Unit 2 Test

When the glucose solution described in part A is connected to an unknown solution via a semipermeable membrane, the unknown solu

gtnhenbr [62]

Answer:

The unknown solution had the higher concentration.

Explanation:

When two solutions are separated by a semi-permeable membrane, depending on the concentration gradient between the two solutions, there is a tendency for water molecules to move across the semi-permeable in order to establish an equilibrium concentration between the two solutions. This movement of water molecules across a semi-permeable membrane in response to a concentration gradient is known as osmosis. In osmosis, water molecules moves from a region of lower solute concentration or higher water molecules concentration to a region of higher solute concentration or lower water molecules concentration until equilibrium concentration is attained.

Based on the observation that when the glucose solution described in part A is connected to an unknown solution via a semipermeable membrane, the unknown solution level rises, it means that water molecules have passed from the glucose solution through the semipermeable membrane into the unknown solution. Therefore, the solution has a higher solute concentration than the glucose solution.

3 0

3 years ago

Calculate the solubility of Mg(OH)2 in water at 25 C. You'll find Ksp data in the ALEKS Data tab. Round your answer to significa

elena-s [515]

Answer:

1.12 × 10⁻⁴ M

Explanation:

Step 1: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 2: Make an ICE chart

We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

I 0 0

C +S +2S

E S 2S

The solubility product constant is:

Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³

S = 1.12 × 10⁻⁴ M

8 0

3 years ago

Write the balanced chemical equation between H2SO4 and KOH in aqueous solution. This is called a neutralization reaction and wil

emmainna [20.7K]

Answer:

0.166M

Explanation:

In a neutralization, the acid, H₂SO₄, reacts with a base, KOH, to produce a salt, K₂SO₄ and water. The reaction is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

To solve this problem, we need to determine moles of H2SO4 and moles of KOH that reacts to find the moles of sulfuric acid that remains after the reaction:

<em>Moles H2SO4:</em>

0.650L * (0.430mol /L) = 0.2795moles H2SO4

<em>Moles KOH:</em>

0.600L * (0.240mol / L) = 0.144 moles KOH

Moles of sulfuric acid that reacts with 0.144 moles of KOH are:

0.144 moles KOH * (1mol H2SO4 / 2 mol KOH) = 0.072 moles of H2SO4 react.

And remain:

0.2795moles H2SO4 - 0.072moles H2SO4 = 0.2075 moles of H2SO4 reamains.

In 0.650L + 0.600L = 1.25L:

Molar concentration of sulfuric acid:

0.2075 moles of H2SO4 / 1.25L =

7 0

3 years ago

Why weren’t scientists initially concerned that the ocean absorbs excess carbon dioxide?

nasty-shy [4]

They only discovered

5 0

2 years ago

In a neutralization reaction, an aqueous solution of an Arrhenius acid reacts with an aqueous solution of an Arrhenius base to p

solmaris [256]

Answer: c. Salt and Water

Explanation:

For example;

When an Arrhenius acid such as; Tetraoxosulphate (VI) acid (H2SO4) reacts with an Arrhenius base such as Potassium hydroxide (KOH), the products formed in this neutralization reaction is a salt known as ''Potassium Sulphate'' (K2SO4) and ''Water'' (H2O).

H2SO4 + KOH -------------> K2SO4 + H2O

5 0

3 years ago