Answer:
11%
Explanation:
1) Calculate van 't Hoff factor:
Δt = i Kf m
0.31 = i (1.86) (0.15)
i = 1.111
2) Calculate value for [H+]:
CCl3COOH ⇌ H+ + CCl3COO¯
total concentration of all ions in solution equals:
(1.11) (0.15) = 0.1665 m
This is a molality, but we will act as if it a molarity since we will assume the density of the solution is 1.00 g/cm3, which makes the molarity equal to the molality.
0.1665 = (0.15 − x) + x + x
x = 0.0165 M
3) Calculate the percent dissociation:
0.0165/ 0.15 = 11 %
I think its the first 1 C i remember answering this question on my school work
Answer:
That means Cu2O is limiting reagent and C is excess reagent
Explanation:
Based on the reaction, 1 mole of Cu2O reacts per mole of C. The ratio of reaction is 1:1.
To solve this question we need to convert the mass of each reactant to moles. The reactant with the lower amount of moles is limiting reactant and the excess reactant is the reactant with the higher number of moles.
<em>Moles Cu2O -Molar mass: 143.09 g/mol-</em>
114.2g Cu2O * (1mol / 143.09g) = 0.798 moles Cu2O
<em>Moles C -Molar mass: 12.01g/mol-</em>
11.1g C * (1mol / 12.01g) = 0.924 moles C
<h3>That means Cu2O is limiting reagent and C is excess reagent</h3>
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The H⁺ ion concentration can be calculated from pH values using the following equation:
![pH=-log[H⁺]](https://tex.z-dn.net/?f=pH%3D-log%5BH%E2%81%BA%5D)
1.) Given pH = 2
Using the above equation, 2 = - log [H⁺]
Therefore, [H⁺] = 10⁻² mol/L
2.) Given pH = 6
Using the same equation, we have 6 = - log [H⁺]
Hence, [H⁺] = 10⁻⁶ mol/L
3.) Taking the ratio of [H⁺] for pH = 2 and pH = 6, we have
= 10⁴
So, there are 10,000 times more H⁺ ions in a solution of pH = 2 than that of pH = 6.
