Answer:
8.70 liters
Explanation:
First we <u>convert 36.12 g of AI₂O₃ into moles</u>, using its <em>molar mass</em>:
- 36.12 g ÷ 101.96 g/mol = 0.354 mol AI₂O₃
Then we <u>convert AI₂O₃ moles into O₂ moles</u>, using the stoichiometric coefficients of the reaction:
- 0.354 mol AI₂O₃ *
= 0.531 mol O₂
We can now use the <em>PV=nRT equation</em> to <u>calculate the volume</u>, V:
- 1.4 atm * V = 0.531 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 280.0 K
Your answer is probably
Vaporization point
N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
mol = mass ÷ molar mass
If mass of hydrazine (N₂H₄) = 5.29 g
then mol of hydrazine = 5.29 g ÷ ((14 ×2) + (1 × 4))
= 0.165 mol
mole ratio of hydrazine to Nitogen is 1 : 1
∴ if moles of hydrazine = 0.165 mol
then moles of nitrogen = 0.165 mol
Mass = mol × molar mass
Since mol of nitrogen (N₂) = 0.165
then mass of hydrazine = 0.165 × (14 × 2)
= 4.62 g
<h3>
Answer:</h3>
<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
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2C + O₂ → 2CO₂
[Given] 0.25 moles O₂
[Solve] moles CO₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol O₂ → 2 mol CO₂
<u>Step 3: Stoichiometry</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
Sodium is a member of the alkali metal family with potassium (K) and Lithium (LI) sodium's big claim to fame is that it's one or two elements in your table salt. when bonded to chlorine (CI) THE two elements make sodium chloride
