B , your products are on the right side of the reaction. The reactants are on the left side
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:
Average atomic mass = 15.86 amu.
Explanation:
Given data:
Number of atoms of Z-16.000 amu = 205
Number of atoms of Z-14.000 amu = 15
Average atomic mass = ?
Solution:
Total number of atoms = 205 + 15 = 220
Percentage of Z-16.000 = 205/220 ×100 = 93.18%
Percentage of Z-14.000 = 15/220 ×100 = 6.82 %
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (93.18×16.000)+(6.82×14.000) /100
Average atomic mass = 1490.88 + 95.48 / 100
Average atomic mass = 1586.36 / 100
Average atomic mass = 15.86 amu.
