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.
Answer:
1 - 3
Explanation:
- Look to see where metals are on the periodic table then look at what group they are in. The group tells you the number of valence electrons. Ex. 1A has one valence electron.
- Hope this helped! If you need a further explanation please let me know.
Final volume is 400 mL
<span>The moles in MgSO4 is 0.00788 </span><span>mL
</span>
The new concentration is 0.197
The molarity of Barium Hydroxide is 0.289 M.
<u>Explanation:</u>
We have to write the balanced equation as,
Ba(OH)₂ + 2 HNO₃ → Ba(NO₃)₂ + 2 H₂O
We need 2 moles of nitric acid to react with a mole of Barium hydroxide, so we can write the law of volumetric analysis as,
V1M1 = 2 V2M2
Here V1 and M1 are the volume and molarity of nitric acid
V2 and M2 are the volume and molarity of Barium hydroxide.
So the molarity of Ba(OH)₂, can be found as,

= 0.289 M
Your drawing looks so good :p