What is the volume of a 3.00M(mol/L) solution made with 122g of LiF?

What is the second hottest layer of the atmosphere?

kherson [118]

Well, I say the answer is Thermosphere! Hope I helped :)

7 0

3 years ago

(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S

Sunny_sXe [5.5K]

These are two questions and two answers

Answer:

Question 1:

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)

Question 2:

0.201 M

Explanation:

Question 1:

The neutralization reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) Word equation:

sulfuric acid + potassium hydroxide → potassium sulfate + water

↑ ↑ ↑ ↑

acid base salt water

2) Skeleton equation (unbalanced)

H₂SO₄ + KOH → K₂SO₄ + H₂O

#) Balanced chemical equation (including phases)

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

Question 2:

1) Mol ratio:

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

2) Moles of H₂SO₄:

V = 0.750 liter
M = 0.480 mol/liter
M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

3) Moles of KOH:

V = 0.700 liter
M = 0.290 mol/liter
M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

4) Determine the limiting reagent:

a) Stoichiometric ratio:

1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄ is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

5) Amount of H₂SO₄ that reacts:

Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

x = 0.203 / 2 = 0.0677 mol of H₂SO₄

6) Concentration of H₂SO₄ remaining:

Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

Total volume = 0.700 liter + 0.750 liter = 1.450 liter

Concetration = M

M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

6 0

3 years ago

In the laboratory, a general chemistry student measured the pH of a 0.486 M aqueous solution of triethanolamine, C6H15O3N to be

STatiana [176]

Answer:

Kb = 6.22x10⁻⁷

Explanation:

Triethanolamine, C₆H₁₅O₃N, is in equilibrium with water:

C₆H₁₅O₃N(aq) + H₂O(l) ⇄ C₆H₁₅O₃NH⁺(aq) + OH⁻(aq)

Kb is defined from concentrations in equilibrium, thus:

Kb = [C₆H₁₅O₃NH⁺] [OH⁻] / [C₆H₁₅O₃N]

The equilibrium concentration of these compounds could be written as:

[C₆H₁₅O₃N] = 0.486M - X

[C₆H₁₅O₃NH⁺] = X

[OH⁻] = X

pH is -log [H⁺], thus, [H⁺] = 10^-pH = 1.820x10⁻¹¹M

Also, Kw = [OH⁻] ₓ [H⁺];

1x10⁻¹⁴ = [OH⁻] ₓ [H⁺]

1x10⁻¹⁴ = [OH⁻] ₓ [1.820x10⁻¹¹M]

5.495x10⁻⁴M = [OH⁻], that means X = 5.495x10⁻⁴M

Replacing in Kb formula:

Kb = [5.495x10⁻⁴M] [5.495x10⁻⁴M] / [0.486M-5.495x10⁻⁴M]

Kb = 6.22x10⁻⁷

5 0

2 years ago

Consider a situation in which the enzyme is operating at optimum temperature and pH, and has been saturated with substrate. What

tekilochka [14]

Answer:Increase the enzyme concentration

Explanation:

If the enzyme is saturated, it simply means that all the enzyme molecules present has been bound to substrates.

Of course, there will still be more substrates left in the system since the enzyme is saturated. The reaction rate can now be increased by increasing the enzyme concentration so that more enzyme molecules can be bound to substrates.

4 0

3 years ago

Convert 1.71 × 1024 atoms of carbon to moles of carbon.

beks73 [17]

You need to use Avogadro's constant to convert from atoms of carbon to moles of carbon.

1.71*10^24 atoms C * (1 mole/6.022*10^23 atoms) = 2.84 moles of carbon

7 0

3 years ago