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SpyIntel [72]
3 years ago
9

Nitric oxide, an important pollutant in air, is formed from the elements nitrogen and oxygen at high temperatures, such as those

obtained when gasoline burns in an automobile engine. At 2000°C, K for the reaction N2(g) + O2(g) 2NO(g) is 0.01. Predict the direction in which the system will move to reach equilibrium at 2000°C if 0.4 moles of N 2, 0.1 moles of O 2, and 0.08 moles of NO are placed in a 1.0-liter container.
Chemistry
1 answer:
Elden [556K]3 years ago
5 0

Answer:

The Equilibrium will shift towards the left.

Explanation:

The reaction for the formation of nitric oxide is follows,

N_2(g)+O_2(g)\leftrightarrow2NO(g)

Expression for reaction quotient is as follows,

Q=\frac{[NO]^2}{[N_2][O_2]}

Putting the values according to the data given and calculating reaction quotient for the reaction

Q=0.16

So, the reaction quotient is 0.16.

and the value of K is 0.01 .

Q>K

Since the value of K is less than Q, therefore the reaction will shift towards left.

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1. In the first step of the mechanism for this process, a phenoxide anion is generated. This phenoxide anion goes on to act as a
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Answer:

See the explanation

Explanation:

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3 0
3 years ago
Determine the temperature change when 150. G block of gold is supplied with 1.00 • 10 ^3 J of heat
IceJOKER [234]

Answer:

∇T = 51.68°C

Explanation:

Mass = 150g

Heat Energy (Q) = 1.0*10³J

Change in temperature ∇T = ?

Q = mc∇T

Q = heat energy

M = mass

C = specific heat capacity of the gold = 0.129j/g°C

∇T = change in temperature

Q = Mc∇T

1.0*10³ = 150 * 0.129 * ∇T

1000 = 19.35∇T

Solve for ∇T

∇T = 1000 / 19.35

∇T = 51.679°C = 51.68°C

The change in temperature of gold was 51.68°C

8 0
3 years ago
What is the final concentration of a solution prepared by diluting 35.0 ml of 12.0 m hcl to a final volume of 1.20 l? what is th
Dafna1 [17]
We know that, M1V1     =     M2V2
                        (Initial)          (Final)
where, M1 and M2 are initial and final concentration of soution respectively. 
V1 and V2 = initial and final volume of solution respectively

Given: M1 = 12 m, V1 = 35 ml and V2 = 1.2 l = 1200 ml

∴ M2 = M1V1/V2 = (12 × 35)/ 1200 = 0.35 m
Final concentration of solution is 0.35 m
8 0
3 years ago
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