The equilibrium concentrations of all chemical species at room temperature are given below.
[H2] = 0.212M
[I2] = 0.212M
[HI] = 1.576M
H2(g) + I2(g) <-----> 2HI(g)
I 1 1 0
c -x -x 2x
E 1-x 1-x 2x
[H2] ⇒ 1-x/1 = 1-x
[I2] ⇒ 1-x/1 = 1-x
[HI] ⇒ 2x
Kc = [HI]^2/[H2][I2]
55.3 = (2x)^2/(1-x)(1-x)
55.3 = (2x/1-x)^2
7.44 = 2x/1-x
7.44*(1-x) = 2x
x = 0.788
[H2] ⇒ 1-x = 1-0.788 = 0.212M
[I2] ⇒ 1-x = 1-0.788 = 0.212M
[HI] ⇒ 2x = 2*0.788 = 1.576M
In a chemical reaction, while each reactant and the products are in a concentration that does not alternate with time any extra, it is said to be in a state of chemical equilibrium. in this kingdom, the price of forwarding response is the same as the rate of backward response.
Equilibrium is whilst the fee of the ahead reaction equals the charge of the opposite reaction. All reactant and product concentrations are consistent at equilibrium.
We are saying that a chemical is in an equilibrium concentration while the products and reactants do not change as time moves on. In different words, chemical equilibrium or equilibrium awareness is a state when the price of an ahead reaction in a chemical response will become equal to the rate of a backward response.
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a)
Consumption of power in one day = 
Consumption of power in one year = 
Since, 
. So,
Given that 
= 
So, for 
:
$ 
= $
Hence, the annual cost is $.
b)
Consumption of power in one day = 
Consumption of power in one year = 
Since, . So,
Given that =
So, for 
:
$ = $
Hence, the annual cost is $.
Answer:
The number of electrons in the outermost shell of a particular atom determines its reactivity, or tendency to form chemical bonds with other atoms. This outermost shell is known as the valence shell, and the electrons found in it are called valence electrons
Explanation:
The correct option is (C) 6.02 X 10²³
A sample of CH₄O with a mass of 32.0 g contains <u>6.02 X 10²³</u> molecules of CH₄O.
To calculate the number of moles;
Molar mass of CH₄O = C + 4(H) + O
= 12.01 + 4(1.008) + 16
= 32.04 g/mol
So, 1 mol of CH₄O = 32.04 g of CH₄O
Given, 32.0 g of CH₄O
According to Avagadro's constant 1 mole of a substance contains 6.022× 10^23 particles (molecules, atoms or ions).
= (32.0 g/1)(1 mol CH₄O/32.04 g CH₄O)(6.02x10²³/1 mol CH₄O)
= 6.02 X 10²³ molecules of CH₄O
Hence, a sample of will contain number of molecules 6.02 X 10²³ molecules.
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