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andrey2020 [161]
3 years ago
12

Consider the following generic reaction for which Kp = 5.51 × 105 at 25°C:2R(g)+A(g)⇌2Z(g)What is the value of Kc for the reacti

on at the same temperature?1.37 × 10^92.25 × 10^41.35 × 10^75.51 × 10^5
Chemistry
1 answer:
IRINA_888 [86]3 years ago
7 0

Answer:

Kc = 1.35x10^7

Explanation:

Let's write the reaction again:

2R + A <------> 2Z    Kp = 5.51x10^5

In order to know the value of Kc, we need to write the expression that relations the Kp with Kc which is:

Kp = Kc * (RT)^Δn (1)

Where:

R: 0.082 L atm/ K mol

Δn: difference between the coefficients of the reaction

Kc: equilibrium constant

T: temperature in K

Now, from the equation (1) we can solve for Kc:

Kc = Kp / (RT)^Δn (2)

Now, before do any calculations, let's do first the value of Δn:

Δn = 2 - (2+1) = -1

Now, replacing all values in (2):

Kc = 5.51x10^5 / (0.082*298)^-1

Kc = 1.346x10^7

The third option is the correct one.

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Answer:C10H10

Explanation:

Use a balanced equation to determined quantity of carbon and hydrogen. Then use simple division to determine the empirical formula before the molecula formula

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3 years ago
Why is the C-14/C-12 ratio used to date once-living organisms?
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3 years ago
Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) → 2SO3(g) Suppose the volume of this system is c
oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be  increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen  (i.e 2 moles +1 mole  =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be P_1

and the total pressure at the final equilibrium be P_2

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝  1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

P_1V_1 = P_2V_2

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

V_2 =  \dfrac{V_1}{\dfrac{3}{2}} ----- (1)

also;

Thus ;

P_1V_1 = P_2(  \dfrac{V_1}{\frac{3}{2}})

P_1V_1 = P_2 * 2  \dfrac{V_1}{3}

3 P_1 V_1 = 2 P_2 V_1

Dividing both sides by V_1

3P_1 = 2P_2

P_2 =P_1 \dfrac{3}{2}  ----- (2)

From ;

P_1V_1 = P_2V_2

P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}

P_2 V_2 = P_1 * \dfrac{3}{2}*   \dfrac{2 }{3}}*V_1

P_2 V_2 = P_1 V_1

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7 0
3 years ago
Assume that the reaction of aqueous hydrobromic acid solution and potassium hydroxide base undergoes a complete neutralization r
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Answer:

a.

HBr + KOH → KBr + H  _{2} O

b.

1000 \: ml \: contains \: 0.685 \: moles \\ 55.4 \: ml \: contains \: ( \frac{55.4 \times 0.685}{1000} ) \\  = 0.038 \:moles \\ 1 \: mole \: of \: hydrobromic \: acid  \: produces \: 1 \: mole \: of \: water \\ 0.038 \: moles \: produce \: (0.038 \times 1) \\  = 0.038 \: moles \\ 1 \: mole \: of \: water \: weighs \: 18 \: g \\ 0.038 \: moles \: weighs \: (0.038 \times 18) \: g \\  = 0.684 \: g

c.

0.042 M

d.

1.4

5 0
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Answer:

The correct option is volume stays constant

Explanation:

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3 0
3 years ago
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