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Mila [183]
3 years ago
8

A blue jay's small ,curved feet hepls it grip on branches.Which type of adption is this?

Chemistry
1 answer:
Bingel [31]3 years ago
5 0
The answer is physical adaption.
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Consider separate 1.0-L samples of He(g) and UF6(g), both at 1.00 atm and containing the same number of moles. What ratio of tem
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V = RMS velocity

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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

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