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Jlenok [28]
3 years ago
14

Which of the following are true statements about equilibrium systems? For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(

s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right. For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right. For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled. For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.
Chemistry
1 answer:
Grace [21]3 years ago
4 0

Answer:

The first, third and fourth statements are correct.

Explanation:

1) For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right.

⇒ Le Chatellier says As the CaCO3 concentration is increased, the system will attempt to undo that concentration change by shifting the balance to the right. <u>This statement is true.</u>

<u />

2) For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right.

⇒ Le chatellier says that if we increase the pressure, the equilibrium will shift to the side with the least number of particles.

Since the molar densities of CaO and CaCO3 are constant, they don't appear in the equilibrium expression. This is why only changes to the pressure (concentration) of CO2 affect the position of the equilibrium.

If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2. <u>This statement is false. </u>

3)For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled.

⇒ Le Chatellier says if we increase the pressure, the equilibrium will shift to the side with the most particles.

In this case we have 2 moles of H2 and 1 mole of O2 on the left side and 2 mole of H2O on the right side. This means on the left side are more particles. So the equilibrium will shift to the left, so <u>this statement is true.</u>

4) For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.

⇒ Le chatellier says if H2 will be removed (this means the left side will get less particles) so the equilibrium will shift to the left, to increase the amount of F2.

⇒Le chatelier says if we increase the temperature of an exotherm reaction , there will be less energy released. The equilibrium will shift to the side of the reactants (the left side).

If we increase the temperature of an endotherm reaction, the equilibrium will shift to the side of the products (the right side). <u>This statement is true.</u>

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A cylinder of oxygen gas has a pressure of 6.00 atm at 25.0
Readme [11.4K]
P1/T1=P2/T2 Gal Lussac's Law
25 C= 298K (just add 273)
0 C= 273 k

6.00atm/298=P2/273

P2=5.50 atm
4 0
3 years ago
If a 32.5 mL of a 15.1 M HCl solution was diluted to 3.25 L with water, what is the final concentration?
user100 [1]

Answer:

The final concentration is 0,151 M.

Explanation:

A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).

We convert the unit of volume in L into ml: 3,25 x 1000= 3250 ml

We use the formula for dilutions:

C1 x V1 = C2 x V2

C2= (C1 xV1)/V2

C2= (32, 5 ml x 15, 1 M)/ 3250 ml

<em>C2=0,151 M</em>

8 0
3 years ago
1.42 g H2 is allowed to react with 10.4 g N2 , producing 2.14 g NH3 . Part A What is the theoretical yield in grams for this rea
Bad White [126]

Taking into account the reaction stoichiometry, the theorical yield for the reaction is 8.0467 grams of NH₃.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 H₂ + N₂ → 2 NH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • H₂: 3 moles
  • N₂: 1 mole
  • NH₃: 2 moles

The molar mass of the compounds is:

  • H₂: 2 g/mole
  • N₂: 28 g/mole
  • NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • H₂: 3 moles ×2 g/mole= 6 grams
  • N₂: 1 mole ×28 g/mole= 28 grams
  • NH₃: 2 moles ×17 g/mole= 34 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 28 grams of N₂ reacts with 6 grams of H₂, 10.4 grams of N₂ reacts with how much mass of H₂?

mass of H_{2} =\frac{10.4 grams of N_{2}x 6 grams of H_{2} }{28 grams of N_{2}}

<u><em>mass of H₂= 2.2286 grams</em></u>

But 2.2286 grams of H₂ are not available, 1.42 grams are available. Since you have less mass than you need to react with 10.4 grams of N₂, H₂ will be the limiting reagent.

<h3>Definition of theorical yield</h3>

The theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Theoretical yield in this case</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 34 grams of NH₃, 1.42 grams of H₂ form how much mass of NH₃?

mass of NH_{3} =\frac{1.42 grams of H_{2} x 34 grams of NH_{3}}{6grams of H_{2} }

<u><em>mass of NH₃= 8.0467 grams</em></u>

Then, the theorical yield for the reaction is 8.0467 grams of NH₃.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

7 0
1 year ago
Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of
shutvik [7]

Answer:

  • <u>79%</u>

Explanation:

<u>1) Balanced chemical equation:</u>

  • 2S + 3O₂ → 2SO₃

<u>2) Mole ratio:</u>

  • 2 mol S : 3 mol O₂ : 2 mol SO₃

<u>3) Limiting reactant:</u>

  • Number of moles of O₂

        n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

  • Number of moles of S:

         n = 7.0 g / 32.065 g/mol = 0.2183 mol S

  • Ratios:

        Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

        Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

<u>4) Calcuate theoretical yield (using the limiting reactant):</u>

  • 0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

  • x = 0.1875 × 2 / 3 mol SO₃ =  0.125 mol SO₃

<u>5) Yield in grams:</u>

  • mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol =  10.0 g

<u>6) </u><em><u>Percent yield:</u></em>

  • Percent yield, % = (actual yield / theoretical yield) × 100
  • % = (7.9 g / 10.0 g) × 100 = 79%
6 0
3 years ago
Explain how carbon’s bonding ability makes it unique.
STatiana [176]

\huge\fcolorbox{red}{pink}{Answer ♥}

The carbon atom is unique among elements in its tendency to form extensive networks of covalent bonds not only with other elements but also with itself. ... Moreover, of all the elements in the second row, carbon has the maximum number of outer shell electrons (four) capable of forming covalent bonds.

Hope it helps uh ✌️✌️✌️

Gud mrng

7 0
3 years ago
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