An occluded front forms when a warm air mass is caught between two cooler air masses.
Answer:
<h2>Industry </h2>
Explanation:
because industry is a building not a weather or related on nature
The amount of Al2O3 in moles= 1.11 moles while in grams = 113.22 grams
<em><u>calculation</u></em>
2 Al + Fe2O3 → 2Fe + Al2O3
step 1: find the moles of Al by use of <u><em>moles= mass/molar mass </em></u>formula
= 60.0/27= 2.22 moles
Step 2: use the mole ratio to determine the moles of Al2O3.
The mole ratio of Al : Al2O3 is 2: 1 therefore the moles of Al2O3= 2.22/2=1.11 moles
Step 3: finds the mass of Al2O3 by us of <u><em>mass= moles x molar mass</em></u><em> </em>formula.
The molar mass of Al2O3 = (2x27) +( 16 x3) = 102 g/mol
mass is therefore= 102 g/mol x 1.11= 113.22 grams
Answer:
4.6 mol Si
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 2.8 × 10²⁴ atoms Si
[Solve] moles Si
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide [Cancel like units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
4.64962 mol Si ≈ 4.6 mol Si