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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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A proton is fired toward a lead nucleus from very far away. How much initial kinetic energy does the proton need to reach a turn
Olegator [25]

Answer:

The electric force is conservative.

Since

ΔK = −ΔU,

Kf − Ki =Ui −Uf.

We have,

Kf = 0

Ui = 0.

Thus Ki =Uf.

<u>so ,at 10 fm Uf = (2×10)−12 J.</u>

6 0
3 years ago
An herbicide contains only C, H, Cl, and N. The complete combustion of a 200.0 mg sample of the herbicide in excess oxygen produ
MAVERICK [17]

Answer:

%C = 56,1%

%H = 5,5%

%Cl = 27,6%

%N = 10,8%

Explanation:

The moles of CO₂ are the same than moles of C in the herbicide.

Moles of H₂O are ¹/₂ of moles of H in the herbicide.

Moles of CO₂ are obtained using:

n = PV/RT

Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K

moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = <em>0,1122 g of C ≡ 112,2mg of C</em>

In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = <em>0,0110 g of H ≡ 11,0mg of H</em>

As you have 55,14 mg of Cl, the mg of N are:

200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N

Thus, precent composition of the herbicide is:

%C = \frac{112,2 mgC}{200,0mg}×100 = 56,1%C

%H = \frac{11,0 mgH}{200,0mg}×100 = 5,5%H

%Cl = \frac{55,14 mgCl}{200,0mg}×100 = 27,6%Cl

%N = \frac{21,66 mgN}{200,0mg}×100 = 10,8%N

I hope it helps!

3 0
3 years ago
2. Under a pressure of 95kPa and a temperature of 25 C, a gas occupies 4.0 liters. What would the new
Alekssandra [29.7K]

Answer:

35

Explanation:

multiply and subtract

3 0
3 years ago
Standard atmospheric pressure corresponds to the normal air pressure at sea level.
jekas [21]
Im not really sure what your asking.... <span>Standard sea-level pressure, by definition, equals 760 mm (29.92 inches) of mercury, </span>14.70 pounds per square inch<span>, 1,013.25 × 10 </span>3<span> dynes per square centimetre, 1,013.25 millibars, one standard atmosphere, or 101.325 kilopascals.

</span><span>""atmospheric pressure | Britannica.com""</span>
6 0
3 years ago
Dinitrogen pentoxide is used in the preparation of explosives. If 7.93 mol of
Tpy6a [65]

The volume of O₂ produced: 84.6 L

<h3>Further explanation</h3>

Given

7.93 mol of  dinitrogen pentoxide

T = 48 + 273 = 321 K

P = 125 kPa = 1,23365 atm

Required

Volume of O₂

Solution

Decomposition reaction of dinitrogen pentoxide

2N₂O₅(g)→4NO₂(g)+O₂ (g)

From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :

= 0.5 x mol N₂O₅

= 0.5 x 7.93

= 3.965 moles

The volume of O₂ :

\tt V=\dfrac{nRT}{P}\\\\V=\dfrac{3.965\times 0.082\times 321}{1.23365}\\\\V=84.6~L

5 0
3 years ago
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