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3 years ago

How many moles of gas Does it take to occupy 520 mL at a pressure of 400 torr and a temperature of 340 k

Chemistry

1 answer:

Lubov Fominskaja [6]3 years ago

3 0

Answer would be B. I provided work on an image attached. Message me if u have any other questions on how to do it

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Calculate the maximum amount of useful work that can be obtained and comment on the spontaneity for the reaction at 25C :

harkovskaia [24]

Answer:

$1.41 *10^{3} kJ/mol$

Explanation:

First, we find in the tables the ΔH of formation of each compound. As you can see in the (image 1)

Then we solve the ecuation for ΔH°reaction

ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(Reactants)

ΔH°reaction= (-2* 393.5 - 2*285.8) - (52.4 + 0) kJ/mol

ΔH°reaction = -1.41 *10^3 kJ/mol

3 0

3 years ago

zloy xaker [14]

Answer:

1. They lack a nucleus

Following me..

5 0

2 years ago

In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher

trasher [3.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of $BC_2$ that should used originally is $C_Z_o = 0.4492M$

Explanation:

From the question we are told that

The necessary elementary step is

$2BC_2 ----->4C + B_2$

The time taken for sixth of 0.5 M of reactant to react $t = 9 hr$

The time available is $t_a = 3.5 hr$

The desired concentration to remain $C = 0.42M$

Let Z be the reactant , Y be the first product and X the second product

Generally the elementary rate law is mathematically as

$-r_Z = kC_Z^2 = - \frac{d C_Z}{dt}$

Where k is the rate constant , $C_Z$ is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

For second order reaction

$\frac{1}{C_Z} - \frac{1}{C_Z_o} = kt$

Where $C_Z_o$ is the initial concentration of Z which a value of $C_Z_o = 0.5M$

From the question we are told that it take 9 hours for the concentration of the reactant to become

$C_Z = C_Z_o - \frac{1}{6} C_Z_o$

$C_Z = 0.5 - \frac{0.5}{6}$

$= 0.4167 M$

$\frac{1}{0.4167} - \frac{1}{0.50} = 9 k$

$0.400 = 9 k$

=> $k = 0.044\ L/ mol \cdot hr^{-1}$

For $C_Z = 0.42M$

$\frac{1}{0.42} - \frac{1}{C_Z_o} = 3.5 * 0.044$

$2.38 - 0.154 = \frac{1}{C_Z_o}$

$2.226 = \frac{1}{C_Z_o}$

$C_Z_o = \frac{1}{2.226}$

$C_Z_o = 0.4492M$

4 0

3 years ago

a 6.7 volume of air, initially at 23 degrees celsius and .98 atm, is compressed to 2.7 L while heated to 125 degrees celsius. Wh

Annette [7]

Data:

V1 = 6.7 liter
T1 = 23° = 23 + 273.15 K = 300.15 K
P1 = 0.98 atm

V2 = 2.7 liter
T2 = 125° = 125 + 273.15 K = 398.15 K
P2 = ?

Formula:

Combined law of ideal gases: P1 V1 / T1 = P2 V2 / T2

=> P2 = P1 V1 T2 / (T1 V2)

P2 = 0.98 atm * 6.7 liter * 398.15 K / (300.15K * 2.7 liter)

P2 = 3.22 atm

6 0

3 years ago

What explains osmosis

Reika [66]

A process by which molecules of a solvent tend to pass through a semipermeable membrane from a less concentrated solution into a more concentrated one thus equalizing the concentrations on each side of the membrane. I hope this helps :D

8 0

2 years ago