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3 years ago

Consider a 0.70 M solution of HOCl. If the molarity was decreased to 0.3 M, what would happen to the percent dissociation?

Chemistry

1 answer:

melomori [17]3 years ago

5 0

Answer:

The percent dissociation would increase

Explanation:

Percent dissociation is the ratio of amount dissociated to the initial concentration.

Decreasing the concentration of the solution leads to an increase in percent dissociation.

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What is the charge of a cation?

kolezko [41]

Answer:

A cation has more protons than electrons, consequently giving it a net positive charge. For a cation to form, one or more electrons must be lost, typically pulled away by atoms with a stronger affinity for them.

Were i found my answer: Cation vs Anion: Definition, Chart and the Periodic Table

Explanation:

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2 years ago

What are some properties of matter that can be identified without testing or measuring

kiruha [24]

Physical change - No change of matter in this phase
chemical change - All types of phase change occur here

8 0

3 years ago

SeF6 1. Lewis Structure 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs

Anarel [89]

The answer is- $SF_{6}$ is octahedral in electronic and molecular geometry with 6 Fluorine atoms bonded to central atom S.

Lewis structures are the diagrams in which the valence electrons of the atoms of a compound are arranged around the atoms showing the bonding between the atom and the lone pair of electrons existing in the molecule.

Determine the molecular geometry of $SF_{6}$ .

Valence Shell Electron Pair Repulsion theory is commonly known as VSEPR theory and it helps to predict the geometry of molecules.
According to this theory, electrons are arranged around the central atom of the molecule in such a way that there is minimum electrostatic repulsion between these electrons.
Now, calculate the total number of valence electrons in $SF_{6}$ .

$Valence\ electrons\ in\ SF_{6}= Valence\ electrons\ in\ S +\ 6(Valence\ electrons\ in\ F)$

Valence electrons of S = 6

Valence electrons of F = 7

Thus, the valence electrons in $SF_{6}$ are-

$Valence\ number\ of\ electrons\ in\ SF_{6} = (6) + 6(7) = 48\ electrons.$

The Lewis structure of $SF_{6}$ is - (Image attached).
In the structure, the number of atoms bonded to central atom (S) = 6.
Number of non-bonding electron pairs on the central atom = 0 (as all the valence electrons are bonded to F).
Electronic geometry in case of 6 bond pairs is octahedral.
Molecular geometry us also octahedral with bond angles 90°.
Central atom is sp3d2 hybridised.
$SF_{6}$ is a non-polar molecule.

To learn more about Lewis structures visit:

brainly.com/question/12307841?referrer=searchResults

#SPJ4

7 0

1 year ago

Hiii pls help me to write out the ionic equation

emmasim [6.3K]

Answer:

STEP I

This is the balanced equation for the given reaction:-

$2KOH_{(aq)} + H_2SO_4{}_{(aq)} \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)}$

STEP II

The compounds marked with (aq) are soluble ionic compounds. They must be

broken into their respective ions.

see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).

On breaking them into their respective ions :-

2KOH -> 2K+ + 2OH-
H2SO4 -> 2H+ + (SO4)2-
K2SO4 -> 2K+ + (SO4)2-

STEP III

Rewriting these in the form of equation

$\underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \: \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O$

STEP IV

Canceling spectator ions, the ions that appear the same on either side of the equation

(note: in the above step the ions in bold have gotten canceled.)

$\boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}}$

This is the net ionic equation.

____________________________

$\\$

$\mathfrak{\underline{\green{ Why\: KOH \:has\: been\: taken\: as\: aqueous ?}}}$

KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.

[Alkali metal hydroxides are the only halides soluble in water ]

4 0

3 years ago

What did Ernest Rutherford s gold foil experiment demonstrate about atoms?

mote1985 [20]

Rutherford performed gold foil experiment to understand that how negative and positive particles could Co exist in an atom. He bombarded alpha particles on a 0.00004 cm thick gold foil.

He proposed a planetary model of the atom and concluded following results and demonstrated that,
1. An atom produces a line spectrum.
2. An Electron revolves around the nucleus without any orbits.
3. Since most of the particles passed through the foil undeflected it means that most of the volume occupied by an atom is empty.
4. An Atom as a whole is neutral.
5. The deflection of few particles on the foil suggested that there is center of positive particles in an atom called the nucleus of the atom.
6. The complete rebounce of few particles on the gold foil suggested that the nucleus is very dense and hard.

6 0

3 years ago