Answer:
<h3>The answer is 0.49 mL</h3>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question
mass = 2.34 g
density = 4.78 g/mL
We have
We have the final answer as
<h3>0.49 mL</h3>
Hope this helps you
According to the reaction, when 355 ml of 1.65 m Hydrochloric acid solution are combined with too much aluminum, 26.03 g of aluminum chloride is created.
An inorganic substance with the formula AlCl3 is aluminum chloride, also referred to as aluminum trichloride. It takes the form of [Al(H2O)6] hexahydrate. Aluminum chloride is ALCl3, which is composed of six water molecules. Aqueous hydrochloric acid, also referred to as muriatic acid, is a type of hydrochloric acid. It is a colourless solution with an overpowering odor. Strong acid is how it is categorized.
2 Al(s) + 6 HCl is the given reaction. ———-> 2 AlCl3 + 3H2
Using the provided data
The amount of HCl in moles is n= Volume * Concentration = 0.58575 mol (1.65 M * 355 ml)/1000 ml
based on the response,
It takes 6 mols of HCl. To neutralize 2 moles of aluminum, 0.58575 mol of hydrochloric acid must be used.
using the cross-multiplication method, 0.19525 moles are obtained by multiplying 0.58575% by 2/ 6.
Weight of the AlCl3 = 0.1952 moles * 133.340538 g/mol = 26.03 g
Learn more about Hydrochloric acid here
brainly.com/question/15102013
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Mole=mass/molecular mass (Mr)
mole=volume/ 22.4
Molecular amount: Mole=amount atoms/ Avogadro's number
Concentration: mole=volume x M
The question is incomplete, here is the complete question:
Calculate the pH at of a 0.10 M solution of anilinium chloride 
. Note that aniline 
is a weak base with a 
of 4.87. Round your answer to 1 decimal place.
<u>Answer:</u> The pH of the solution is 5.1
<u>Explanation:</u>
Anilinium chloride is the salt formed by the combination of a weak base (aniline) and a strong acid (HCl).
To calculate the pH of the solution, we use the equation:
![pH=7-\frac{1}{2}[pK_b+\log C]](https://tex.z-dn.net/?f=pH%3D7-%5Cfrac%7B1%7D%7B2%7D%5BpK_b%2B%5Clog%20C%5D)
where,
= negative logarithm of weak base which is aniline = 4.87
C = concentration of the salt = 0.10 M
Putting values in above equation, we get:
![pH=7-\frac{1}{2}[4.87+\log (0.10)]\\\\pH=5.06=5.1](https://tex.z-dn.net/?f=pH%3D7-%5Cfrac%7B1%7D%7B2%7D%5B4.87%2B%5Clog%20%280.10%29%5D%5C%5C%5C%5CpH%3D5.06%3D5.1)
Hence, the pH of the solution is 5.1
