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Papessa [141]
3 years ago
7

What is the molarity of a solution made by adding :1.565 moles of PbN03 to 500 mL

Chemistry
1 answer:
jeyben [28]3 years ago
6 0

Answer:

This question is a bit vague. We're not told if 500ml is the volume of the solution and additional info is not given to calculate the volume of pbNo3 being added.

If you solve with 500ml as the volume of solution ... You'll have the answer below.

Molarity=moles of solutes/Volume of solvent(in litres)

moles of solute given=1.565moles

Volume = 500ml. There's 1000ml to a litre. Dividing by 1000ml to convert to Litres... You have 0.5L

Molarity = 1.565/0.5

=3.13M

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Read 2 more answers
The equilibrium constant Kc for the decomposition of phosgene, COCl2, is 4.63x10^-3 at 527 C. Calculate the equilibrium partial
Studentka2010 [4]

<u>Answer:</u> The partial pressure of the CO,Cl_2\text{ and }COCl_2 are 0.352 atm, 0.352 atm and 0.408 atm respectively.

<u>Explanation:</u>

We are given:

K_c=4.63\times 10^{-3}

p_{COCl_2}=0.760atm

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

Where,

K_p = equilibrium constant in terms of partial pressure = ?

K_c = equilibrium constant in terms of concentration = 4.63\times 10^{-3}

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 527^oC=527+273=800K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-1=1

Putting values in above equation, we get:

K_p=4.63\times 10^{-3}\times (0.0821\times 800)^{1}\\\\K_p=0.304

The chemical reaction for the decomposition of phosgene follows the equation:

                   COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

At t = 0          0.760            0     0  

At t=t_{eq}      0.760-x          x      x

The expression for K_p for the given reaction follows:

K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}

We are given:

K_p=0.304\\p_{COCl_2}=0.760-x\\p_{CO}=x\\p_{Cl_2}=x

Putting values in above equation, we get:

0.304=\frac{x\times x}{0.760-x}\\\\x=0.352,-0.656

Negative value of 'x' is neglected because partial pressure cannot be negative.

So, the partial pressure for the components at equilibrium are:

p_{COCl_2}=0.760-0.352=0.408atm\\\\p_{CO}=0.352atm\\\\p_{Cl_2}=0.352atm

Hence, the partial pressure of the CO,Cl_2\text{ and }COCl_2 are 0.352 atm, 0.352 atm and 0.408 atm respectively.

6 0
3 years ago
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