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:
Tests for gases
Hydrogen, oxygen, carbon dioxide, ammonia and chlorine can be identified using different tests.
Hydrogen. A lighted wooden splint makes a popping sound in a test tube of hydrogen.
Oxygen. A glowing wooden splint relights in a test tube of oxygen.
hope it will help
BaO, Barium Oxide.
Na2SO4, Sodium Sulfate.
CuO, Copper (II) Oxide.
P2O5, Diphosphorus Pentoxide.
HNO3, Nitric Acid.
CO32-, Molecular Formula.
Hope this helps. :)
<h2>Giant impact and metalcore.</h2>
Explanation :
- Mercury has a large core of liquid metal.
- The metal core is of iron metal.
- The core is surrounded by a mantle which is made up of silica and a solid outer crust.
- In the case of Mercury, the total core percentage is 42% of the planet while the Earth's core is only 17% of the planet.
- Mercury has lost part of its mantle and crust that left the mercury with a large metal core.
We know that, M1V1 = M2V2
(Initial) (Final)
where, M1 and M2 are initial and final concentration of soution respectively.
V1 and V2 = initial and final volume of solution respectively
Given: M1 = 12 m, V1 = 35 ml and V2 = 1.2 l = 1200 ml
∴ M2 = M1V1/V2 = (12 × 35)/ 1200 = 0.35 m
Final concentration of solution is 0.35 m