A<em>ns</em><em>w</em><em>e</em><em>r</em>
<h2>
<em>K </em><em>eq</em><em>=</em><em>1</em><em>.</em><em>3</em><em>3</em></h2>
<em>please </em><em>see</em><em> the</em><em> attached</em><em> </em><em>picture.</em><em>.</em><em>.</em><em>.</em>
<em>Hope </em><em>it</em><em> helps</em><em>.</em><em>.</em>
<em>good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>
<span>The amount of a component or solute that dissolves in a given volume of solvent at a certain temperature. </span>
Answer:
The correct options are A, and C.
Explanation:
Osmosis: It is defined as the movement of solvent with the help of selectively semipermeable membrane into a region of where high solute concentration is present to equalize the concentration of solute on the both compartments.
Reverse osmosis: It is defined as the movement of the high concentration solvent is forced onto the lighter concentration side with the help of mechanical pressure.
Answer:
0.504 M
Explanation:
Step 1: Write the balanced neutralization reaction
2 KOH + H₂SO₄ ⇒ K₂SO₄ + 2 H₂O
Step 2: Calculate the reacting moles of KOH
55.2 mL (0.0552 L) of 0.500 M KOH react. The reacting moles of KOH are:
0.0552 L × 0.500 mol/L = 0.0276 mol
Step 3: Calculate the moles of H₂SO₄ that reacted with 0.0276 moles of KOH
The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of H₂SO₄ are 1/2 × 0.0276 mol = 0.0138 mol
Step 4: Calculate the concentration of H₂SO₄
0.0138 moles of H₂SO₄ are in 27.4 mL (0.0274 L). The molarity of H₂SO₄ is:
[H₂SO₄] = 0.0138 mol/0.0274 L = 0.504 M
