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Phantasy [73]
4 years ago
9

A 1.00 L flask was filled with 2.00 mol gaseous SO2 and 2.00 mol gaseous NO2 and heated. After equilibrium was reached, it was f

ound that 1.30 mol gaseous NO was present. Assume that the reaction occurs under these conditions. Calculate the value of the equilibrium constant, K, for the following reaction. SO2(g) NO2(g) equilibrium reaction arrow SO3(g) NO(g)
Chemistry
1 answer:
Nataly_w [17]4 years ago
6 0

Answer:

Explanation:

SO3 (g) + NO (g) U SO2 (g) + NO2 (g)

occurs under these conditions. Calculate the value of the equilibrium constant, Kc, for the above reaction.

SO3 (g) + NO (g) U SO2 (g) + NO2 (g)

Initial (M) 2.00 2.00 0 0

Change (M) −x −x +x +x

Equil (M) 2.00 − x 2.00 − x x x

2 2 c 3

2

c 2

[SO ][NO ]

[SO ][NO]

(2.00 )

=

= −

K

x K

x

Since the problem asks you to solve for Kc, it must indicate in the problem what the value of x is. The concentration of

NO at equilibrium is given to be 1.30 M. In the table above, we have the concentration of NO set equal to 2.00 − x.

2.00 − x = 1.30

x = 0.70

Substituting back into the equilibrium constant expression:

2c 2 2c 2

(2.00 )

(0.70)

(2.00 0.70)

= − = −

x KxK

Kc = 0.290

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3 years ago
UGRENT! Please help showing all work
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Answer:

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6 0
3 years ago
Please answer ASAP!!
Alenkasestr [34]

Answer:

C.0.28 V  

Explanation:

Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:

The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:

Cu²⁺ + 2e⁻ → Cu E° = 0.52V

Ag⁺ + 1e⁻ → Ag E° = 0.80V

As the Cu will be oxidized:

Cu → Cu²⁺ + 2e⁻

The cell potential is:

E°Cell = E°cathode(reduced) - E°cathode(oxidized)

E°cell = 0.80V - (0.52V)

E°cell = 1.32V

Right answer is:

<h3>C.0.28 V </h3>

<h3 />

4 0
3 years ago
A scooter maintains a pace of 1.45 minutes per mile, with fuel economy of 85 miles per gallon. How many seconds can it travel on
vlada-n [284]

Answer:

281s

Explanation:

Given parameters:

Speed of the scooter  = 1.45min per mile

Fuel economy of engine  = 85miles per gallon

Quantity of fuel  = 145mL

Unknown:

Time of travel with the volume of fuel given  = ?

Solution:

To solve this problem, we need to find the distance the fuel will last.

 Rate of fuel consumption by the engine  = 85miles per gallon

Convert 145mL to gallons;

                3785.41mL  = 1 gallon

               145mL will therefore give \frac{145}{3785.41}    = 0.038gallons

So;

   Distance covered  = 85miles per gallon  x 0.038gallons  = 3.23miles

From;

             Rate of travel  = \frac{time }{distance}  

       Time  = rate of travel x distance  = 1.45 minutes per mile x  3.23miles

     Time  = 4.7min

             1 min  = 60s

            4.7min  = 4.7 x 60 = 281s

6 0
3 years ago
5.00 L of air at 750 mmHg pressure was compressed into a 3.00 L steel cylinder. What is the final pressure? (round to significan
kozerog [31]

Answer:

P2 = 1250mmHg

Explanation:

V1 = 5.0L

P1 = 750mmHg

V2 = 3.0L

P2 = ?

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

P = k/V k = P*V

P1*V1 = P2*V2 = P3*V3 =........=Pn*Vn

P1 *V1 = P2 * V2

Solve for P2

P2 = (P1 * V1) / V2

P2 = (750 * 5.0) / 3.0

P2 = 3750 / 3

P2 = 1250mmHg

The final pressure of the gas is 1250mmHg

5 0
3 years ago
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