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2 years ago

List the following compounds in decreasing electronegativity difference. cl2 hcl nacl

1 answer:

8 0

Based on Pauling Scale, electro negativity of Cl = 3.2, Na = 0.9 and H = 2.1

Thus, Electronegativity difference  in $Cl_{2}$ = 3.2 -3.2 = 0
Electronegativity difference  in NaCl = 3.2-0.9 = 2.3
Similarly, Electronegativity difference  in HCl = 3.2 - 2.1 = 1.1

Thus, among the listed molecules following is the decreasing order of electronegativity difference: NaCl> HCl > $Cl_{2}$

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Jane made this picture to represent a chemical reaction: Two circles, one white and the other gray are shown on the left. A smal

7nadin3 [17]

It represents a decomposition reaction because one reactant breaks apart and forms two products.

Explanation:

The type of chemical reaction depicted by Jane represents a decomposition reaction because on reactant breaks apart and forms two products.

In this reaction XY breaks apart to form X and Y. This is a decomposition reaction.

XY → X + Y

Decomposition or cracking is the formation of two or more products from a single reactant.
The break down of a compound into individual atoms or molecules falls into this category.
The extreme instability of a compound drives such a reaction.
The products is usually stable or it can dissociate further till it is stable.

Find the image attached for more explanation:

Learn more:

Thermal decomposition brainly.com/question/11181911

#learnwithBrainly

8 0

2 years ago

Read 2 more answers

What vitamin a compounds bind with opsin to form rhodopsin??

irina [24]

The organic compound retinal binds with opsin and forms rhodopsin. Retinal is part of the molecule that is responsible for its color. This part is called chromophore. On the other hand, opsins are the proteins in photoreceptor cells. Retinal bounds with these opsins and forms rhodopsin: the basis of the human vision. Rhodopsin is also a protein.It is the pigment in the retinas of humans and animals.

7 0

2 years ago

Draw the product obtained when trans-2-butene is treated first with br2 in ch2cl2, second with nanh2 in nh3, and then finally wi

oee [108]

<span>In order to do this, you have change the alkene into an alkyne. That is the aim of Br2/CH2Cl2 trailed by NaNH2. The Br2 with form a vic dihalide (3,4-dibromo octane). Adding of NaNH2 will execute two E2 reactions. -NH2 will eliminate an H from carbons 3 and 4. This double elimination will make the alkyne. Then handling the alkyne with H2/Lindlar will form the cis alkene. The final product will be CIS-3-octene.</span>

3 0

2 years ago

Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to

Delvig [45]

Answer:

H2O<en<phen

Explanation:

The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.

7 0

2 years ago

Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit

luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

$K~=~10^{-pK}$