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aev [14]
3 years ago
15

A solution of hcl in water conducts electricity, but a solution of hcl in chloroform, chcl3, does not. what does this observatio

n tell you about how hcl exists in water and how it exists in chloroform?
Chemistry
2 answers:
Serhud [2]3 years ago
6 0

Explanation:

A polar covalent bond is formed due to unequal sharing of electrons between the combining atoms.

For example, HCl is a polar covalent compound.

This is because of the difference in electronegativities of both hydrogen and chlorine, partial opposite charges tend to develop on the hydrogen and chlorine atoms.

Whereas CHCl_{3} is a non-polar compound and when HCl is added to it then HCl being polar in nature is unable to dissolve.

On the other hand, water is a polar compound and when HCl is dissolved in it then it will dissociate into hydrogen and chlorine ions.

As electricity is the flow of ions or electrons. Therefore, HCl in water is able to conduct electricity because it dissociates into ions.

Therefore, we can conclude that HCl in water exists as H^{+} and Cl^{-} ions. And, HCl in CHCl_{3} exists as HCl.

PSYCHO15rus [73]3 years ago
4 0
Hydrochloric acid (HCl) dissociate on positive ions or cations of hydogen (H⁺) and negative ions or anions of chlorine (Cl⁻). When there is free cations and ions, water solution can conduct electricity. Chloroform do not dissociate on ions in water, because that is not polar molecule, so chloroform is in molecular form in solution.
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Will NaCl be soluble or insoluble
attashe74 [19]

NaCl is salt to it is obviously SOLUBLE :)

6 0
3 years ago
A certain weak acid, ha, has a ka value of 6.0×10−7. calculate the percent ionization of ha in a 0.10 m solution.
joja [24]
For the purpose, we will use the equation for determining the dissociation constant from concentration and <span>percent of ionization:

Kd = c </span>× α²

α = √(Kd/c) × 100%

Kd = 6.0×10⁻⁷

c(HA) = 0.1M

α = √(6.0×10⁻⁷/0.1)  × 100% =  0.23%

So, in the solution, the acid <span>percent of ionization will be just 0.23%.</span>

5 0
3 years ago
A 0.288 g sample of an unknown monoprotic acid is dissolved in water and titrated with a 0.115 M NaOH solution. After the additi
Sholpan [36]

Answer:

74.0 g/mol

Explanation:

Step 1: Write the generic neutralization reaction

HA + NaOH ⇒ NaA + H₂O

Step 2: Calculate the reacting moles of NaOH

At the equivalence point, 33.83 mL of 0.115 M NaOH react.

0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol

Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH

The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.

Step 4: Calculate the molar mass of the acid

3.89 × 10⁻³ moles of HA have a mass of 0.288 g.

M = 0.288 g / 3.89 × 10⁻³ mol = 74.0 g/mol

4 0
3 years ago
What happens to the number of valence electrons as you move from element 1 to element 18 on the periodic table?
BARSIC [14]

Answer:

B

Explanation:

the group number is=valence electrons. element 1 is in group 1 element 18 is in group 8. 1<8

6 0
3 years ago
Read 2 more answers
How does a chemist count the number of particles in a given number of moles of a substance?
jolli1 [7]
The chemist the count the number of particles (Atoms, Molecules or Formula Unit) in a given number of moles of a substance by using following relationship.

                              Moles  =  # of Particles / 6.022 × 10²³

Or,

                              # of Particles  =  Moles × 6.022 × 10²³

So, from above relation it is found that 1 mole of any substance contains exactly 6.022 × 10²³ particles. Greater the number of moles greater will be the number of particles.
8 0
3 years ago
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