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2 years ago

How does a covalent bond form?

Chemistry

1 answer:

Zepler [3.9K]2 years ago

4 0

Answer: A covalent bond forms when the difference between the electronegativities of two atoms is too small for an electron transfer to occur to form ions.

Explanation:

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Why is it not possible to just consume only pure ATP and thus eliminate some steps in

xeze [42]

Our digestive has the steps of digestion and absorption associated with it. So all the compounds taken in as food are first broken down into simpler components before absorption.

<h3>Explanation:</h3>

ATP or Adenosine Triphosphate is a compound containing a molecule of adenine as nitrogen base, a ribose sugar and three molecules of phosphate attached with the adenine in a chain. Its unable to get absorbed into the system as ATP itself. In intestine, it will be broken down into Adenine, ribose, and phosphates and then absorbed individually into blood.

Thus even if we consume raw ATP, we can't bypass the process of ATP formation. So, there's no significance. And secondly, ATP is very expensive and not suitable for consumption itself.

4 0

3 years ago

In a laboratory, 1.55mg of an organic compound containing carbon, hydrogen, and oxygen is burned for analysis. This combustion r

Serjik [45]

Answer:

CH₃O

Step-by-step explanation:

1. Calculate the mass of each element

Mass of C = 1.45 mg CO₂ × (12.01 mg C/44.01 mg CO₂) = 0.3957 mg C

Mass of H = 0.89 mg H₂O × (2.016 mg H/18.02 mg H₂O) = 0.0996 mg H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.55 – 0.3957 – 0.0996) mg = 1.055 mg

2. Calculate the moles of each element

Moles of C = 0.3957 mg C × 1mmol C/12.01 mg C = 0.03295 mol C

Moles of H = 0.0996 mg H × 1 mmol H/1.008 mg H = 0.0988 mol H

Moles of O = 1.055 mg O × 1 mmol O/ 16.00 mg O = 0.06592 mol O

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

C: 0.032 95/0.032 95 = 1

H: 0.0988/0.032 95 = 2.998

O: 0.065 92/0.032 95 = 2.001

4. Round the ratios to the nearest integer

C:H:O = 1:3:2

5. Write the empirical formula

The empirical formula is CH₃O.

4 0

4 years ago

A solid piece of aluminum (51.0 g) was added to a solution of sodium hydroxide (84.1 g) in water, A balanced equation for this r

EastWind [94]

$M_{Al}=26,98\frac{g}{mol}\\
m=51g\\\\
n=\frac{m}{M_{Al}}=\frac{51g}{26,97\frac{g}{mol}}\approx1,89mol\\\\\\
M_{NaOH}=39,4\frac{g}{mol}\\
m=84,1g\\\\
n=\frac{m}{M_{NaOH}}=\frac{84,1g}{39,4\frac{g}{mol}}\approx2,14mol$

2 NaOH(aq)+ 2 Al(s)+ 2 H₂O → 2 NaAlO₂(aq)+ 3 H₂(g)
 2mol : 2mol : 3mol
2,14mol : 1,89mol : 2,835mol
remains completely consumed
2,14-1,89=0,25mol

A) Al

B)

$M_{NaOH}=39,4\frac{g}{mol}\\
n=0,25mol \Rightarrow \ \ \ m=n*M_{NaOH}=0,25mol*39,4\frac{g}{mol}=9,85g$

C)
$n=2,835mol\\
M_{H_{2}}=2,02\frac{g}{mol} \Rightarrow \ \ \ m=n*M_{H_{2}}=2,835mol*2,02\frac{g}{mol} \approx 5,73g$

7 0

3 years ago

Consider the decomposition of red mercury(II) oxide under standard state conditions. )H0 T SFE ڮ( H M 0 H (a) Is the decompositi

sertanlavr [38]

The question is incomplete, complete question is :

Consider the decomposition of red mercury(II) oxide under standard state conditions.

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

Given :

$\Delta S_{HgO}^o=70.29 J/mol K$

$\Delta S_{Hg}^o=75.9 J/mol K$

$\Delta S_{O_2}^o=205.2 J/mol K$

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer:

a) The decomposition is spontaneous under standard state conditions.

b)The reaction will spontaneous above -419.69 Kelvins.

Explanation:

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

Given :

$\Delta S_{HgO}^o=70.29 J/mol K$

$\Delta S_{Hg}^o=75.9 J/mol K$

$\Delta S_{O_2}^o=205.2 J/mol K$

Entropy change of the reaction ; ΔS

$\Delta S=[2\times \Delta S_{Hg}^o+1\times \Delta S_{O_2}^o]-[2\times \Delta S_{HgO}^o]$

$=[2\times 75.9 J/mol K+1\times 205.2 J/molK]-[2\times 70.29 J/molK]=216.42 J/mol K$

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K

1 kJ = 1000 J

At standard condition the value of temperature = T = 298 K

ΔG = ΔH - TΔS

ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol

ΔG < 0 ( spontaneous)

The decomposition is spontaneous under standard state conditions.

b) Above what temperature does the reaction become spontaneous

Let the ΔG = 0

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K

ΔG = ΔH - TΔS

0 = -90830 J/mol K - T × 216.42 J/mol K

T = -419.69 K

The reaction will spontaneous above -419.69 Kelvins.

6 0

3 years ago

Which of the following best describes weight? The amount of space an object takes up The amount of matter in an object The force

Alenkasestr [34]

The force of an object due mass and gravity.it can also be defined as the force exerted by gravity on a body

4 0

3 years ago