Answer:
[H⁺] = 0.000048936M
pH = 4.31
Explanation:
Acetic acid, CH₃COOH, dissociates in water as follows:
CH₃COOH(aq) ⇄ H⁺(aq) + CH₃COO⁻(aq)
And Ka is defined as:
Ka = 1.76x10⁻⁵ = [H⁺] [CH₃COO⁻] / [CH₃COOH]
<em>Where [] are equilibrium concentrations of the species.</em>
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The 0.000185M of acetic acid will decreases X, and X of [H⁺] and [CH₃COO⁻] will be produced. That means Ka is:
1.76x10⁻⁵ = [X] [X] / [0.000185 - X]
3.256x10⁻⁹ - 1.76x10⁻⁵X = X²
3.256x10⁻⁹ - 1.76x10⁻⁵X - X² = 0
Solving for X:
X = -0.000066M → False solution. There is no negative concentrations.
X = 0.000048936
As [H⁺] = X,
[H⁺] = 0.000048936M
And pH = -log [H⁺]
<h3>pH = 4.31</h3>
Answer:
Mass = 20 g
Explanation:
Given data:
Number of moles of He = 5 mol
Mass of He = ?
Solution:
Formula:
Number of moles = mass/ molar mass
Molar mass = 4 g/mol
by putting values,
5 mol = Mass / 4 g/mol
Mass = 5 mol × 4 g/mol
Mass = 20 g
Answer:
253.85 L
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
C₃H₈ + 5O₂ —> 3CO₂ + 4H₂O
From the balanced equation above, 4 moles of water vapor were produced.
Finally, we shall determine the volume of the water vapor produced as follow:
Mole water vapor (n) = 4 moles
Pressure (P) = 1 atm
Temperature (T) = 500 °C = 500 °C + 273 = 773 K
Gas constant (R) = 0.821 atm.L/Kmol
Volume (V) =?
PV = nRT
1 × V = 4 ×0.0821 × 773
V = 253.85 L
Thus, the volume of water vapor obtained is 253.85 L
Answer:
0.533 mol O2
Explanation:
4 Fe3O4 + O2 -> 6 Fe2O3
1 mol O2 -> 6 mol Fe2O3
x -> 3.2 mol Fe2O3
x = (3.2 mol Fe2O3 * 1 mol O2)/ 6 mol Fe2O3
x= 0.533 mol O2
