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3 years ago

PLEASE HELP :-)

1 answer:

6 0

6.4 x 10^-7 = [CO]^2[O2]/ [CO2}^2 = ( 2.0 x 10^-3)^2 ( 1.0 x 10^-3)/ [CO2]^2 =
=4.0 x 10^-9 / [CO2]^2 

[CO]= sq.rt ( 4.0 x 10^-9)/ 6.4 x 10^-7=7.9 x 10^-2 M 

2.6 x 10^-3 = [I]^2 / [I2] = [I]^2 / 0.95 
[I]= sq.rt ( 2.6 x 10^-3 x 0.95)=5.0 x 10^-2 M 

Ksp = [Ba2+][CO32-] = ( 1.1 x 10^-4)^2=1.2 x 10^-8

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3 years ago

4.41 g of propane gas (C3H8) is injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below

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Answer : The heat of the reaction is -221.6 kJ

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Heat released by the reaction = Heat absorbed by the calorimeter

$q_{rxn}=-q_{cal}$

$q_{cal}=c_{cal}\times \Delta T$

where,

$q_{rxn}$ = heat released by the reaction = ?

$q_{cal}$ = heat absorbed by the calorimeter

$c_{cal}$ = specific heat of calorimeter = $97.1kJ/^oC=97100J/^oC$

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=(27.282-25.000)=2.282^oC$

Now put all the given values in the above formula, we get:

$q_{cal}=(97100J/^oC)\times (2.282^oC)$

$q_{cal}=221582.2J=221.6kJ$

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3 years ago

Phosgene (COCl2) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: CO(g) + Cl2

lapo4ka [179]

Answer: Concentration of $CO$ = 0.328 M

Concentration of $Cl_2$ = 0.328 M

Concentration of $COCl_2$ = 0.532 M

Explanation:

Moles of $CO$ and $Cl_2$ = 0.430 mole

Volume of solution = 0.500 L

Initial concentration of $CO$ and $Cl_2$ = $\frac{moles}{volume}=\frac{0.430}{0.500}=0.860M$

The given balanced equilibrium reaction is,

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

Initial conc. 0.860M 0.860M 0

At eqm. conc. (0.860-x) M (0.860-x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[COCl_2]}{[CO][Cl_2]}$

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

$4.95=\frac{x}{(0.860-x)^2}$

By solving the term 'x', we get :

x = 0.532 M

Thus, the concentrations of $CO,Cl_2\text{ and }COCl_2$ at equilibrium are :

Concentration of $CO$ = (0.860-x) M =(0.860-0.532) M = 0.328 M

Concentration of $Cl_2$ = (0.860-x) M = (0.860-0.532) M = 0.328 M

Concentration of $COCl_2$ = x M = 0.532 M

3 0

3 years ago