Answer:
[H⁺] = 1.58x10⁻⁶M; [OH⁻] = 6.31x10⁻⁹M.
pH = 8.23; pOH = 5.77
Explanation:
pH is defined as <em>-log [H⁺]</em> and also you have <em>14 = pH + pOH </em>
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Thus, for a solution of pH = 5.80.
5.80 = -log [H⁺] → [H⁺] = 10^-(5.80) = 1.58x10⁻⁶M
pOH = 14-5.80 = 8.20 → [OH⁻] = 10^-(8.20) = 6.31x10⁻⁹M
Thus, for a solution of [H⁺] = 5.90x10⁻⁹M and pH = -log 5.90x10⁻⁹M = 8.23
And pOH = 14-8.23 = 5.77
Which ever penny that your using will be hot if you use heat lamp the only difference is the direction.
Oxygen and Hydrogen would most likely form a covalent bond that is polar, or a polar covalent bond. Due to the electronegativity difference between the 2 elements, unequal sharing of the valence electrons will occur, electrons being in closer proximity to Oxygen and farther away from Hydrogen. Resulting in the characteristic partial positive and negative charges to appear for the respective elements.
The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O applies to an acid-base titration.
moles NaOH = c · V = 0.2423 mmol/mL · 32.23 mL = 7.809329 mmol
moles H2SO4 = 7.809329 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.9046645 mmol
Hence
[H2SO4]= n/V = 3.9046645 mmol / 37.21 mL = 0.1049 M
The answer to this question is [H2SO4] = 0.1049 M
Answer:
1.68 mol Ca(NO₃)₂
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Compounds
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 275 g Ca(NO₃)₂
[Solve] mol Ca(NO₃)₂
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Ca: 40.08 g/mol
[PT] Molar Mass of N: 14.01 g/mol
[PT] Molar Mass of O: 16.00 g/mol
Molar Mass of Ca(NO₃)₂: 40.08 + 2[14.01 + 3(16.00)] = 164.10 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.67581 mol Ca(NO₃)₂ ≈ 1.68 mol Ca(NO₃)₂
