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
I think the answer would be trenches but I’m sorry if I’m wrong
Answer:
3
Explanation:
the three elements involved in this compound are Li, S, O.
lithium, sulfur, and oxygen. Which create the ionic compound, "Lithium Sulfate."
7 atoms total, since there are two lithium, four oxygen, and one sulfate atom. this is a white inorganic salt.
Answer:
Solution's mass = 200.055 g
[PbSO₄] = 275 ppm
Explanation:
Solute mass = 0.055 g of lead(II) sulfate
Solvent mass = 200 g of water
Solution mass = Solvent mass + Solution mass
0.055 g + 200 g = 200.055 g
ppm = μg of solute / g of solution
We convert the mass of solute from g to μg
0.055 g . 1×10⁶ μg/ 1g = 5.5×10⁴μg
5.5×10⁴μg / 200.055 g = 275 ppm
ppm can also be determined as mg of solute / kg of solution
It is important that the relation is 1×10⁻⁶
Let's verify: 0.055 g = 55 mg
200.055 g = 0.200055 kg
55 mg / 0.200055 kg = 275 ppm