There are 3 moles of
<span>per 1 mole of salt and 1 mole of
</span>per mole of salt, the total ionic concentrations must be
of
, and
of
Answer:
Propane
Explanation:
From the question given, we were told that 0.1240 kg of propane reacted with excess oxygen to produce 0.3110kg of carbon dioxide.
Since the reaction took place in the presence of excess oxygen, therefore, propane is the limiting reactant as all of it is used up in the presence of excess oxygen.
Answer: 1.414x10^24 molecules in 94.4g MgO
Explanation: molar mass MgO 40.204
molecules in 40.204 g MgO = avogadro number
molecules in 94.4 g MgO = (94.4/40.204)*avogadro number
(94.4/40.204)*6.02214076*10^23 = 14.14x10^23
Answer:
an increase of acceleration or decrease of acceleration
Explanation:
By using the formula, mass = density x volume, we
calculate mass in grams
20.0 mL CH₃COOH x (1.05 g / mL) = 21.0
g CH₃COOH
To find the moles, molar mass of CH₃COOH = 60.05g/mol<span>
21.0 g </span>CH₃COOH x (1 mole CH₃COOH / 60.05 g CH₃COOH)
= 0.350 moles CH₃COOH
To find molarity,<span>
[</span>CH₃COOH] = moles CH₃COOH / L of solution = 0.350 /
1.40 = 0.250 M CH₃COOH<span>
When </span>CH₃COOH is dissolved in water, it produces
small and equal amounts of H₃O⁺+ and C₂H₃O₂⁻.
<span>
Molarity , </span>CH₃COOH<span> + H</span>₂O <==> H₃O⁺ + C₂H₃O₂⁻
<span>
<span>Initial 0.250 0 0 </span>
Change -x x x
Equilibrium 0.250-x x x
K</span>ₐ = [H₃O⁺][C₂H₃O₂⁻] / [HC₂H₃O₂] = (x)(x) /
(0.250-x) = 1.8 x 10⁻⁵
<span>Since K</span>ₐ is relatively small, we can neglect the -x
term after 0.250 to simplify
<span>x</span>² / 0.250 = 1.8 x 10⁻⁵
x² = 4.5 x 10⁻⁶
<span>
x = 2.1 x 10</span>⁻³<span> = [H</span>₃O⁺]
pH = -log [H₃O⁺] = -log (2.1 x 10⁻³) = 2.68
