How much 5.50M NaOH must be added to 680.0 mL of a buffer that is 0.0215 M acetic acid and 0.0270 M sodium acetate to raise the
1 answer:
5.50M NaOH must be added to 680.0 mL of a buffer, The amount of NaOH is mathematically given as
V=2.107mL
<h3>What is the volume of 5.50M NaOH?</h3>Generally, the equation for Chemical Reaction is mathematically given as
CH3COOH+OH^{-} ⇄ CHCOO^{-} + H2O
Therefore
pka=-log(1.8*10{-5})
pKa=5-0.25527
pKa=4.3447
Hence
x=0.0115898
Volume of NaOH
V=0.0021072L
V=2.107mL
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sp²
Explanation:
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Answer:
510 g NO₂
General Formulas and Concepts:
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
- Reading the Periodic Table
- Writing Compounds
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)
<u>Step 2: Define conversions</u>
Avogadro's Number
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol
<u>Step 3: Use Dimensional Analysis</u>
<u /> = 511.901 g NO₂
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules.</em>
511.901 g NO₂ ≈ 510 g NO₂