A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
will be needed to produce 725 mL of a solution that has a concentration of Na ions of 0.700 M
2 answers:
Explanation:
Below is an attachment containing the solution.
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.
You might be interested in
Answer :
Explanation:
- (b) a ligand is an ion or molecule that binds to the central metal atom to form a cordinate complex.
- (c)
<h3> above</h3>
Answer:
i cant see the whats in the picture
1.785714286 moles
The number of moles (n) for nitrogen is: [ n=50.0÷28.0 ] = 1.785714286 moles.
Answer:
Cl-
Explanation:
Neutral (Cl) have 7 electron, but Cl- have 8 electron due to gain of 1 electron
