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:
The specific heat of sodium is 1,23J/g°C
Explanation:
Using the atomic weight of sodium (23g/mol) and the atomic weight definition, we have that each mole of the substance has 23 grams of sodium.
starting from this, we use the atomic weight of sodium to convert the units from J / mol ° C to J / g ° C
This is just addition. Put 2140.56 on top, line up 9.3456 under it appropriately. Doing this will give you the answer: 2149.9056
Answer:
<em>Mg = 24.30 g/mol) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) Hint: 1 mole of gas at STP occupies 22.4 L</em>
I think is False! Because it is not example of a disaccharide it is not saccharin-aspartame molecule.
