<h3>
Answer:</h3>
0.127 mol Au
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 25.0 g Au
[Solve] moles Au
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.126923 mol Au ≈ 0.127 mol Au
Answer:
If there reacted 1.5 moles of O2, there will be produced 1.0 mol of Fe2O3
Explanation:
Step 1: Data given
Number of moles oxygen reacted = 1.5 moles
Step 2: The balanced equation
4Fe + 3O2 → 2Fe2O3
Step 3: Calculate moles of Fe2O3
For 4 moles Fe consumed, we need 3 moles of O2 to produce 2 moles of Fe2O3
For 1.5 moles O2 consumed, we'll have 2/3 * 1.5 = 1.0 mol of Fe2O3
If there reacted 1.5 moles of O2, there will be produced 1.0 mol of Fe2O3
Answer:
Magnesium
0.003mole
Explanation:
The problem here entails we find the metal in the carbonate.
For group 2 member, let the metal = X;
The carbonate is XCO₃;
If we sum the atomic mass of the elements in the metal carbonate, we should arrive at 84g/mol
Atomic mass of C = 12g/mol
O = 16g/mol
Atomic mass of X + 12 + 3(16) = 84
Atomic mass of X = 84 - 60 = 24g/mol
The element with atomic mass of 24g is Magnesium
B.
Number of moles in 0.3g of CaCO₃:
Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100g/mol
Number of moles = 
Number of moles = 
= 0.003mole
Answer:
The process of elemental stratification relies on the diffusion velocity, which causes the migration of the different chemical elements within stars.
Explanation:
