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zhuklara [117]
3 years ago
8

The production of ammonia (NH3) is exothermic and is in equilibrium with its breakdown (which is endothermic):

Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
8 0

Answer:

The equilibrium will shift to produce less ammonia

Explanation:

According to Le Chatelier's principle, the reaction will try to oppose anything that is done on it, if it was at equilibrium.

When the concentration of H2 is decreased, you are decreasing the concentration of hydrogen so the reaction tries to increase the concentration of hydrogen by breaking down the ammonia on the products side. This will decrease the output of ammonia

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2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

No. of moles of CO = 0.800 mol

No. of moles of H_2 = 2.40 mol

Volume = 8.00 L

Concentration = \frac{Moles}{Volume\ in\ L}

Concentration of CO = \frac{0.800}{8.00} = 0.100\ mol/L

Concentration of H_2 = \frac{2.40}{8.00} = 0.300\ mol/L

                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium H_2O (x) = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium H_2 = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium CH_4 = 0.0386 M

Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}

Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90

8 0
3 years ago
When most liquids freeze, explain what happens to the motion and the space between the atoms.
KiRa [710]

Answer:

atoms relative motion slow down and begin to vibrate in place

8 0
3 years ago
In the world of chemistry, a 'mole' is a number of something... a very large number of something.
kipiarov [429]

Given:

No of atoms present= 8.022 x 10^23 atoms

Now we know that 1 mole= 6.022 x 10^23 atoms


Hence number of moles present in 8.022 x 10^23 atoms is calculated as below.

Number of moles

= 8.022 x 10^23/6.022x 10^23

=1.3 moles.

Hence we have 1.3 moles present.

3 0
3 years ago
When a connector is marked with "al-cu," the connector is suitable for use with copper, copper-clad aluminum, and aluminum condu
IgorLugansk [536]

When connectors are marked with a combination of metals, it can be used as a connector of one of the metals or an alloy of the two metals. So in this case, since the marking is “Al – Cu” where Al is aluminium and Cu is copper, therefore the answer is:

<span>Yes, it is suitable for use with copper, copper-clad aluminum, and aluminum conductors.</span>

6 0
3 years ago
The specific heat of copper is 0.385 j/g°c which equation would you use
cestrela7 [59]
Since there's specific heat, you should use Q=mc△T. Depends on if this question also involves phase change or not, you might will need Lf (latent heat of fusion) or Lv (latent heat of vaporisation).
8 0
3 years ago
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