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bija089 [108]
3 years ago
11

Given the equation representing a system at equilibrium:

Chemistry
2 answers:
marissa [1.9K]3 years ago
8 0

Answer:

Option (4) increasing the temperature

Explanation:

PCl5(g) + energy ⇌ PCl3(g) + Cl2(g)

The reaction above clearly indicates endothermic reaction since heat is required for the reaction to proceed to product.

As the heat is supplied, the temperature of reaction increases thereby making the reaction to proceed forward at a much faster rate and hence the equilibrium position will shift to the right. This is in accordance with Le Chatelier's principle will explained that for an endothermic reaction, an increase in temperature will cause the equilibrium position to shift to the right.

Yuliya22 [10]3 years ago
4 0

Answer:

increasing the temperature

Explanation:

Now look carefully at the reaction equation, notice the inclusion on an energy term on the left hand side;

PCl5(g) + energy ⇌ PCl3(g) + Cl2(g)

The inclusion of an energy term means that the reaction is endothermic. Energy is absorbed as the reaction goes from left to right.

Since energy is absorbed, increasing the temperature (supplying energy in the form of heat) will favour the forward reaction over the reverse reaction in accordance with Le Chatelier's principle. Hence the answer.

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The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate
Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

6 0
3 years ago
Enter a balanced equation for the reaction between solid nickel(II)(II) oxide and carbon monoxide gas that produces solid nickel
velikii [3]

Answer: A balanced equation for the given reaction is NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g).

Explanation:

The reaction equation will be as follows.

NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)

Number of atoms on the reactant side is as follows.

  • O = 2
  • C = 1

Number of atoms on the product side is as follows.

  • Ni = 1
  • O = 2
  • C = 1

Since number of atoms on both the reactant and product sides are equal. Hence, the reaction equation is balanced.

Thus, we can conclude that a balanced equation for the given reaction is NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g).

7 0
3 years ago
Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,
8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

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