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:
B is the answer because it doesn't matches the num of valence electrons
Answer:
The answer is: <u>Al2O3</u>
Explanation:
The data they give us is:
To find the empirical formula without knowing the grams of the compound, we find it per mole:
- 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
- 0.485 g O * 1 mol O / 16 g O = 0.03 mol O
Then we must divide the results obtained by the lowest result, which in this case is 0.02:
- 0.02 mol Al / 0.02 = 1 Al
- 0.03 mol O / 0.02 = 1.5 O
Since both numbers have to give an integer, multiply by 2 until both remain integers:
Now the answer is given correctly:
The scale of most metal characteristics goes from the bottom left-hand corner.
The least metallic is the top right-hand.
So then that means that
Calcium-YES, second column
Germanium-No, to far, in the middle
Arsenic-Non-metal,
Bromine, same for this
Calcium
