What is the structure of epithelial cells?

Phosphorus pentachloride decomposes to phosphorus trichloride at high temperatures according to the equation:

weqwewe [10]

Answer:

D) [PCl5]=0.00765M,[PCl3]=0.117M,and [Cl2]=0.0.117M

Explanation:

Based on the reaction:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

And knowing:

Kc = [PCl₃] [Cl₂] / [PCl₅] = 1.80

When you add PCl₅ into a flask, this gas will react producing PCl₃ and Cl₂ until [PCl₃] [Cl₂] / [PCl₅] = 1.80

This could be written as:

[PCl₃] = X

[Cl₂] = X

[PCl₅] = 0.125M - X

<em>Where X represents the moles of PCl₅ that react, </em><em>reaction coordinate.</em>

Replacing in Kc expression:

[PCl₃] [Cl₂] / [PCl₅] = 1.80

[X [X] / [0.125 - X] = 1.80

X² = 0.225 - 1.80X

0 = -X² -1.80X + 0.225

Solving for X:

X = -1.9M → False solution, there is no negative concentrations

X = 0.11735M → Right solution.

Replacing, concentrations in equilibrium are:

[PCl₃] = X

[Cl₂] = X

[PCl₅] = 0.125M - X

[PCl₃] = 0.117M

[Cl₂] = 0.117M

[PCl₅] = 0.00765M

And right option is:

7 0

3 years ago

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r

rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

= 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

$K=Ae^{\frac{-E}{RT}}$

If $K_1=Ae^{\frac{-E_1}{RT}} -------equation 1$ &

$K_2=Ae^{\frac{-E_2}{RT}} -------equation 2$

$\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}$

$E_1= E_2-RT*In(\frac{K_1}{K_2})$

$E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})$

$E_1 = 28957.39292$ $J/mol$

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0

2 years ago

What is the acronym of DNA??

g100num [7]

Abbreviation. DNA, which stands for deoxyribonucleic acid, is defined as a nucleic acid that contains the genetic code.

3 0

3 years ago

Use the balanced equation from number four above to determine the limiting reactant if we had a mixture of 5.0 moles of Fe and 4

Tom [10]

Answer:

2Fe + 3O₂ → 2Fe₂O₃

Limiting reactant: O₂

2Fe + O₂ → 2FeO

Limiting reactant: Fe

Explanation:

2Fe + 3O₂ → 2Fe₂O₃

2Fe + O₂ → 2FeO

These are the possible reactions:

In first case, 2 moles of Fe need 3 mol of oyxgen to react

If I have 5 moles of Fe, I will need (5 .3)/2 = 7.5 moles of O₂

Then, the oxygen is my limiting ( I only have 4 moles)

3 moles of O₂ need 2 moles of Fe

If I have 4 moles of O₂, I will need (4 .2)/3 = 2.66 moles (I have 5)

Fe, is the reactant in excess.

For second case, 2 moles of Fe need 1 mol of O₂, to react.

If I have 5 moles of Fe, I will need ( 5 .1) / 2 =2.5 moles of O₂

I have 4 moles of oxygen, so now it is my excess.

1 mol of O₂ need 2 moles of Fe, to react

If I have 4 moles of O₂, I will need the double of amount, 8 moles of Fe.

I have 5 moles, then the Fe is my limtiing reactant.

8 0

3 years ago

Why am i not good enough for my mom

Lisa [10]

You are good enough in general. As long as you know and feel that you are good enough within yourself, you won’t have to worry about meeting anyone else’s expectations. Maybe it’s your mom.. not you.

8 0

3 years ago

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