All categories

4 years ago

[HELP PLEASE]How is electronegativity related to covalent bonding and molecular polarity?

Chemistry

2 answers:

Anuta_ua [19.1K]4 years ago

8 0

Answer:

The larger the value of the electronegativity, the greater the atom's strength to attract a bonding pair of electrons. ... Electronegativities give information about what will happen to the bonding pair of electrons when two atoms bond. A bond in which the electron pair is equally shared is called a nonpolar covalent bond.

Explanation:

taurus [48]4 years ago

3 0

Electrons in a polar covalent bond are shifted toward the more electronegative atom; thus, the more electronegative atom is the one with the partial negative charge. The greater the difference in electronegativity, the more polarized the electron distribution and the larger the partial charges of the atoms.

You might be interested in

Mechanism of aspirin by salicyclic acid reacts with acetic acid ? Mechanism of benzoic acid from benzamide?

nexus9112 [7]

Answer:

See explanation

Explanation:

The synthesis of aspirin is shown in the reaction scheme attached to this answer. The production of aspirin involves the reaction of acetyl salicyclic acid and acetic anhydride. HA in the reaction mechanism refers to an acid that is used in the reaction

Similarly benzamide, is converted to the carbonyl carbocation by reaction with acid. The -NH2 group is protonated and subsequently departs as a leaving group. An OH^- completes the mechanism forming a carboxylic acid. The movement of electrons is shown in the image attached.

Again, the diazotization of benzamide using sodium nitrate and concentrated HCl yields benzene diazonium chloride. Addition of water to benzene diazonium chloride yields the benzoic acid and nitrogen gas.

3 0

3 years ago

How many quarters fit in a 1.75 liter bottle?

WARRIOR [948]

1.85 quarts can fit into a 1.75 liter bottle

8 0

3 years ago

Read 2 more answers

Does the Caribbean have high or low viscosity lava?

butalik [34]

High now I need to write 20 characters

8 0

3 years ago

E85 fuel is a mixture of 85% ethanol and 15% gasoline by volume. What mass in kilograms of E85 (d = 0.758 g/mL) can be contained

alexandr402 [8]

The answer is: mass is 40.17 kilograms.

d = 0.758 g/mL; density of fuel.

V = 14.0 gal; volume.

A gallon is a unit of volume in both the US customary and imperial systems of measurement. The US gallon is defined as 231 cubic inches (3.785 liters).

1 gal = 3785.41 mL.

V = 14 gal · 3785.41 mL:

V = 52995.74 mL.

m = 52995.74 mL · 0.758 g/mL.

m = 40170.77 g; mass of fuel.

m = 40170.77 g ÷ 1000 g/kg.

m = 40.17 kg.

7 0

4 years ago

Read 2 more answers

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

5 0

3 years ago