Answer:
0.425M NaOH assuming the volume of KHP was 25.50mL and the volume of the NaOH solution was 30.0mL
Explanation:
The KHP reacts with NaOH as follows:
KHP + NaOH → KNaP + H₂O
<em>Where 1 mole of KHP reacts per mole of KNaP</em>
<em />
That means, the moles of KHP added to the NaOH solution = Moles NaOH at equivalence point. With the moles of NaOH and the volume in liters we can find the molar concentration of NaOH.
<em>Assuming the volume added of KHP was 25.50mL and the solution of NaOH contains 30.0mL (0.0300L), the concentration of the NaOH is:</em>
<em />
<em>Moles KHP = Moles NaOH:</em>
25.50mL = 0.02550L * (0.500mol / L) = 0.01275 moles KHP = Moles NaOH
<em>Molarity NaOH:</em>
0.01275 moles NaOH / 0.0300L =
<h3>0.425M NaOH assuming the volume of KHP was 25.50mL and the volume of the NaOH solution was 30.0mL</h3>
Y EQUALS X - 1 I MIGHT BE WRONG
D. The particles are tightly packed together
Answer:
[NO] = 1.72 x 10⁻³ M.
Explanation:
<em>2NO(g) ⇌ N₂(g)+O₂(g),</em>
Kc = [N₂][O₂] / [NO]².
- At initial time: [NO] = 0.171 M, [N₂] = [O₂] = 0.0 M.
- At equilibrium: [NO] = 0.171 M - 2x , [N₂] = [O₂] = x M.
∵ Kc = [N₂][O₂] / [NO]².
∴ 2400 = x² / (0.171 - 2x)² .
<u><em>Taking the aquare root for both sides:</em></u>
√(2400) = x / (0.171 - 2x)
48.99 = x / (0.171 - 2x)
48.99 (0.171 - 2x) = x
8.377 - 97.98 x = x
8.377 = 98.98 x.
∴ x = 8.464 x 10⁻².
<em>∴ [NO] = 0.171 - 2(8.464 x 10⁻²) = 1.72 x 10⁻³ M. </em>
<em>∴ [N₂] = [O₂] = x = 8.464 x 10⁻² M.</em>
