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goblinko [34]
3 years ago
8

At standard pressure, the difference between the freezing point and the boiling point of water, in Kelvin degrees, is D. 373 A.

100 B. 180 C. 273
Chemistry
1 answer:
Sergio [31]3 years ago
7 0
The answer to this question is A. 100
 

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Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe
Illusion [34]

Answer:

a. Kp=1.4

P_{A}=0.2215 atm

P_{B}=0.556 atm

b.Kp=2.0 * 10^-4

P_{A}=0.495atm

P_{B}=0.00995 atm

c.Kp=2.0 * 10^5

P_{A}=5*10^{-6}atm

P_{B}=0.9999 atm

Explanation:

For the reaction  

A(g)⇌2B(g)

Kp is defined as:

Kp=\frac{(P_{B})^{2}}{P_{A}}

The conditions in the system are:

          A                    B

initial   0                1 atm

equilibrium x       1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.    

Replacing these values in the expression for Kp we get:

Kp=\frac{(1-2x)^{2}}{x}

Working with this equation:

x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a} (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128

x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215

With x1 we get a partial pressure of:

P_{A}=1.128 atm

P_{B}=1-2*1.128 = -1.256 atm

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

P_{A}=0.2215 atm

P_{B}=1-2*0.2215 = 0.556 atm

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

P_{A}=0.505atm

P_{B}=1-2*0.505 = -0.01005 atm

Not sense.

With x2

P_{A}=0.495atm

P_{B}=1-2*0.495 = 0.00995 atm

X2 is the solution for this equation.  

Kp=2*10^5

X1=50001

X2=5*10^{-6}

With x1

P_{A}=50001atm

P_{B}=1-2*50001=-100001atm

Not sense.

With x2

P_{A}= 5*10^{-6}atm

P_{B}=1-2*5*10^{-6}= 0.9999 atm

X2 is the solution for this equation.  

8 0
3 years ago
When the white part is on the left the moon is.??​
Kazeer [188]

Answer:

Last Quarter also called Third Quarter.

Explanation:

6 0
3 years ago
I went for a walk the other day. I went four blocks east, then seven blocks south, then one block west and finally
nalin [4]

Answer:

a) distance is 4+7+1+8=20 blocks

b) displacement is 10 blocks

Explanation:

find displacement: x and y

x axis displacement = 4-1 = 3 blocks

y axis displacement = -7+8= 1 block

displacement = the square root of 3^2 + 1^2

= 9+1 = 10 blocks.

You can find the angle of displacement with respect to the initial position using trig identities, if you wish.

4 0
3 years ago
Which is false HCI + CaCO3
tekilochka [14]

Answer:

CaCO3 is false

Explanation:

Because HCl is hydrongen chloride

7 0
3 years ago
Please help, with step by step work
natali 33 [55]

\qquad ☀️\pink{\bf{ {Answer  = \: \:   85.57g }}}

Molar mass of \bf Cu_2O

\qquad \twoheadrightarrow\sf 63.546 \times 2 +16

\qquad \pink{\twoheadrightarrow\bf 143.092 g}

<u>As we know</u>–

1 mol =\bf 6.02×10^{23} formula units

1 mol\bf Cu_2O = 143.092 g = \bf 6.02×10^{23}formula units

Henceforth –

\bf 3.60×10^{23} formula units \bf Cu_2O–

\qquad \sf :\implies \dfrac{143.092 \times3.60×10^{23  }}{6.02×10^{23}}

\qquad \sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23  }}}{6.02×\cancel{10^{23}}}

\qquad \pink{:\implies\bf 85.57 g}

5 0
2 years ago
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