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shusha [124]
3 years ago
9

For the reaction PCl5(g) + heat PCl3(g) + Cl2(g), what will happen when the pressure is increased? There will not be a shift in

equilibrium. There will be a shift toward the products. There will be a shift toward the reactants. Not enough information is given.
Chemistry
2 answers:
Gnesinka [82]3 years ago
8 0

Answer:

There will be a shift toward the reactants.

Explanation:

According to Le Chatelier’s Principle, when a stress is applied on a system in dynamic equilibrium, the equilibrium shifts in such a way to minimize or counteract the stress. In the given reaction:

PCl₅(g) –> PCl₃(g) + Cl₂(g)

both the reactants and products are gases. But there are higher moles of gaseous particles on the product side.  As the pressure is increased the equilibrium will shift in a direction to reduce the pressure. So the equilibrium will shift towards the reactant side as that side has less moles of gaseous substance.

Tamiku [17]3 years ago
6 0
When the pressure is increased, the equilibrium will shift to the left to offset the pressure increase. Equilibrium shifting to the left side is favored because the left side has fewer moles of gas than the number of moles of gas on the right-hand side and because it exerts less pressure. Therefore, the answer is there will be a shift toward the reactants. 
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Answer: The molar mass of the given compound is 1.11\times 10^2 g/mol

Explanation:

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The atomic mass of the calcium = 40.07 g/mol

The atomic mass of chlorine = 35.5 g/mol

Molar mass of the guven compound is :

40.07 g/mol + 2\times 35.5 g/mol=111.07 g/mol\approx 1.11\times 10^2 g/mol

The molar mass of the given compound is 1.11\times 10^2 g/mol

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Answer:

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A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles
alexgriva [62]

Answer : The pH of the solution is, 3.41

Explanation :

First we have to calculate the moles of HF.

\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}

\text{Moles of HF}=0.250M\times 1.50L=0.375mol

Now we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (6.8\times 10^{-4})

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pK_a=3.17

The reaction will be:

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Initial moles     0.375     0.100   0.375

At eqm.   (0.375-0.100)      0     (0.375+0.100)

                     = 0.275                    = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[F^-]}{[HF]}

Now put all the given values in this expression, we get:

pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]

pH=3.41

Thus, the pH of the solution is, 3.41

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