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Angelina_Jolie [31]
3 years ago
13

What is covalent bond????​

Chemistry
2 answers:
Zepler [3.9K]3 years ago
5 0

Answer:

<em><u>A covalent 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.</u></em>

<em><u>Covalent bonds can be single, double, and triple bonds. Single bonds occur when two electrons are shared and are composed of one sigma bond between the two atoms.</u></em>

<em><u>A</u></em><em><u>R</u></em><em><u>I</u></em><em><u>A</u></em><em><u>♡</u></em>

adell [148]3 years ago
4 0

Explanation:

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms

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One mole of an atom of a substance is equal to
solong [7]

Answer:

One mole of a substance is equal to 6.022 × 10²³ units of that substance (atoms, molecules, or ions)

Explanation:

This number is Avogadro's number. The concept of the mole can be used to convert between mass and number of particles. its used to compare very large numbers.

4 0
3 years ago
what pressure is generated when 5 mol of ethane is stored in a volume of 8 dm 3 at 25°C? Base calculations on each of the follow
Goshia [24]

Answer:

a) 40,75 atm

b) 30,11 atm

Explanation:

The Ideal Gas Equation is an equation that describes the behavior of the ideal gases:

                                     PV = nRT

where:

  • P = pressure [atm]
  • V = volume [L]
  • n = number of mole of gas [n]
  • R= gas constant = 0,08205 [atm.L/mol.°K]
  • T=absolute temperature [°K]

<em>Note: We can express this values with other units, but we must ensure that the units used are the same as those used in the gas constant.</em>

The truncated virial equation of state, is an equation used to model the behavior of real gases. In this, unlike the ideal gas equation, other parameters of the gases are considered as the <u>intermolecular forces</u> and the <u>space occupied</u> by the gas

\frac{Pv}{RT} = 1 + \frac{B}{v}

where:

  • v is the molar volume [L/mol]
  • B is the second virial coefficient [L/mol]
  • P the pressure [atm]
  • R the gas constant = 0,08205 [atm.L/mol.°K]

a) Ideal gas equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

We clear pressure of the idea gas equation and replace the data:

PV = nRT ..... P = nRT/V = 5 * 0,08205 * 298/3 =40,75 atm

b) Truncated virial equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

B = -156,7*10^-6 m3/mol = -156,7*10^-3 L/mol

We clear pressure of the idea gas equation and replace the data:

\frac{Pv}{RT} = 1 + \frac{B}{v} ...... P = (1 + \frac{B}{v}) \frac{RT}{v}

and v = 3 L/5 moles = 0,6 L/mol

P = (1 + \frac{-156,7*10^{-3} }{0,6} ) \frac{0,08205*298}{0,6} = 30,11 atm

5 0
4 years ago
You measure the absorbance of your extracted curcumin solution and realize that the solution is too concentrated. What made you
zepelin [54]

If the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high.

Spectrometry measures the interaction of light with molecules. The absorbance refers to how much light that interacts with molecules of the substance. The more the concentration of the substance the higher the absorbance of the solution.

Hence, if  the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high. An unusually high absorbance tells us that the solution is too concentrated.

Learn more: brainly.com/question/13440572

8 0
2 years ago
) Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2is 7
Travka [436]

Answer:

Rate of the reaction is 0.2593 M/s

-0.5186 M/s is the rate of the loss of ozone.

Explanation:

The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.

2O_3\rightleftharpoons 3O_2

Rate of formation of oxygen : 7.78\times 10^{-1} M/s

Rate of the reaction(R) =\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}

R=\frac{1}{3}\frac{d[O_2]}{dt}

Rate of formation of oxygen=3 × (R)

7.78\times 10^{-1} M/s=3\times R

Rate of the reaction(R): 0.2593 M/s

Rate of the reaction is 0.2593 M/s

Rate of disappearance of the ozone:

R=-\frac{1}{2}\frac{d[O_3]}{dt}

\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s

-0.5186 M/s is the rate of the loss of ozone.

6 0
3 years ago
Find the volume of 56.0 grams of O2
N76 [4]

Answer:

39.2 L at STP

Explanation:

Convert the grams to moles first by dividing 56.0 by the molar mass of O2 (32.0) then convert to volume by multiplying by 22.4.

= 39.2 L

5 0
3 years ago
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