<h3>Answer:</h3>
a) Moles of Caffeine = 1.0 × 10⁻⁴ mol
b) Moles of Ethanol = 4.5 × 10⁻³ mol
<h3>Solution:</h3>
Data Given:
Mass of Caffeine = 20 mg = 0.02 g
M.Mass of Caffeine = 194.19 g.mol⁻¹
Molecules of Ethanol = 2.72 × 10²¹
Calculate Moles of Caffeine as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 0.02 g ÷ 194.19 g.mol⁻¹
Moles = 1.0 × 10⁻⁴ mol
Calculate Moles of Ethanol as,
As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.
The relation between Moles, Number of Particles and Avogadro's Number is given as,
Number of Moles = Number of Molecules ÷ 6.022 × 10²³
Putting values,
Number of Moles = 2.72 × 10²¹ Molecules ÷ 6.022 × 10²³
Number of Moles = 4.5 × 10⁻³ Moles
Answer:
Neon
Explanation:
The most stable Atoms are from the elements on group 18 of the predictable.
Answer:
10.71%
Explanation:
The dissociation of acetic acid can be well expressed as follow:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:
Then:
The I.C.E Table is expressed as follows:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Initial 0.0014 0 0
Change - x +x +x
Equilibrium (0.0014 - x) x x
Recall that:
Ka for acetic acid CH₃COOH = 1.8×10⁻⁵
∴
![K_a = \dfrac{[x][x]]}{[0.0014-x]}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cdfrac%7B%5Bx%5D%5Bx%5D%5D%7D%7B%5B0.0014-x%5D%7D)
![1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}](https://tex.z-dn.net/?f=1.8%2A10%5E%7B-5%7D%20%3D%20%5Cdfrac%7B%5Bx%5D%5Bx%5D%5D%7D%7B%5B0.0014-x%5D%7D)
![1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}](https://tex.z-dn.net/?f=1.8%2A10%5E%7B-5%7D%20%3D%20%5Cdfrac%7B%5Bx%5D%5E2%7D%7B%5B0.0014-x%5D%7D)
By rearrangement:
Multiplying through by (-) and solving the quadratic equation:
x = 0.00015 or x = -0.000168
We will only consider the positive value;
so x=[CH₃COO⁻] = [H⁺] = 0.00015
CH₃COOH = (0.0014 - 0.00015) = 0.00125
However, the percentage fraction of the dissociated acetic acid is:
= 10.71%
A civilização egípcia se desenvolveu ao longo do rio Nilo em grande parte porque as enchentes anuais do rio garantiam um solo rico e confiável para o cultivo.
Ok first, we have to create a balanced equation for the dissolution of nitrous acid.
HNO2 <-> H(+) + NO2(-)
Next, create an ICE table
HNO2 <--> H+ NO2-
[]i 0.139M 0M 0M
Δ[] -x +x +x
[]f 0.139-x x x
Then, using the concentration equation, you get
4.5x10^-4 = [H+][NO2-]/[HNO2]
4.5x10^-4 = x*x / .139 - x
However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,
assume 0.139-x ≈ 0.139
4.5x10^-4 = x^2/0.139
Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.
We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.
Then to find percent dissociation, you do final concentration/initial concentration.
0.007M/0.139M = .0503 or
≈5.03% dissociation.
