Considering the definition of Kc, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].
The balanced reaction is:
H₂(g) +Cl₂(g) ⇆ 2 HCl(g)
Equilibrium is a state of a reactant system in which no changes are observed as time passes, despite the fact that the substances present continue to react with each other. In other words, reactants become products and products become reactants and they do so at the same rate.
In other words, chemical equilibrium is established when there are two opposite reactions that take place simultaneously at the same speed.
The concentration of reactants and products at equilibrium is related by the equilibrium constant Kc. Its value in a chemical reaction depends on the temperature and the expression of a generic reaction aA + bB ⇄ cC is
[tex]K_{c} =\frac{[C]^{c} x[D]^{d} }{[A]^{a} x[B]^{b} }[/tex]
That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
In this case, the constant Kc can be expressed as:
[tex]K_{c} =\frac{[HCl]^{2} }{[H_{2} ]x[Cl_{2} ] }[/tex]
You know that in an equilibrium mixture of HCl, Cl₂, and H₂:
the concentration of H₂ is 1.0×10⁻¹¹ [tex]\frac{mol}{L}[/tex]the concentration of Cl₂ is 2.0×10⁻¹⁰ [tex]\frac{mol}{L}[/tex]Kc=4×10¹⁸Replacing in the expression for Kc:
[tex]4x10^{18} =\frac{[HCl]^{2} }{[1x10^{-11} ]x[2x10^{-10} ] }[/tex]
Solving:
[tex]4x10^{18} =\frac{[HCl]^{2} }{2x10^{-21} }[/tex]
[tex]4x10^{18} x 2x10^{-21}=[HCl]^{2}[/tex]
[tex]8x10^{-3} =[HCl]^{2}[/tex]
[tex]\sqrt[2]{8x10^{-3}} =[HCl][/tex]
0.0894 [tex]\frac{mol}{L}[/tex]= [HCl]
Finally, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].
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Answer:
Explanation:
Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. ... After the sample has been applied on the plate, a solvent or solvent mixture (known as the mobile phase) is drawn up the plate via capillary action.
Doing Labs at home
I’m a junior and I’m staying home for this semester and I have to take chemistry and a lot of my work is Labs but I don’t know how to do them since I don’t have the materials at home to do the labs. Someone please help!!!
Answer:
go get the stuff.
Explanation:
Oxidation unit test
Please help ASAP!!!
Which statement correctly describes the oxidation number of the manganese atom (Mn) in Mnl2 and MnO2?
O Manganese has an oxidation number of +4 in Mnl2 and +2 in MnO2.
o Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.
o Manganese has an oxidation number of +4 in both Mnl2 and MnO2.
Manganese has an oxidation number of +2 in both Mnl2 and MnO2.
In this case, according to the given information about the oxidation numbers and the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.
Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.
Next, we can write the following [tex]x[/tex], since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:
[tex]Mn^xI_2^-\\\\Mn ^xO_2^{-2}[/tex]
Next, we multiply each anion's oxidation number by the subscript, to obtain the following:
[tex]Mn^xI_2^-\rightarrow x-2=0;x=+2\\\\Mn ^xO_2^{-2}\rightarrow x-4=0;x=+4[/tex]
Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.
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What identifies the number of protons in the nucleus of an atom?
Answer: Atomic number
Explanation:
I hope this helps you!
a. Calculate the change in enthalpy when 20.0 grams of aluminum metal is heated from 298 K to 573 K at constant pressure of 1 atm.
b. Calculate the change in enthalpy when 20.0 grams of metallic lead is taken through the same process. In both cases assume the heat capacity values predicted by equipartition are valid through the temperature range stated.