You must use 1880 mL of O₂ to react with 4.03 g Mg.
A_r: 24.305
2Mg + O₂ ⟶ 2MgO
<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg
<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂
STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.
<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL
Answer:
Explanation:
Ksp(BaSO4)=1.07×10−10
BaSO₄ → Ba²⁺ + SO₄²⁻
1.07×10⁻¹⁰ = ( Ba²⁺) × ( SO₄²⁻)
but Ba²⁺ = 1.3×10⁻² M
1.07×10⁻¹⁰ = 1.3×10⁻² M × ( SO₄²⁻)
( SO₄²⁻) = 1.07×10⁻¹⁰ / 1.3×10⁻² = 0.823 × 10⁻⁸ M
while Ksp(CaSO4)=7.10×10−5
CaSO₄ → Ca²⁺ + SO₄²⁻
7.10×10⁻⁵ = 2.0×10⁻² × ( SO₄²⁻)
( SO₄²⁻) = 7.10×10⁻⁵ / 2.0×10⁻² = 3.55 × 10⁻³ M
comparing the concentration of sulfate ions, Ba²⁺ cation will precipitate first because the Ba²⁺ requires 0.823 × 10⁻⁸ M sodium sulfate which less compared the about needed by CaSO₄
The rows are called Periods.
Isotope - the same element, but different atomic mass so 1)35 Cl and 37 Cl
Hi, P.S. The atomic number will always be the same as the number of protons...
