<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
First, we need the no.of moles of O2 = mass/molar mass of O2
= 55 g / 32 g/mol
= 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2
So we can get the no.of moles of H2O = 2 * moles of O2
= 2 * 1.72 mol
= 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT
when P = 12.4 atm & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K
12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
∴ V ≈ 8.2 L
Answer: CaC12
Explanation: Calcium chloride not 100% sure tho
Answer:
the correct answer is b antifreeze protects the radiator during summer in winter.
Explanation:
Antifreeze is (typically mixed with water) to help manage the engine during extreme temperatures. As the temperature fluctuates between hot and cold coolant is distributed throughout the engine block to sustain even operating temperature.