How many molecules of propane were in the erlenmeyer flask? Avogadro's number is 6. 022 × 10^23 molecules/mol
1 answer:
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 ×
- 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.
- 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.
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Answer:
A)
<u>4, 7, 4, 6</u>
B)
<u>12 moles</u>
Explanation:
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8.00 mol | 14.00 mol
________________
You can turn this into a system of variables which are solvable.
To do this, create variables for the coefficients of each compound in the reaction respectively.
Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.
a corresponds to the coefficient of the first compound, b corresponds to the coefficient of the second compound, c corresponds to the coefficient of the third compound, and d corresponds to the coefficient of the fourth compound.
(Reactant = Product)
Reactant: 1a [N] Product: 1c.
Reactant: 3a [H] Product: 2d.
Reactant: 2b [O] Product: 2c + 1d.
Thus the system is:
1a = 1c
3a = 2d
2b = 2c + 1d.
Then just use the substitution methods to solve.
Answer:
a) [H₃O⁺] = 1.8x10⁻⁵ M
b) pH = 4.75
c) % rxn = 3.5x10⁻³ %
Explanation:
a) The dissociation reaction of HCN is:
HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)
0.5 M - x x x
The dissociation constant from the above reactions is given by:
By solving the above quadratic equation we have:
x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]
Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.
b) The pH is equal to:
Then, the pH of the HCN solution is 4.75.
c) The % reaction is the % ionization:
Therefore, the % reaction or % ionization is 3.5x10⁻³ %.
I hope it helps you!