Answer:
147.6 mL .
Explanation:
sodium carbonate , Na₂CO₃
Molecular weight = 106
3.55 gm of sodium carbonate = 3.55 / 106
= .0335 moles
Let the volume of litre required = V
V litre of .227 M solution will contain
V x .227 moles of sodium carbonate . So
V x .227 = .0335
V = .1476 L
= 147.6 mL .
Answer:
5.158 × 10²³ atoms K
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Explanation:
<u>Step 1: Define</u>
33.49 g K
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of K - 39.10 g/mol
<u>Step 3: Convert</u>
<u />
= 5.15797 × 10²³ atoms K
<u>Step 4: Check</u>
<em>We are given 4 sig figs. Follow sig figs and round.</em>
5.15797 × 10²³ atoms K ≈ 5.158 × 10²³ atoms K
Its true it affects the pH balance