<h3>
Answer:</h3>
6.26 g C
<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>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 3.14 × 10²³ atoms C
[Solve] grams C
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of C - 12.01 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>
6.26227 g C ≈ 6.26 g C
First we calculate the concentration of HCl:
Moles = mass / Mr
= 25 / 36.5
= 0.685 mol
Concentration = 0.685/1.5 = 0.457 mol / dm³
For a strong monoprotic acid, the concentration of hydrogen ions is equal to the acid concentration.
pH = -log[H+]
pH = -log(0.457)
= 0.34
Answer:
1) HBr; 2) NaCCH; 3) O3; 4) H2O
Explanation:
The first step is formation of alkyl halide followed by reaction with sodium acetylide, to form 3-methylbutene, which is then followed by oxidation reaction with O3& H2O to 3-methylbutanal
Glucose is a simple sugar with the molecular formula C6H12O6. It is a carbohydrate.