<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>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂
[Given] 20 mol Al₂O₃
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Al₂O₃ → 4 mol Al
<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 1 sig fig.</em>
Since our final answer already has 1 sig fig, there is no need to round.
Answer:
Mass = 255 g
Explanation:
Given data:
Number of moles of nitrogen = 7.5 mol
Mass of ammonia formed = ?
Solution:
Chemical equation:
3H₂ + N₂ → 2NH₃
Now we will compare the moles of nitrogen and ammonia.
N₂ : NH₃
1 : 2
7.5 : 2/1×7.5 = 15
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 15 mol × 17 g/mol
Mass = 255 g
Oil is sucked up through wide floating heads and pumped into storage tanks. Although suction skimmers are generally very efficient, one disadvantage is that they are vulnerable to becoming clogged by debris and ice and require constant skilled observation.
Answer: The value of 
for chloroform is 
when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
Explanation:
Given: Moles of solute = 0.793 mol
Mass of solvent = 0.758
As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.
Now, the values of 
is calculated as follows.
where,
i = Van't Hoff factor = 1 (for chloroform)
m = molality
= molal boiling point elevation constant
Substitute the values into above formula as follows.
Thus, we can conclude that the value of for chloroform is when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
