Answer:
B) Observations and experimentation
Explanation:
this is the most suitable words to describe how is Science used in describing the world. we collect the knowledge in organized way and provide it with explanations and start to predict about the future world
ِAny Inventions are based on Science and no Inventions without Observations and experimentation
No one of us have went to a chemistry or a physics labs without using Observations and experimentation
<h3>
Answer:</h3>
0.127 mol Au
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 25.0 g Au
[Solve] moles Au
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</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>
0.126923 mol Au ≈ 0.127 mol Au
Answer:
You will never know the exact volume with charles law
Explanation:
Doubling the temperature of gas doubles its volume, so long as the pressure and quantity of the gas are unchanged.
1860-х годах, а кульминацией — распространение поточного производства и поточных линий. В 1860—1870-х годах технологическая революция быстро охватила Западную Европу, США, Российскую империю и Японию.
The answer is: H₃PO₄.
A phosphoric acid is three protic acid, which means that in water release tree protons.
Phosphoric acid ionizes in three steps in water.
First step: H₃PO₄(aq) ⇄ H₂PO₄⁻(aq) + H⁺(aq).
Second step: H₂PO₄⁻(aq)⇄ HPO₄²⁻(aq) + H⁺(aq).
Third step: HPO₄²⁻(aq) ⇄ PO₄³⁻(aq) + H⁺(aq).
Species that are present: H₃PO₄, H₂PO₄⁻, HPO₄²⁻, PO₄³⁻ and H⁺.
A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.
Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable.