answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Juli2301 [7.4K]
2 years ago
5

Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) If the VOLUM

E of the equilibrium system is suddenly decreased at constant temperature: The value of Kc A. increases. B. decreases. C. remains the same. The value of Qc A. is greater than Kc. B. is equal to Kc. C. is less than Kc. The reaction must: A. run in the forward direction to reestablish equilibrium. B. run in the reverse direction to reestablish equilibrium. C. remain the same. It is already at equilibrium. The number of moles of Cl2 will: A. increase. B. decrease. C. remain the same.
Chemistry
1 answer:
7nadin3 [17]2 years ago
3 0

Answer:

The value of Kc C. remains the same.

The value of Qc C. is less than Kc.

The reaction must: A. run in the forward direction to reestablish equilibrium

The number of moles of Cl2 will  B. decrease.

Explanation:

Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.

A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.

But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: <u><em>The value of Kc remains the same.</em></u>

<u><em> </em></u>As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:

PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)

The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, <u><em>The reaction must: A. run in the forward direction to reestablish equilibrium</em></u>.

By decreasing the volume, and so that Kc remains constant, being:

Kc=\frac{[PCl_{5} ]}{[PCl_{3}]*[Cl_{2}  ]}=\frac{\frac{nPCl_{5} }{Volume} }{\frac{nPCl_{3}}{Volume}*\frac{nCl_{2} }{Volume}  } =\frac{nPCl_{5}}{nPCl_{3}*nCl_{2}} *Volume

 where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂

so,  the number of moles of Cl₂ should decrease.<u><em>The number of moles of Cl2 will  B. decrease.</em></u>

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system will evolve to the right, the direct reaction prevailing, to increase the concentration of products. So in this case, if the reaction moves to the right, <em><u>the value of Qc C. is less than Kc.</u></em>

You might be interested in
6.0 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 13
vodomira [7]

<u>Answer:</u>

<em>The molecular formula of X is given as C_7 H_6 O_3</em>

<em></em>

<u>Explanation:</u>

Moles $C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol}$\\\\moles $\mathrm{C}=$ moles $\mathrm{CO}_{2}=0.304 \mathrm{mol}$

mass $C=$ moles $\times$ molar mass $=0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g$\\\\moles $\mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol}$\\\\moles $\mathrm{H}=2 \times$ moles $\mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol}$\\\\Mass $\mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$

mass O = Total mass of the compound - (mass of C + mass of H)

=6.0 g - ( 3.65 + 0.262 ) g

=2.09 g

moles $O=\frac{2.09 g}{16 g \text { per mole }}=0.131 \mathrm{mol}$

Least moles is for O that is 0.131mol and dividing all by the least we get

$\begin{aligned} C &=\frac{0.304}{0.131}=2.3 \\\\ H &=\frac{0.260}{0.131}=2 \\\\ O &=\frac{0.131}{0.131}=1 \end{aligned}$

Since 2.3 is a fraction it has to be converted to a whole number so we multiply all the answers by 3

\\$C 2.3 \times 3=7$\\\\$H 2 \times 3=6$\\\\$O 1 \times 3=3$

So the empirical formula is C_7 H_6 O_3

Empirical formula mass

=(7 \times 12) +(6\times1.008)+(3\times16)=138.048g

$n=\frac{\text { molar mass }}{\text { empirical formula mass }}=\frac{138}{138.048}=1$

Molecular formula =n × empirical formula

=1 \times C_7 H_6 O_3

Compound X  = C_7 H_6 O_3  is the Answer

8 0
2 years ago
A NiCd battery uses nickel and cadmium to produce a potential difference. Using these equations, answer the following questions.
Dahasolnce [82]

Answer : The correct chemical reaction within the galvanic cell is,

(3) Cd(s)+2NiO(OH)(s)+2H_2O(l)\rightarrow 2Ni(OH)_2(s)+Cd(OH)_2(s)

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the electrochemical cell.

The redox reaction occurs between the nickel and cadmium.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

The balanced two-half reactions will be,

Oxidation half reaction : Cd(s)+2OH^-(aq)\rightarrow Cd(OH)_2(aq)+2e^-

Reduction half reaction : NiO(OH)(aq)+H_2O(l)+e^-\rightarrow Ni(OH)_2(aq)+OH^-(aq)

Thus the overall reaction will be,

Cd(s)+2NiO(OH)(s)+2H_2O(l)\rightarrow 2Ni(OH)_2(s)+Cd(OH)_2(s)

6 0
2 years ago
Identify factors that account for the high phosphoryl‑transfer potential of the nucleoside triphosphates, NTPs.
lesantik [10]

Answer:

B) stabilization by hydration

C) resonance stabilization

E) increase in entropy

Explanation:

The high phosphoryl potential of ATP results from structural differences that exist between ATP and it's product of hydrolysis. There is higher phosphoryl transfer potential from ATP than glycerol 3-phosphate.

There are some factors associated to the high phosphoryl-transfer potential of ATP which are;

1.)Electrostatic repulsion

2.) Resonance stabilization

3.) Increase in entropy.

4. Stabilization by hydration.

ATP has a phosphoryl-transfer potential which lyes between high phosphoryl-potential compounds that is a derivation of fuel molecules and acceptor molecules that needs the adequate addition of a phosphoryl group for cellular needs.

8 0
2 years ago
In a different experiment, a student places a piece of pure Ba(s) in a beaker containing 250.mL of 6.44MHCl(aq) and observes tha
kkurt [141]

Answer: The Ionization energy  of Barium is lower and hence it can lose its electrons easily.

Explanation:

Ionization energy is defined as the energy required to remove an electron from an isolated gaseous atom. It is represented as E_i

This energy will be higher for fully filled and half-filled electronic configuration than partially filled electronic configuration. This is so because half filled and fully filled configurations are stable.

Magnesium is the 12th element of the periodic table having electronic configuration of = [Ne]3s^2

Barium is the 56th element of the periodic table having electronic configuration of [Xe]6s^2

Both magnesium and barium belong to the same group as they have similar valence electrons. But as Barium is larger in size, the valence electrons lie farther from the nucleus and lesser energy is required to remove the electron and thus ionization energy will be lower for barium and thus will be more reactive.

6 0
2 years ago
Choose chlorous acid, hclo2, from the list above and press the equilibrate button. when the 0.10 m solution of hclo2 is allowed
Arturiano [62]
Chlorous Acid Ionizes as,

                                        HClO₂   ⇆   H⁺  +   ClO₂⁻

                                        Ka  =  [H⁺] [ClO₂⁻] / [HClO₂]

 Ka of Chlorous Acid  =  1.1 × 10⁻²

The concentration of H⁺ ions at equilibrium are calculated as,

Initial                               0.1 M     ⇆    0          0
At Equilibrium                 0.1-X      ⇆    X         X
So,
                                       Ka  =  [X] [X] / [0.1-X]
Putting value of Ka,
                                1.1 × 10⁻²  =  X² / 0.1-X
Solving for X,
(www.cymath.com)
                                 X  =  0.028 M

Hence at equilibrium concentration of H⁺ and ClO₂⁻ is 0.028 M.

Percentage Ionization is calculated as,

                                 =  [H⁺] / [HA] × 100

                                 =  (0.028 / 0.1) × 100

                                 =  0.28 × 100

                                 =  28 %     (Percentage Ionization)
6 0
2 years ago
Other questions:
  • A gem has a mass of 4.50 g. When the gem is placed in a graduated cylinder 12.00 mL of water, the water level rises to 13.45 mL.
    12·1 answer
  • In which region are most particles moving the fastest?
    12·2 answers
  • Determine the number of atoms in 1.85 ml of mercury. (the density of mercury is 13.5 g/ml.)
    15·1 answer
  • A bottle of concentrated aqueous sulfuric acid, labeled 98.0 wt% h2so4, has a concentration of 18.0 m. (a) how many milliliters
    10·1 answer
  • Identify the limiting reactant when 1.22 g of O2 reacts with 1.05 g H2 to produce water.
    5·1 answer
  • Which of the following statements are True about the experimental process used in the Diels Alder reaction?
    15·1 answer
  • how many grams of calcium oxide will be produced in a closed vessel containing 20.0 kg of calcium and 20.0 kg of oxygen gas if t
    11·1 answer
  • A balloon filled with 1.22 L of gas at 286 K is heated until the
    10·1 answer
  • When collecting temperature as a function of time for the reaction of KOH with HCL, which time is most significant
    8·1 answer
  • g When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!