To determine the moles in 40 grams of magnesium, we need the atomic weight. This can easily be found on a periodic table. For this problem, let's use 24.305 grams/mole.
We are going to set up an equation to determine this problem. In this equation, we want all our units to cancel out except for 'moles.'
In this, we can see that the unit 'grams' will cancel out to leave us with moles.
In solving the equation, we determine that there are approximately 1.65 moles of Magnesium.
Answer:
83.20 g of Na3PO4
Explanation:
1 mole of Na3PO4 contains 3 moles of Na+.
Mole of Na ion to be prepared = Molarity x volume
= 0.700 x 725/1000
= 0.5075 mole
If 1 mole of Na3PO4 contains 3 moles of Na ion, then 0.5075 Na ion will be contained in:
0.5075/3 x 1 = 0.1692 mole of Na3PO4
mole of Na3PO4 = mass/molar mass = 0.1692
Hence, mass of Na3PO4 = 0.1692 x molar mass
= 0.1692 x 163.94
= 83.20 g.
83.20 g of Na3PO4 will be needed.
The mass of Zr deposited in the process is 41.4 g.
<h3>What is electrolytic cell?</h3>
An electrolytic cell is a chemical cell which produces electrical energy by non-spontaneous chemical processes.
From the question;
Zr^4+(aq) + 4e ------> Zr(s)
We know that;
91 g of Zr is deposited by 4(96500) C
xg of Zr is deposited by (7.92 × 6.16 × 60 × 60) C
xg = 91 g × (7.92 × 6.16 × 60 × 60) C/4(96500) C
x g = 41.4 g
Learn more about electrolysis: brainly.com/question/12054569
