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iogann1982 [59]
2 years ago
11

Consider this equilibrium: N2(g) + 3H2(g) 2NH3(g) + energy. Certain conditions provide less than 10% yield of NH3 at equilibrium

. Which statement describes this equilibrium?
The Keq is large and products are favored.
The Keq is small and products are favored.
The Keq is large and reactants are favored.
The Keq is small and reactants are favored.
The conditions of the reaction must be known to answer the question.
Chemistry
1 answer:
alisha [4.7K]2 years ago
5 0
The correct option is this: THE KEQ IS SMALL AND THE REACTANT ARE FAVOURED.
Since we are told in the question that the reaction give rise to only 10% product, this means that much product are not been produced, thus the reactants are favoured. Keq stands for the equilibrium constant of the reaction and it measures the rate at which the reactants are converted to products. For this reaction, the equilibrium constant is small and the backward reaction [reactant] is favoured.
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A 5-column table with 2 rows. Column 1 is labeled number of protons, with entries A and D; column 2 is number of neutrons, with
pogonyaev

Answer:

A = 8

B = 8

C = Oxygen (O)

D = 26

E = 30

F = Iron (Fe)

Explanation:

protons    neutrons    atomic number    mass number     element

A               7                 B                           15                        C

D               E                 26                         56                       F

mass number = protons + neutrons

E = 56 - 26 = 30

A = 15 - 7 = 8

protons = atomic number

B = 8

D = 26

From atomic number:

C = Oxygen (O)

F = Iron (Fe)

5 0
2 years ago
In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a
drek231 [11]

The enthalpy change of the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction is

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To find the enthalpy change we need to apply the following equation

\Delta H =\frac{Q}{n}

To find the heat (Q):

Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J

Now, to find the number of moles that react (n):

n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles

Having these two values we can plug in the first equation:

\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole

Have a nice day!

5 0
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Write a balanced half-reaction describing the reduction of aqueous chromium(IV) cations to solid chromium.
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<span> Chromium(IV) cations gain four electrons and became solid chromium with neutral charge.
</span>Reduction is lowering oxidation number because element or ions gain electrons. 

Oxidation reaction is increasing of oxidation number of element, because element or ion lost electrons in chemical reaction.

<span>
</span>

3 0
2 years ago
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The final temperature of the lead-water system will be lower than the final temperature of the copper-water system.
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When heating a flammable or volatile solvent for a recrystallization, which of these statements are correct? More than one answe
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Explanation:

A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.

Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.

Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.

And, you should ensure that no one else is using an open flame near your experiment.

Thus, we can conclude that following statements are correct:

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