3.74×
3.74 × molecules of propane were in the erlenmeyer flask.
number of moles of propane can be calculated as moles of propane.
mass of propane = 0.274 g
molar mass of propane = 44.1
So this gives us the value of 6.21×
moles of propane
No one mole of propane As a 6.0-2 × 
so, 6.21 × × 6. 022 × 10^23
= 3.74 ×
Therefore, molecules of propane were in the erlenmeyer flask is found to be 3.74 ×
<h3>What is erlenmeyer flask?</h3>
- A laboratory flask with a flat bottom, a conical body, and a cylindrical neck is known as an Erlenmeyer flask, sometimes known as a conical flask or a titration flask.
- It bears the name Emil Erlenmeyer after the German chemist.
<h3>What purpose does an Erlenmeyer flask serve?</h3>
- Liquids are contained in Erlenmeyer flasks, which are also used for mixing, heating, chilling, incubating, filtering, storing, and other liquid-handling procedures.
- For titrations and boiling liquids, their sloped sides and small necks make it possible to whirl the contents without worrying about spills.
To learn more about calculating total molecules visit:
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Answer:
28.93 g/mol
Explanation:
This is an extension of Graham's Law of Effusion where 
We're only talking about molar mass and time (t) here so we'll just concentrate on 
. Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.
Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.
The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.
M2= 0.22 x 131
M2= 28.93 g/mol
Using hydrogen and Lindlar catalyst the triple bond will be hydrogenated to a double one with a cis conformation.
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
