All the following can result in the disruption of a protein's ability to function, except A. mechanic agitations. B. maintaining
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Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K =
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M
The chemical compound sodium sulfide has the formula Na2S. It is written as . It has a crystalline state. Na is followed by a two or 2. this indicates that the molecule has two of the specified element's atoms.
There is just one of these because there is no number following the S.
In all, this molecule consists of one sulfur atom and two sodium atoms.
A compound's chemical formula, which uses symbols for combining atoms, reveals the composition of the complex. We require the valency of the combining parts in order to write the formula.
The sodium sulfide chemical formula must be written down. So sodium and sulfur make up the two elements in the chemical. Sodium has a valency of +1, while sulfur has a +2.
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Answer:
Explanation:
Moles of = 1 mole
Moles of = 1 mole
Volume of solution = 1 L
Initial concentration of = 1 M
Initial concentration of = 1 M
The given balanced equilibrium reaction is,
Initial conc. 1 M 0M 1 M
At eqm. conc. (1-2x) M (2x) M (1+x) M
The expression for equilibrium constant for this reaction will be,
The =
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
Concentration of at equilibrium= (2x) M =