N(Mg) : n(O)
1 : 1
the ratio is one to one
<h3>a) <u>Answer;</u></h3>
A- 7.30 M
<h3><u>Explanation;</u></h3>
3.65 Sodium sulfate
Na2(SO4) dissociates to give sodium ions and sulfate ions.
Na2(SO4) → 2Na+ + SO₄²₋
Therefore, twice the concentration of sodium ions as far as molarity is concerned.
<em><u>Concentration of sodium ions = 3.65 × 2= 7.3 M</u></em>
<h3>b) <u> Answer;</u></h3>
B- 2.76 M
<h3><u>Explanation;</u></h3>
b) 1.38 M sodium carbonate
Sodium carbonate dissociates completely to yield Sodium ions and carbonate ions
Na₂CO₃ → 2Na+ + CO₃²₋
The concentration of sodium ions will be twice the concentration of initial compound since it has a ratio of two.
<em>Concentration of sodium ions = 1.38 ×2 </em>
<em> = 2.76 M</em>
<h3>
c) <u>
Answer;</u></h3>
<em>0.785 M</em>
<h3><u>
Explanation;</u></h3>
b) 0.785 sodium bicarbonate
Sodium bicarbonate dissociates completely to sodium ions and a bicarbonate ions.
NaHCO₃ →Na+ + HCO₃⁻
In this case the concentration of Na+ will be equal to the concentration of the original compound since they share the same ratio.
Thus; <em><u>Na+ concentration = 0.785 M</u></em>
Because gases are less dense than a liquid and a solid
Note that
1 mi = 1609.34 m = 1609.34 x 10⁶ um
1 h = 3600 s = 3600 x 10⁹ ns
Therefore the speed of the bug is
(3.0 mi/h)*(1609.34x10⁶ um/mi)*(1/3600x10⁹ h/ns)

Answer:
Answer:
1). 107.2L
Explanation:
The idea here is that if all the gases thaction are kept under te same conditions for pressure and temperature, then you can treat the mole ratios that exist between them as volume ratios.