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
A.) Phosphate ion or Orthophosphate
d.) Hydroxide
D.) Ammonium
e.) Iron
C.) Nitrate
f.) Sulfur dioxide
Answer:
148 grams of relative atomic mass
Explanation:
magnesium atomic mass : 24
nitrogen : 14
oxygen : 16
24 × 1
14 × 1 × 2
16 × 3 × 2
24 + 28 + 80 = 148 grams
Answer:
pOH = 5.961
Explanation:
To find the pH of a weak base we can use Henderson-Hasselbalch equation for weak bases:
pOH = pKb + log [(CH₃)₃NHCl] / [(CH₃)₃N]
<em>Where pKb is -log Kb = 4.187 and [] could be taken as moles of each specie.</em>
<em />
<em>Moles (CH₃)₃NHCl:</em>
0.0441L * (0.15mol/L) = 6.615x10⁻³moles
<em>Moles (CH₃)₃N:</em>
0.0233L * (0.16mol/L) = 3.728x10⁻³moles
And pOH is:
pOH = pKb + log [(CH₃)₃NHCl] / [(CH₃)₃N]
pOH = 4.187 + log [6.615x10⁻³moles] / [3.728x10⁻³moles]
<h3>pOH = 5.961</h3>
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