For the equation to balance out, you would need 2 H2O on the product. That will give a total of 1 Carbon, 4 Hydrogens and 4 Oxygens on each side.
As mentioned above, phosphoric acid has 3 pKa values, and after 3 ionization it gives 3 types of ions at different pKa values:
H₃PO₄(aq)
+ H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq) pKₐ₁
<span>
</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq) pKₐ₂
HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq) pKₐ₃
At the highest pKa value (12.4) of phosphoric acid, the last OH group will lose its hydrogen. On the picture I attached, it is shown required protonated form of phosphoric acid before reaction whose pKa value is 12.4.
the numbers are going to be small so like a power but its at the bottom
NH3, H2O2, NHO2
Answer:
D
Explanation:
Because the silk took away the electrons to allow the rod to become positive. They cant be destroyed because of the law of conservation of charges. Also protons cannot move only electrons can.
Answer:
9.8 × 10²⁴ molecules H₂O
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Organic</u>
<u>Stoichiometry</u>
- Analyzing reaction rxn
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 130 g CH₄
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[RxN] 1 mol CH₄ → 2 mol H₂O
[PT] Molar Mass of C: 12.01 g/mol
[PT] Molar Mass of H: 1.01 g/mol
Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O