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scoundrel [369]
3 years ago
11

In the covalent bond between hydrogen and chlorine in the HCl molecule, where will the majority of shared electrons be found?

Chemistry
2 answers:
dusya [7]3 years ago
7 0
B.near the chlorine atom

H→Cl
Lerok [7]3 years ago
5 0

<u>Answer:</u> The correct answer is Option B.

<u>Explanation:</u>

A covalent bond is formed when sharing of electrons takes place between the atoms combining. The atom which attracts the shared pair of electron towards itself is more electronegative and the atom which does not attracts the shared pair of electron towards itself is less electronegative.

In HCl molecule, more electronegative atom is chlorine atom having an electronegativity of 3.16 and hydrogen has an electronegativity of 2.1

So, majority of shared electrons will be found near to the chlorine atom in hydrochloric molecule.

Hence, the correct answer is Option B.

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Which are different structures of the eye? Select four options.
Setler [38]

Answer:

Pupil, Cornea, Retina, and Lens

Explanation:

5 0
3 years ago
Read 2 more answers
You have a saturated solution of BaSO4, a slightly soluble ionic compound. What happens if you add Ba(OH)2, NaNO3, and CuSO4 to
igomit [66]

Answer:

- Addition of Ba(OH)2: favors the formation of a precipitate.

- Undergo a chemical reaction forming soluble species.

- Addition of CuSO4 : favors the formation of a precipitate.

Explanation:

Hello,

In this case, since the dissociation reaction of barium sulfate is:

BaSO_4(s)\rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq)

We must analyze the effect of the common ion:

- By adding barium hydroxide, more barium ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).

- By adding sodium nitrate, the following reaction will undergo:

BaSO_4(s)+NaNO_3(aq)\rightarrow Ba(NO_3)_2(aq)+Na_2SO_4(aq)

So the precipitate will turn into other soluble species.

- By adding copper (II) sulfate, more sulfate ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).

All of this is supported by the Le Chatelier's principle.

Best regards.

7 0
3 years ago
What is the electron configuration for Ba​
babunello [35]

Full electron configuration of barium: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s2

8 0
3 years ago
In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an ex
Kay [80]

The question is incomplete. The complete question is :

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an explanation that tells how much NaOH needs to be added to a beaker of HCl to cause the color to change. Your explanation can be something like: The color change will occur when [some amount] of NaOH is added because the color change occurs when [some condition]. The goal for your explanation is that it describes the outcome of this example, but can also be used to predict the outcome of other examples of this phenomenon. Here's an example explanation: The color of the solution will change when 40 ml of NaOH is added to a beaker of HCl because the color always changes when 40ml of base is added. Although this explanation works for this example, it probably won't work in examples where the flask contains a different amount of HCl, such as 30ml. Try to make an explanation that accurately predicts the outcome of other versions of this phenomenon.

Solution :

Consider the equation of the reaction between NaOH and $HCl$

  NaOH (aq) + HCl (aq) → NaCl(aq) + $H_2O (l)$

The above equation tells us that $1 \text{mole}$ of $NaOH$ reacts with $1 \text{mole}$ of $HCl$.

So at the equivalence point, the moles of NaOH added = moles of $HCl$present.

If the volume of the $HCl$ taken = $V_1$ mL and the conc. of $HCl$ = $M_1$  mole/L

The volume of NaOH added up to the color change = $V_2 \text{  and conc of NaOH = M}_2$ mole/L

Moles of $HCl$ taken = $V_1 \ mL \times M_1 \ mol/100 \ mL = V_2M_2 \times 10^{-3}$  moles.

The color change will occur when the moles of NaOH added is equal to the moles of $HCl$ taken.

Thus when $V_1 M_1 \times 10^{-3} = V_2M_2 \times 10^{-3}$

or   when    $V_1M_1 = V_2M_2$

or $V_2=\frac{V_1M_1}{M_2}$  mL of NaOH added, we observe the color change.

Where $V_1, M_1$ are the volume and molarity of the $HCl$ taken.

$M_2$ is the molarity of NaOH added.

When both the NaOH and $HCl$ are of the same concentrations, i.e. if $M_1=M_2$, then $V_2=V_1$

Or the 40 mL of $HCl$ will need 40 mL of NaOH for a color change and

30 mL of $HCl$ would need 30 mL of NaOH for the color change (provided the concentration $M_1=M_2$)

7 0
3 years ago
In general, for a gas at a constant volume, _____.
densk [106]

It is kept constant

There is the answer if it helped

5 0
3 years ago
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