I am almost positive that the answer is: Plasmas. Let me know if im right or not please!! :)
62.23 = 1512.5001499999998 moles
1) You need to use the atomic mass of copper.
You can find it in a periodic table. It is 63.546 amu.
2) The atomic mass is the weigthed mass of the different isotopes.
This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:
=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 * atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.
3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x
=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278
=> 63.546 = 62.9296x + 64.9278 - 64.9278x
=> 64.9278x - 62.9296 = 64.9278 - 63.546
=> 1.9982x = 1.3818
=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%
=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%
Answer:
Cu-63 69.15%;
Cu-65 : 30.85%
<h3>
Answer:</h3>
1.69 g Mg₃N₂
<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>
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Unbalanced] Mg + N₂ → Mg₃N₂
[RxN - Balanced] 3Mg + N₂ → Mg₃N₂
[Given] 1.22 g Mg
[Solve] grams Mg₃N₂
<u>Step 2: Identify Conversions</u>
[RxN] 3 mol Mg → Mg₃N₂
[PT] Molar Mass of Mg - 24.31 g/mol
[PT] Molar Mass of N - 14.01 g/mol
Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol
<u>Step 3: Stoich</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>
1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂
Answer:
The equation for percent composition is (mass of element/molecular mass) x 100.
please mark me as a brainlist