Answer:
Explanation:
1) A fulcrum is a pivot point that plays a central role (not necessarily located at the center) in a lever. The fulcrum of the attached picture has been circled (in blue).
2) The object placed on this lever's measurement tray is balanced by placing it at the center of the tray. This is the standard way of placing objects on any balance.
Answer:
The balanced single displacement reaction between aluminum and copper sulphate is as follows.

Explanation:
The given ions are as follows.
- Aluminium ion and copper sulfate ion.
Single displacement reaction occurs one element replaces another element.
Hence, the balanced chemical reaction is as follows.

<u>Oxidation half reaction:</u>

Aluminium loses three electrons.
<u>Reduction half reaction:</u>
<u>
</u>
Copper gain three electrons.
<h3>
Answer:</h3>
10.6 mol NO
<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] 4NH₃ + 5O₂ → 4NO + 6H₂O
[Given] 13.2 mol O₂
<u>Step 2: Identify Conversions</u>
[RxN] 5 mol O₂ → 4 mol NO
<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>
10.56 mol NO ≈ 10.6 mol NO
Answer:
1.The electrode on the right is positive
2. 0.058V
Explanation:
The above cell is a concentration cell.
A concentration cell is an electrolytic cell that is made of two half-cells with the same electrodes, but differs in concentrations of the solutions. A concentration cell functions by diluting the more concentrated solution and concentrating the more dilute solution, creating a voltage as the cell reaches an equilibrium thereby transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.
In the above cell, electrons flow from the left electrode (less concentrated) to the right electrode (more concentrated). Therefore, the right electrode is the positive electrode (cathode).
Part 2: Please, see the attachment below for the calculations.