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Sauron [17]
2 years ago
12

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibriu

m is established: 2NO(g)+2H2(g)⥫⥬==N2(g)+2H2O(g) At equilibrium [NO]=0.062M. Calculate the equilibrium concentrations of H2, N2, and H2O.
Calculate the equilibrium concentration of N2
Chemistry
1 answer:
tatyana61 [14]2 years ago
5 0

Answer:

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

                  2NO(g)         +    2H2(g)    ⇒        N2(g)      +         2H2O(g)

i  mol            0.10                   0.050                                             0.10

c mol            -0.038                -0.038                +0019                +0.038                                                

e mol            0.062                 0.012                  00.019               0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] =   0.062 M

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

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Answer:

B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2

Explanation:

Hello,

In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.

Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.

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8 0
2 years ago
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F = ma.

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QED!

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