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dalvyx [7]
3 years ago
13

Calculate the pH at of a 0.10 Msolution of anilinium chloride . Note that aniline is a weak base with a of . Round your answer t

o decimal place. Clears your work. Undoes your last action. Provides information about entering answers.
Chemistry
1 answer:
notsponge [240]3 years ago
8 0

The question is incomplete, here is the complete question:

Calculate the pH at of a 0.10 M solution of anilinium chloride (C_6H_5NH_3Cl) . Note that aniline (C6H5NH2) is a weak base with a pK_b 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]

where,

pK_b = 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

Hence, the pH of the solution is 5.1

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1) A potassium carbonate hydrate has a formula K2CO3.XH2O. 10g of the hydrate leave 7.83g of anhydrous salt upon heating. Deduce
OLEGan [10]

The formula of the hydrated potassium carbonate salt is K₂CO₃.2H₂O

Based on the calculated mass ratio of carbon and oxygen in carbon dioxide, carbon has a fixed composition.

<h3>What are hydrated compounds?</h3>

Hydrated compounds are compounds that contain one or more molecules of water physically combined with a molecule of the compound.

The formula of the hydrated potassium carbonate salt is determined as follows:

Mass of the hydrated sample = 10.0 g

Mass of anhydrous salt = 7.83

mass of water = 10 - 7.83

mass of water = 2.17 g

Molar mass of water = 18.0 g

Molar mass of anhydrous potassium carbonate = 138 g

moles of anhydrous potassium carbonate in sample = 7.83/138

moles of anhydrous potassium carbonate = 0.056 moles

moles of water in the hydrated salt = 2.17/18

moles of water in the hydrated salt = 0.12

Mole ratio of water to anhydrous salt = 0.12/0.056

Mole ratio of water to anhydrous salt = 1 : 2

Formula of hydrated salt = K₂CO₃.2H₂O

The mass ratio of carbon to oxygen in the compounds is given below:

Sample 1:

Mass ratio = 3.62 / (13.26 - 3.62)

Mass ratio = 0.38 : 1

Sample 2:

Mass ratio = 5.91 / (21.66 - 5.91)

Mass ratio = 0.38 : 1

Sample 3:

Mass ratio = 7.07 / (25.91 - 7.07)

Mass ratio = 0.38 : 1

Carbon has a fixed composition.

Learn more about hydrated compounds at: brainly.com/question/11112492

#SPJ1

6 0
1 year ago
How is an isotope different from the standard form of a chemical element?
velikii [3]

Answer:

  • The standard form of a chemical element is the natural mixture of several isotopes of the same element, which is atoms with the same number of protons but different number of neutrons, while an isotope is a particular kind of atom with a definite number of neutrons.

Explanation:

A <em>chemical element</em> is a pure substance formed by atoms with the same atomic number (number of protons). This is because it is the number of protons what identifies an element.

For example: oxygen is a chemical element, so oxygen is formed by only atoms of oxygen, and the atomic number of those atoms is 8, because every oxygen atom has 8 protons.

Nevertheless, some atoms of oxygen, may have different number of neutrons. Isotopes are different kind of atoms of the same element, which only differ in the number of neutrons. So, some atoms of oxygen will have 8 neutrons, other 9 neutrons, and other 10 neutrons (those are the stable isotopes of oxygen).

That difference in neutrons, is generally accepted that, does not modifiy substantially the chemical properties of the element, but the mass number. So, the isotopes with more neutrons wil be heavier, and the isotopes with less neutrons will be lighter.

  • Mass number = number of protons + number of neutrons.

In general a chemical element is formed by a mixutre of isotopes of the same element.

4 0
3 years ago
Read 2 more answers
By the reaction of carbon &amp; oxygen , a mixture of CO &amp;CO2 is obtained. What is the composition by mass of the mixture ob
Hatshy [7]
M(O₂)=20g
M(O₂)=32.0 g/mol
n(O₂)=20/32.0=0.625 mol

m(C)=12 g
M(C)=12.0 g/mol
n(C)=12/12.0=1.0 mol

   2C     +     O₂      →    2CO
1 mol    0.625 mol        1 mol
         0.625-0.5=0.125 mol

      2CO    +         O₂       →        2CO₂
0.250 mol       0.125 mol       0.250 mol

n(CO)=1 mol - 0.250 mol = 0.750 mol
M(CO)=28.0 g/mol
m(CO)=0.750*28.0=21.0 g

n(CO₂)=0.250 mol
M(CO₂)=44.0 g/mol
m(CO₂)=0.250*44.0=11.0 g
4 0
3 years ago
In the coordination compound [Cr(NH3)2(en)Cl2]Br2, the coordination number (C.N.) and oxidation number (O.N.) of the metal atom,
erma4kov [3.2K]

The coordination number (C.N.) is 6 and oxidation number (O.N.) is +4 of the metal atom.

C.N. = 6; O.N. = +4

<h3><u>What </u><u>is Co-ordination number ?</u></h3>
  • The coordination number, also known as ligancy, of a central atom in a molecule or crystal in chemistry, crystallography, and materials science refers to the quantity of atoms, molecules, or ions bound to it.
  • A ligand is the ion, molecule, or atom that surrounds the center ion, molecule, or atom.
  • The number of atoms, ions, or molecules that a central atom or ion in a complex, coordination compound, or crystal holds as its closest neighbors.
<h3><u>What is Oxidation number ?</u></h3>
  • The total number of electrons that an atom acquires or loses to establish a chemical connection with another atom is known as the oxidation number, also known as the oxidation state.
  • Each atom involved in an oxidation-reduction reaction has an oxidation number that represents how many electrons it may take, give, or share.

To view more questions of coordination number, refer to:

brainly.com/question/8717978?referrer=searchResults

#SPJ4

7 0
1 year ago
4. Write the chemical formulas for the following compounds:
IRISSAK [1]
Lithium oxide ( Li2O)
Carbon monoxide ( CO )
Carbon tetrachloride ( CCl4 )
Nitrogen trifluoride ( NF3 )
Calcium chloride ( CaCl2)
4 0
2 years ago
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