Answer:
Less.
Explanation:
Since there was an increase of concentration of NaOH, there would be more molecules of NaOH present. There would be no need to use the same amount of NaOH because there was already more in the solution.
<h3>
Answer:</h3>
12 mol NH₃
<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] N₂ + 3H₂ → 2NH₃
[Given] 6 moles N₂
[Solve] moles NH₃
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol N₂ → 2 mol NH₃
<u>Step 3: Stoich</u>
- [DA] Set up conversions:
- [DA] Multiply/Divide [Cancel out units]:
Answer:
This is the partial pressure of Ar in the flask. A 1.00 L flask is filled with 1.20g of argon at 25 degrees C. A sample of ethane vapor is added to the same flask until the total pressure is 1.400atm .
Explanation:
Answer:
Cold Front. A side view of a cold front (A, top) and how it is represented on a weather map (B, bottom). ...
Warm Front. ...
Stationary Front. ...
Occluded Front.