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3 years ago

Describe the characteristics of the different sections (families) of the periodic table

1 answer:

6 0

Answer: The vertical columns on the periodic table are called groups or families because of their similar chemical behavior. All the members of a family of elements have the same number of valence electrons and similar chemical properties. The horizontal rows on the periodic table are called periods

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(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

How does an increase in reactant concentration affect the rate of reaction?

Trava [24]

Option A is the correct option.

In terms of the collision theory, increasing the concentration of a reactant increases in the number of collisions between the reacting species per second and therefore increases the reaction rate.

7 0

3 years ago

Which goal do all environmental regulations for industry serve?

ololo11 [35]

Answer:

To protect people and the environment.

Explanation:

4 0

2 years ago

Read 2 more answers

Does flammability increase the size of the molecule

dem82 [27]

Answer:

Yes, the size of the mass increases which increases size.

Explanation:

Hope this helps! Have a great day! :)

8 0

3 years ago

A solution contains 482.6 g

mariarad [96]

Answer:

$\large \boxed{\text{102.57 $\, ^{\circ}$C}}$

Explanation:

Data:

m₂ = 482.6 g sucrose

m₁ = 0.2804 kg water

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{\text{b}} = iK_{\text{b}}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For sucrose, i = 1.

1. Moles of sucrose

$n = \text{482.6 g} \times \dfrac{\text{1 mol}}{\text{342.30 g}} = \text{1.410 mol}$

2. Molal concentration of sucrose

$b = \dfrac{\text{1.410 mol}}{\text{0.2804 kg}} = \text{5.029 mol/kg}$

3. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 5.029 \cdot mol \cdot kg^{-1} = 2.574 \, ^{\circ}\text{C}$

4. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 2.574 \, ^{\circ}\text{C} = \mathbf{102.57 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{102.57 \, ^{\circ}C}}$}$

7 0

3 years ago