I’m guessing it would be A because the river discharge tends to gather near coastlines where the river ends
Answer: Complete ionic equations dissociate all aqueous solutions into ions. Net ionic equations show the change that occurs in chemical reactions and do not show spectator ions that are the same in reactants and products. (b) If no spectator ions were present then complete and net ionic equations would be identical.
Explanation:
M(H₂O) = 97,2 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 97,2 g ÷ 18 g/mol.
n(H₂O) = 5,4 mol.
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 5,4 mol · 6,023·10²³ 1/mol.
N(H₂O) = 3,25·10²⁴ molecules of water.
n - amount of substance.
Na - Avogadro number.
Answer:
1. 6.116x1024 Molecules of H2O
2. 13400 L
3. 8.001x1024 Molecules of Mg3(PO4)2
4. 572 g.
5. 1.017x1025 Molecules of N2
6. 7.24 g
.7. 6980 g. of Al(OH)3
8. 3H2 + N2 => 2NH3
9. S8 + 8O2 => 8SO2
10. Ni(ClO3)2→ NiCl2 + 3O2
11. C2H4 + 3O2→ 2CO2 + 2H2O
12. 2KClO3→ 2KCl + 3O2
13. Cu(OH)2 + 2HC2H3O2→ Cu(C2H3O2)2 + 2H2O
14. C3H8 + 5O2→ 3CO2 + 4H2O
15. 191 g of CO
Answer:
19 g
Explanation:
Data Given:
Sodium Chloride (table salt) = 50 g
Amount of sodium (Na) = ?
Solution:
Molecular weight calculation:
NaCl = 23 + 35.5
NaCl = 58.5 g/mol
Mass contributed by Sodium = 23 g
calculate the mole percent composition of sodium (Na) in sodium Chloride.
Since the percentage of compound is 100
So,
Percent composition of sodium (Na) = 23 / 58.5 x 100
Percent of sodium (Na) = 39.3 %
It means that for ever gram of sodium chloride there is 0.393 g of Na is present.
So,
for the 50 grams of table salt (NaCl) the mass of Na will be
mass of sodium (Na) = 0.393 x 50 g
mass of sodium (Na) = 19 g
