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SCORPION-xisa [38]
3 years ago
6

Phosphorus pentachloride decomposes to phosphorus trichloride at high temperatures according to the equation:

Chemistry
1 answer:
weqwewe [10]3 years ago
7 0

Answer:

D) [PCl5]=0.00765M,[PCl3]=0.117M,and [Cl2]=0.0.117M

Explanation:

Based on the reaction:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

And knowing:

Kc = [PCl₃] [Cl₂] / [PCl₅] = 1.80

When you add PCl₅ into a flask, this gas will react producing PCl₃ and Cl₂ until [PCl₃] [Cl₂] / [PCl₅] = 1.80

This could be written as:

[PCl₃] = X

[Cl₂] = X

[PCl₅] = 0.125M - X

<em>Where X represents the moles of PCl₅ that react, </em><em>reaction coordinate.</em>

Replacing in Kc expression:

[PCl₃] [Cl₂] / [PCl₅] = 1.80

[X [X] / [0.125 - X] = 1.80

X² = 0.225 - 1.80X

0 = -X² -1.80X + 0.225

Solving for X:

X = -1.9M → False solution, there is no negative concentrations

X = 0.11735M → Right solution.

Replacing, concentrations in equilibrium are:

[PCl₃] = X

[Cl₂] = X

[PCl₅] = 0.125M - X

[PCl₃] = 0.117M

[Cl₂] = 0.117M

[PCl₅] = 0.00765M

And right option is:

<h3>D) [PCl5]=0.00765M,[PCl3]=0.117M,and [Cl2]=0.0.117M</h3>
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4 0
3 years ago
In a single replacement reaction between 2.57 moles Aluminum and 3.59 moles of Hydrochloric acid, how many moles of Hydrogen can
Alenkasestr [34]

Answer:

1.795 mole of H2.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2Al + 6HCl —> 2AlCl3 + 3H2

Step 2:

Determination of the limiting reactant.

From the balanced equation above,

2 moles of Al reacted with 6 moles

Therefore, 2.57 moles of Al will react with = (2.57 x 6)/2 = 7.71 moles of HCl.

From the calculation made above, it will require a higher amount of HCl than what was given to react completely with 2.57 moles of Al. Therefore, HCl is the limiting reactant and Al is the excess reactant.

Step 3:

Determination of the number of mole H2 produced from the reaction.

Here, we shall be using the limiting reactant because it will produce the maximum yield of the reaction since all of it were consumed by the reaction.

The limiting reactant is HCl and the amount of H2 produce can be obtained as follow:

From the balanced equation above,

6 moles of HCl reacted to produce 3 moles of H2.

Therefore, 3.59 moles of HCl will produce = (3.59 x 3)/6 = 1.795 mole of H2.

From the calculations made above, 1.795 mole of H2 is produced from the reaction.

5 0
3 years ago
The ease with which the charge distribution in a molecule can be distorted by an external electric field is called the___. A) el
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Question 1

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Question 2

C) London dispersion forces

Question 3:

D)Kr

Question 4:

E) strong enough to hold molecules relatively close together but not strong enough to keep molecules from moving past each other

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3 years ago
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8 0
3 years ago
The half-life for the radioactive decay of C-14C-14 is 5730 years. You may want to reference (Pages 598 - 605) Section 14.5 whil
KIM [24]

<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life of the reaction = 5730 years

Putting values in above equation, we get:

k=\frac{0.693}{5730yrs}=1.21\times 10^{-4}yrs^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant  = 1.21\times 10^{-4}yr^{-1}

t = time taken for decay process = ? yr

[A_o] = initial amount of the sample = 100 grams

[A] = amount left after decay process =  (100 - 25) = 75 grams

Putting values in above equation, we get:

1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{75}\\\\t=2377.9yrs

Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

8 0
3 years ago
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