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2 years ago

Describe a covalent bond?

1 answer:

Natasha2012 [34]2 years ago

7 0

A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

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Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

2 years ago

What 2 ways that humans have contributed to desertifaction

sdas [7]

Deforestation and inappropriate agriculture otherwise known as over farming are factors of and causes of desertification.

8 0

2 years ago

Check my answer plz!! 100 points. Tell me if they are wrong! 1. You have three elements, A, B, and C, with the following electro

Artyom0805 [142]

look good w small changes below:

Answer:

AB is an ionic compound. The electronegativity difference between A and B is greater.

AC is an ionic compound. The electronegativity difference between A and C is greater.

BC is a covalent compound because the electronegativity difference between C and B is small.

everything else look good!

6 0

2 years ago

Read 2 more answers

A container of gas has a volume of 280 mL at a temperature of 22 Celsius if the pressure remains constant what is the volume 44

d1i1m1o1n [39]

Answer:

300.9mL

Explanation:

Given parameters:

V₁ = 280mL

T₁ = 22°C

T₂ = 44°C

Unknown:

V₂ = ?

Solution:

To solve this problem, we apply Charles's law;

it is mathematically expressed as;

$\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }$

We need to convert the temperature to kelvin;

T₁ = 22°C = 22 + 273 = 295K

T₂ = 44°C = 44 + 273 = 317K

Input the parameters and solve;

$\frac{280}{295}$ = $\frac{V_{2} }{317}$

V₂ x 295 = 280 x 317

V₂ = 300.9mL

5 0

2 years ago

Balance the following equation:

Brut [27]

Answer:

Explanation:There are no results for balance the following equation:ni(cio), →nici,+0when the equation is properly balanced, the correctcoefficients are, respectively

6 0

2 years ago