms
Therefore, there are atoms of fluorine were present.
1. Go to periodic table and find<span> the atomic </span>mass<span> (top little number) of each element.
2. Multiply each atomic </span>mass<span> by the number of atoms in the </span>formula<span>.
3. Add up the results together. </span>