ASAP!<br> I have 20 min to finish this i need help

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

6 0

3 years ago

Calculate the number of calories needed to increase the temperature of 50.0 g of copper metal from 21.0 degrees C to 75.0 degree

KonstantinChe [14]

<h3>Answer:</h3>

1031.4 Calories.

<h3>Explanation:</h3>

We are given;

Mass of the copper metal = 50.0 g

Initial temperature = 21.0 °C

Final temperature, = 75°C

Change in temperature = 54°C

Specific heat capacity of copper = 0.382 Cal/g°C

We are required to calculate the amount of heat in calories required to raise the temperature of the copper metal;

Quantity of heat is given by the formula,

Q = Mass × specific heat capacity × change in temperature

= 50.0 g × 0.382 Cal/g°C × 54 °C

= 1031.4 Calories

Thus, the amount of heat energy required is 1031.4 Calories.

4 0

3 years ago

You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with K

seropon [69]

Answer:

HA + KOH → KA + H₂O

Explanation:

The unknown solid acid in water can release its proton as this:

HA + H₂O → H₃O⁺ + A⁻

As we have the anion A⁻, when it bonded to the cation K⁺, salt can be generated, so the reaction of HA and KOH must be a neutralization one, where you form water and a salt

HA + KOH → KA + H₂O

It is a neutralization reaction because H⁺ from the acid and OH⁻ from the base can be neutralized as water

7 0

3 years ago

This substance consists of two or more different elements chemically combined

kow [346]

The substance would be referred to as a compound substance.

8 0

3 years ago

The concentrations of the products at equilibrium are [pcl3] = 0.180 m and [cl2] = 0.250 m. what is the concentration of the rea

sineoko [7]

We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula:
Kc = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 = 0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07

4 0

3 years ago

ASAP! I have 20 min to finish this i need help