aluminum, sulfur, and chlorine are elements found on the periodic table. tests involving these three elements would show that at
1 answer:
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Explanation:
It is known that value of acetic acid is 4.74. And, relation between pH and
is as follows.
pH = pK_{a} + log
= 4.74 + log
So, number of moles of NaOH = Volume × Molarity
= 71.0 ml × 0.760 M
= 0.05396 mol
Also, moles of = moles of
= Molarity × Volume
= 1.00 M × 1.00 L
= 1.00 mol
Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.
Initial : 1.00 mol 1.00 mol
NaoH addition: 0.05396 mol
Equilibrium : (1 - 0.05396 mol) 0 (1.00 + 0.05396 mol)
= 0.94604 mol = 1.05396 mol
As, pH = pK_{a} + log
= 4.74 + log
= 4.69
Therefore, change in pH will be calculated as follows.
pH = 4.74 - 4.69
= 0.05
Thus, we can conclude that change in pH of the given solution is 0.05.
Answer: 0.002 m³
Explanation:
We can use our unit conversions to find the volume in m³.
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