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postnew [5]
3 years ago
6

A catalyst lowers the activation energy for both the forward and the reverse reactions in an equilibrium system, so it has no ef

fect on the equilibrium position of a system. True False

Chemistry
2 answers:
rosijanka [135]3 years ago
5 0

Answer:

True.

Explanation:

  • Catalyst increases the rate of the reaction without affecting the equilibrium position.
  • Catalyst increases the rate via lowering the activation energy of the reaction.
  • This can occur via passing the reaction in alternative pathway (changing the mechanism).
  • The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).
  • in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.
  • with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.
  • Kindly, see the attached image.

stich3 [128]3 years ago
5 0
<h2>Answer: </h2>

<u>The statement is</u><u> true </u>

<h2> Explanation: </h2>

When we add a catalyst to a system, both the forward and reverse reaction rates will speed up equally at the same time thereby allowing the system to reach equilibrium faster. But one thing that is very important to keep in mind that the addition of a catalyst has no effect on the final equilibrium position of the reaction because it only makes the reaction faster and nothing else.

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The permanent electric dipole moment of the water molecule 1H2O2 is 6.2 * 10-30 C m. What is the maximum possible torque on a wa
stellarik [79]

Answer:

3.1 x 10⁻²¹ Nm

Explanation:  

When placed in an external electric filed, an electric dipole experiences a torque. and this torque is represented mathematically with the equation:

torque (τ) = dipole moment vector (P) x electric field vector (E)

τ = P. E . sin θ

where θ is the angle between the water molecule and the electric field, which in this case is 90° (because this is where the torque is maximum)

τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . sin90

τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . 1

solve for τ

τ = 3.1 x 10⁻²¹ Nm

the maximum possible torque on the water molecule is therefore 3.1 x 10⁻²¹ Nm

7 0
3 years ago
What else is produced during the combustion of butane, C4H10?<br><br> 2C4H10 + 13O2 → <br> + 10H2O
Neko [114]
<h3>Answer:</h3>

8CO₂

<h3>Explanation:</h3>

We are given;

  • Butane, C₄H₁₀
  • Butane is a hydrocarbon in the homologous series known as alkane.

We are required to determine the other product produced in the combustion of butane apart from water.

  • We know that the complete combustion of alkane yields carbon dioxide and water.
  • Therefore, combustion of butane will yield carbon dioxide and water.
  • The balanced equation for the complete combustion of butane will be;

       2C₄H₁₀ + 13O₂ →  8CO₂ + 10H₂O

8 0
3 years ago
Read 2 more answers
What is the correct prefix for 3 <br> tri-<br> di-<br> mono-<br> tetra-
Ivenika [448]

Answer:

tri-

Explanation:

Examples could be Tri-angle, Tri-cycle, Tri-ceratops

5 0
2 years ago
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Earth is approximately 1.5 x 10^8 km from the sun the. How far is earth from the sun meters
Zielflug [23.3K]

Answer:

{ \tt{1 \: km = 1000m}} \\ { \tt{1.5  \times {10}^{8} km =  (\frac{1.5 \times  {10}^{8} \times 1000 }{1} )m}} \\  = 1.5 \times  {10}^{11}  \: metres

4 0
2 years ago
2.0L of hydrogen gas is mixed with 3.0L of nitrogen gas at STP in a rigid 5.0L vessel. A reaction occurs, producing ammonia gas
IrinaK [193]

Answer:

Moles NH₃: 0.0593

0.104 moles of N₂ remain

Final pressure: 0.163atm

Explanation:

The reaction of nitrogen with hydrogen to produce ammonia is:

N₂ + 3 H₂ → 2 NH₃

Using PV = nRT, moles of N₂ and H₂ are:

N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂

H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂

The complete reaction of N₂ requires:

0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = <em>0.402 moles H₂</em>

That means limiting reactant is H₂. And moles of NH₃ produced are:

0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = <em>0.0593 moles NH₃</em>

Moles of N₂ remain are:

0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = <em>0.104 moles of N₂</em>

And final pressure is:

P = nRT / V

P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L

<em>P = 0.163atm</em>

7 0
2 years ago
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