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3 years ago

Which type of bonding is most important in ch3ch2ch2ch2ch2ch3?

Chemistry

1 answer:

saveliy_v [14]3 years ago

7 0

The compound $CH_3CH_2CH_2CH_2CH_2CH_3$ is formed only by sharing of electrons between the atoms. The structure of the compound is shown in the image.

Each line between two atoms represents the sharing of an electron pair which results in the formation of a single bond. Since, carbon has 4 electrons in its valence shell and hydrogen has 1 electron in its valence shell so in order to complete the octet ( to have 8 electrons in their valence shell, noble gas configuration) to attain stability carbon needs 4 more electrons and hydrogen needs 1 electron. So, sharing of electron will occur as shown in the image and the formed compound is stable in nature.

Since, the bond that is formed by sharing of electrons between atoms is known as covalent bond. So, covalent bonding is most important in $CH_3CH_2CH_2CH_2CH_2CH_3$ .

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You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

3 years ago

What is meant by solvation?

Kamila [148]

Solvation describes the interaction of solvent with dissolved molecules.

6 0

3 years ago

Fill in the chart to identify three parts of Dalton's model of the atom.

mamaluj [8]

The first part of Dalton's model is that all matter is made of atoms, and these atoms are indivisible.

The second part of this model is that atoms of an element have identical masses and identical propertes.

At last, the third part of Dalton's model is that all compounds are formed by two or more different type of atoms combined.

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1 year ago

Fe2O3(s)+3C(s)→2Fe(s)+3CO(g) how many grams of CO are produced when 32.0 grams of C react

krok68 [10]

Answer:

74.6 g

Explanation:

The computation of the number of grams produced at the time when 32.0 grams of C react is shown below:

As we know that

But before that we need to find out the moles of c i.e carbon which is

$= \frac{Grams }{C\ molar\ mass}$

$= \frac{32\ grams}{12.0107 g/mol}$

= 2.66 mol

As we know that

1 mol of C produces 1 mol of CO

So, for 2.66 mol, the mass of CO is

$= 2.66 \times28.0101 g/mol$

= 74.6 g

Basically we applied the above equation so that we can reach to the answer

8 0

3 years ago

The energy released by combustion of 1 g of a substance is called the ________ of the substance. the energy released by combusti

Anastaziya [24]

Enthalpy of the substance

5 0

3 years ago

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