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geniusboy [140]
3 years ago
8

Which equilibrium reaction will experience a shift towards the products in equilibrium position when the concentration of Ni2+ i

s increased? View Available Hint(s) Which equilibrium reaction will experience a shift towards the products in equilibrium position when the concentration of is increased? Ni(OH)2(s)⇌Ni2+(aq)+2OH−(aq) [Ni(H2O)6]2+(aq)+3en(aq)⇌[Ni(en)3]2+(aq)+6NH3(aq) Ni2+(aq)+6NH3(aq)⇌[Ni(NH3)6]2+ NiS(s)⇌Ni2+(aq)+S2−(aq)
Chemistry
1 answer:
shutvik [7]3 years ago
7 0

Explanation:

According to Le Chatelier's principle, any disturbance occurring in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

As the given reactions are as follows.

  •  Ni(OH)_{2}(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^{-}(aq)

When concentration of Ni^{2+} is increased in this reaction then reaction will shift in a direction that will be opposing the change. This means that the reaction will shift in backward direction.

  • [Ni(H_{2}O)_{6}]^{2+}(aq) + 3en(aq) \rightleftharpoons [Ni(en)_{3}]^{2+}(aq) + 6NH_{3}(aq)

In this reaction, both water and ethylenediamine are neutral molecules. Hence, charge on nickel on each side will be equal to +2. Therefore, reaction will remain in equilibrium.

  • [Ni^{2+}(aq) + 6NH_{3}(aq) \rightleftharpoons [Ni(NH_{3})_{6}]^{2+}(aq)

When we increase the concentration of Ni^{2+} then equilibrium reaction will shift in the forward direction.

  • [NiS(aq) \rightleftharpoons Ni^{2+}(aq) + S^{2-}(aq)

Here, on increasing the concentration of Ni^{2+} will shift the equilibrium in backward direction.

Thus, we can conclude that out of the given options [Ni^{2+}(aq) + 6NH_{3}(aq) \rightleftharpoons [Ni(NH_{3})_{6}]^{2+}(aq) is the chemical reaction in which equilibrium reaction will experience a shift towards the products in equilibrium position when the concentration of Ni^{2+} is increased.

 

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dlinn [17]

Answer:

A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.

Step by Step Explanation?

Boyle's law states that in constant temperature the variation volume of gas is inversely proportional to the applied pressure.

The formula is,

P₁ x V₁ = P₂ × V₂

Where,

P₁ is initial pressure = 1 atm

P2 is final pressure = ? (Not Known)

V₁ is initial volume = 10 L

V₂ is final volume = 15 L

Now put the values in the formula,

\begin{gathered}\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67\end{gathered]

Therefore, the answer is 0.67 atm.

5 0
2 years ago
Read 2 more answers
Write the balanced NET IONIC equation for the reaction that occurs when ammonium nitrate and potassium hydroxide are combined. N
vova2212 [387]

Answer:

Net Ionic equation

NH₄⁺ + OH⁻ → NH₃ + H₂O

Option B is correct.

Weak Acid Strong Base

Check Explanation for the extent of the reaction.

Explanation:

Ammonium nitrate = NH₄NO₃

Potassium Hydroxide = KOH

Ammonium salts combine with alkalis to liberate NH₃ and form water.

The two reactants combine to give

NH₄NO₃ + KOH → KNO₃ + NH₃ + H₂O

In ionic form,

- NH₄NO₃ exists as NH₄⁺ and NO₃⁻

- KOH exists as K⁺ and OH⁻

- KNO₃ as K⁺ and NO₃⁻

And NH₃ and H₂O stay as they are, as per covalent compounds.

So, we have

NH₄⁺ + NO₃⁻ + K⁺ + OH⁻ → K⁺ + NO₃⁻ + NH₃ + H₂O

Eliminating the ions that exist on both sides, we have the net ionic equation to be

NH₄⁺ + OH⁻ → NH₃ + H₂O

which shows that this reaction is essentially a neutralization reaction in which the Bronsted Lowry acid, NH₄⁺, loses its proton to the base, OH⁻ and gives conjugate base, NH₃ and conjugate acid, H₂O.

This reaction is classified as a Weak acid versus Strong Base reaction as NH₄⁺ is from a Weak acid and OH⁻ is from a strong base.

Since this reaction is between a Weak base and a strong acid, the ionization isn't expected to be 100%, Hence, the extent of this reaction will be any option that is not 100%, a couple pieces of information might be required for the correct estimate, but above 50% seems correct.

Hope this Helps!!!

8 0
3 years ago
Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

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3 years ago
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The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands
Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

E=\frac{R_y}{n^2}

R_y=2.18\times 10^{-18} J =  Rydberg energy

n =  principal quantum number of the orbital

Energy of 11th orbit = E_{11}

E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J

Energy of 10th orbit = E_{10}

E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J

=E'=0.38\times 10^{-20} J

\lambda =\frac{hc}{E'} (Planck's' equation)

\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}

\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

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2 years ago
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