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coldgirl [10]
3 years ago
5

How do catalysts increase the rate of a chemical reaction?

Chemistry
1 answer:
kramer3 years ago
4 0
<span>They lower the activation energy.</span>
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What is the coefficient in a chemical equation
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A coefficient is the number or letter before the variable. The variable is the unknown value.
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If the pKa of Lysine is 5, then at a pH of 7, it is most likely going to be a) Neutral. b) Negatively charged. c) Positively cha
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C. Positively charged
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The equilibrium constant Kc for the reaction below is 0.00584 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentra
Vinil7 [7]

Answer:

Explanation:

Given that:

The chemical equation for the reaction is:

             Br2(g)    ⇌  2Br(g)

Initially  0.0345M  0.0416M

Q_C = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416)^2}{(0.0345)}= 0.05016

Q_C =0.05016 >>> K_c(0.00584)

Thus, the given reaction will proceed in the  backward direction

 The I.C.E table is as follows:

                       Br2(g)    ⇌  2Br(g)

I               0.0345                 0.0416

C                 +x                        -2x

E             (0.0345+x)            (0.0416 -2x)

K_c = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416-2x)^2}{(0.0345+x)} = 0.00584

= 0.00173056 - 0.0832x - 0.0832x + 4x² = 0.00584 (0.0345 +x)

= 0.00173056 - 0.166x + 4x² = 2.0148× 10⁻⁴ + 0.00584x

= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²

= 0.00152908  - 0.17184x + 4x²

Solving by using Quadratic formula

x = 0.03038 or 0.0126

For x = 0.03038

At equilibrium

[Br₂] = (0.0345 + 0.03038) = 0.06488 M

[Br] =  (0.0416 -2(0.03038)) = - 0.01916 M

Since we have a negative value for [Br], we discard the value for x

For x = 0.0126

At equilibrium

[Br₂] = (0.0345 + 0.0126) = 0.0471 M

[Br] =  (0.0416 -2(0.0126)) = 0.0164 M

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3 years ago
Write a story of your life when you were hurted by someone whom you trusted blindly...​
SOVA2 [1]

Answer:

Sis I think it happened with me but I am not able to remember if u want u can share if it happened with u

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What minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 155.0 ml of a
Blababa [14]
First, we write the balanced equation for this reaction:

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From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:

Moles = volume (in L) * molarity

We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:

M₁V₁ = 2M₂V₂

V₁ = 2M₂V₂/M₁
V₁ = 2 * 0.112 * 0.155 / 0.2
V₁ = 0.1736 L


The volume required is 173.6 mL
5 0
3 years ago
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