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Kruka [31]
2 years ago
14

For the following equilibrium system, which of the following changes will form more CaCO3? CO2(g) + Ca(OH)2(s) ⇌ CaCO3(s) + H2O(

l) ΔH o rxn = −113 kJ a. Decrease temperature at constant pressure (no phase change). b. Increase volume at constant temperature. c. Increase partial pressure of CO2. d. Remove one-half of the initial CaCO3.
Chemistry
1 answer:
Talja [164]2 years ago
8 0

Answer: d. Remove one-half of the initial CaCO3.

Explanation: Le Chatelier's principle states that changes on the temperature, pressure, concentration and volume of a system will affect the reaction in an observable way. So in the reaction above:

A decrease in temperature will shift the equilibrium to the left because the reaction is exothermic, which means heat is released during the reaction. In other words, when you decrease temperature of a system, the equilibrium is towards the exothermic reaction;

A change in volume or pressure, will result in a production of more or less moles of gas. A increase in volume or in the partial pressure of CO2, the side which produces more moles of gas will be favored. In the equilibrium above, the shift will be to the left.

A change in concentration will tip the equilibrium towards the change: in this system, removing the product will shift the equilibrium towards the production of more CaCO3 to return to the equilibrium.

So, the correct answer is D. Remove one-half of the initial CaCO3.

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Which statement about gases is true? A. They are made up of particles that always move very slowly. B. They are made up of parti
rosijanka [135]

Answer: D. They are made up of hard spheres that are in random motion.

Explanation:

A gas is a <u>state of aggregation of matter</u> in which, under certain conditions of temperature and pressure, <u>its molecules interact weakly with each other, without forming molecular bonds</u>, adopting the shape and volume of the container that contains them and tending to separate everything possible because of its <u>high concentration of kinetic energy</u>.

The molecules of a gas are practically <u>free</u> and have the ability to be distributed throughout the space in which they are contained because <u>the gravitational forces and attraction between them are practically negligible</u> compared to the speed at which they move. .

Therefore, gas molecules do not travel specific trajectories or vibrate in a stationary position, instead <u>they move quickly and randomly through the entire space of the container that contains them.</u>

7 0
2 years ago
Read 2 more answers
The specific rotation of (R) carvone is (+) 61°. The optical rotation of a sample of a mixture of R &amp;S carvone is measured a
shusha [124]

Answer:

See explanation

Explanation:

% optical purity = specific rotation of mixture/specific rotation of pure enantiomer  * 100/1

specific rotation of mixture = 23°

specific rotation of pure enantiomer = 61°

Hence;

% optical purity = 23/61 * 100 = 38 %

More abundant enantiomer = 100% - 38 % = 62%

Hence the pure  (S) carvone is (-) 62° is the more abundant enantiomer.

Enantiomeric excess = 62 - 50/50 * 100 = 24%

Hence

(R) - carvone  =  38 %

(S) - carvone = 62%

7 0
1 year ago
Read 2 more answers
What mass of carbon dioxide could be made from 100 tonnes of calcium carbonate?
MA_775_DIABLO [31]
Total mass of CaCO3 = 40 amu of Ca + 12amu of C + 16×3 amu of oxygen = 100amu of CaCO3




i.e 100 tonnes of CaCO3 .


mass of CO2 = 12amu of C + 2× 16amu of O = 44 amu of CO2




mass % of CO2 in CaCO3 = (44/100)×100 =44%


i.e
44% of 100 tonnes is CO2.
=44 tonnes of CO2.

therefore, 44% of CO2 is present in CaCO3.








3 0
2 years ago
Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b
frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}?

(a) Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}

(b) 2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}

(c)Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}

(d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Answer: (d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}

Aluminum's half-equation is oxidation:

Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}

For Lead, half-equation is reduction:

Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}

Multiply first half-equation for 2 and second half-equation by 3:

2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}

3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}

Adding them:

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

The balanced redox reaction with cell notation Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)} is

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

6 0
2 years ago
What pressure (in atm) would be exerted by 76 g of fluorine gas (f2) in a 1.50 liter vessel at -37oc? (a) 26 atm(b) 4.1 atm(c) 1
nirvana33 [79]

<span>Let's assume that the F</span>₂ gas has ideal gas behavior. 

<span> Then we can use ideal gas formula,
PV = nRT

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span>⁻¹ K⁻<span>¹) and T is temperature in Kelvin.</span>


Moles = mass / molar mass


Molar mass of F₂ = 38 g/mol

Mass of F₂  = 76 g

Hence, moles of F₂ = 76 g / 38 g/mol = 2 mol

<span>
P = ?
V = 1.5 L = 1.5 x 10</span>⁻³ m³

n = 2 mol

R = 8.314 J mol⁻¹ K⁻<span>¹
T = -37 °C = 236 K

By substitution,
</span>

P x 1.5 x 10⁻³ m³ = 2 mol x 8.314 J mol⁻¹ K⁻¹ x 236 K

                         p = 2616138.67 Pa

                         p = 25.8 atm = 26 atm


Hence, the pressure of the gas is 26 atm.

Answer is "a".

<span>

</span>
5 0
2 years ago
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