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marissa [1.9K]
3 years ago
12

Nitrogen monoxide, NO, is formed in automobile exhaust by the reaction of nitrogen and oxygen in the air: N2 (g) + O2 (g) ↔ NO (

g) BALANCE THE EQUATION The equilibrium constant Kc is 0.0025 at 2127 oC. If an equilibrium mixture at this temperature contains 0.023 M N2 and 0.031 M O2, what is the equilibrium concentration of NO? __________ M NO
Chemistry
1 answer:
Solnce55 [7]3 years ago
6 0

Answer:

The equilibrium concentration of NO is 0.001335 M

Explanation:

Step 1: Data given

The equilibrium constant Kc is 0.0025 at 2127 °C

An equilibrium mixture contains 0.023M N2 and 0.031 M O2,

Step 2: The balanced equation

N2(g) + O2(g) ↔ 2NO(g)

Step 3:  Concentration at the equilibrium

[N2] = 0.023 M

[O2] = 0.031 M

Kc = 0.0025 = [NO]² / [N2][O2]

Kc = 0.0025 = [NO]² / (0.023)(0.031)

[NO] = 0.001335 M

The equilibrium concentration of NO is 0.001335 M

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Read 2 more answers
Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.20 MJ 1
mars1129 [50]

Answer : The mass of liquid sodium needed are, 3.17\times 10^3g

Explanation :

First we have to calculate the number of moles of liquid sodium.

q=n\times c\times (\Delta T)

where,

q = heat absorb = 1.20 MJ = 1.20 × 10⁶ J

n = number of moles of liquid sodium = ?

c = specific heat capacity = 30.8J/mol.K

\Delta T = change in temperature = 10.0^oC=273+10.0=283K

Now put all the given values in the above formula, we get:

1.20\times 10^6J=n\times 30.8J/mol.K\times 283K

n=137.7mol

Now we have to calculate the mass of liquid sodium.

Molar mass of Na = 23 g/mol

\text{Mass of Na}=\text{Moles of Na}\times \text{Molar mass of Na}

\text{Mass of Na}=137.7mol\times 23g/mol=3167.1g=3.17\times 10^3g

Therefore, the mass of liquid sodium needed are, 3.17\times 10^3g

3 0
4 years ago
I need help asap please
monitta

unbalanced

there has to be 4 hydrogens in total on both sides

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3 years ago
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