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3 years ago

Describe the electron distribution in a polar-covalent bond and it's effect on the partial charges of the compound?

Chemistry

1 answer:

Artemon [7]3 years ago

5 0

Answer:

In polar Covalent bonds, the electrons which are in bonded shifts towards an atom which has more valance electrons.

<u>Explanation:</u>

We know if an atom takes the electron it acquires a negative charge whereas if it gives an electron it acquires a positive charge in the ionic bond. But here we are talking about covalent bonds. Covalent bonds are those in which atoms share the electron instead of completely giving off the electron. If the atoms are identical in case of covalent bond that is 2 hydrogen atoms then this type of bonding is called pure covalent bonds but if the atoms linked in covalent bonds are different then it is called polar covalent bonds.

In this, the bonding electrons will shift towards an atom which has more valence electron thereby acquiring the partial negative charges and the other atom will acquire a partial positive charge. For example, HCl. In this the Chlorine atom is having more valence electron than hydrogen atom, and hence Chlorine atom has a partial negative charge and Hydrogen atom has a partial positive charge.

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A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

3 0

3 years ago

Read 2 more answers

What is meant by Vulcanicity?

scZoUnD [109]

The quality or state of being volcanic.

3 0

2 years ago

47.0ml of a HBr solution were titrated with 37.5ml of a 0.215M LiOH solution to reach the equivalence point. what is the molarit

faltersainse [42]

Hello!

The molarity of the HBr solution is 0,172 M.

Why?

The neutralization reaction between LiOH and HBr is the following:

HBr(aq) + LiOH(aq) → LiBr(aq) + H₂O(l)

To solve this exercise, we are going to apply the common titration equation:

$M1*V1=M2*V2$

$M1=\frac{M2*V2}{V1}= \frac{0,215 M * 37,5 mL}{47 mL}=0,172 M$

Have a nice day!

4 0

3 years ago

6 How many moles are in a 321 g sample of magnesium?

Tatiana [17]

Answer:

13.4mol of Mg

Explanation:

Given parameters:

Mass of magnesium = 321g

Unknown:

Number of moles = ?

Solution:

The number of moles of a substance is given as;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Mg = 24g/mol

Insert the parameters and solve;

Number of moles = $\frac{321}{24}$ = 13.4mol of Mg

3 0

2 years ago

Draw the aromatic compound toluene (methylbenzene). show all hydrogen atoms, including those on the ring.

Otrada [13]

Aromatic compound has continuous cyclic structure with( 4n+2)π electrons (Huckels rule), where n = 0,1,2…

Here number of pi electron are 6, where 4 from two double bond and 2 from nitrogen non-bonding electrons, hence it has total 6 pi electrons, therefore

6= ( 4n+2)π

4 = 4n

n =1

Hence it is an aromatic compound

7 0

3 years ago