<h3>
Answer:</h3>
28 mol CaF
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 1.7 × 10²⁵ molecules CaF
[Solve] moles CaF
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<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 2 sig figs.</em>
28.2298 mol CaF ≈ 28 mol CaF
Answer:
John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? xdsz.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts? John is buying shirts for his softball team. He will pay a one-time processing fee of $27.50 and $12.75 per each shirt ordered. Which equation can be used to find y, the total cost to buy xshirts?
Explanation:
The correct answer attached in file, Thank you for joining brainly community.
Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles =
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O =
= 0.13moles
Number of moles of NH₃ =
= 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g