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Gennadij [26K]
3 years ago
8

For which of the following elements (in their normal, stable, forms) would it be correct to describe the bonding as involving a

"sea of electrons"? a. hydrogen b. helium c. sulfur d. iodine e. lithium
Chemistry
1 answer:
Dvinal [7]3 years ago
4 0

Answer: The correct option is e.

Explanation: In the given question it is asked to tell the element which has 'Sea of electrons' while bonding.

This means that the element should have extra electrons so that it can bond easily or this also means which element can easily donate its electron to form a bond.

  • Electronic configuration of Hydrogen = 1s^1

Hydrogen cannot loose electrons easily because the electron is present in 1s sub-shell which is the closest to the nucleus and hence will be tightly bonded to it.

  • Electronic configuration of Helium = 1s^2

This element will not loose electrons easily because the 1s sub-shell is fully filled and hence will not be available for bonding.

  • Electronic configuration of Sulfur = 3s^23p^4

This element lacks 2 electrons from attaining stable configurations. Hence, it will gain 2 electrons rather than donating for bonding.

  • Electronic configuration of Iodine = 5s^25p^5

This element lacks 1 electron to attain stable configuration. Hence, it will gain 1 electron rather than donating for bonding.

  • Electronic configuration of Lithium = 1s^22s^1

This element has an extra electron in its valence shell and can be easily lost in order to form bond.

Hence, the correct option is e.

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Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is C₁₂H₂₆.
padilas [110]

The calculated enthalpy of formation of kerosene is 365.4 kJ and heat produced is 78650.3 kJ

For this, we need the normal enthalpy of formation given below

ΔH∘f(CO2)=−393.5kJ/molΔH∘f(H2O)\s=−241.8kJ/molΔH∘f(O2)=0kJ/mol

We shall now determine the enthalpy of kerosene formation:

H rxn = 24 mol H f (CO2) + 26 mol H f (H2O) + 2 mol H f (C12H26) + 37 mol H f (O2) + 1.50 104 kJ = 9444 kJ + 6286.8 kJ + 1500 kJ 2 mol H f (C12H26) = 730.8 kJ H f (C12H26) = 365.4 kJ

Kerosene has a density of 0.74 g/mL.

Kerosene volume (V) equals 0.63 gallons, or 0.63 x 3785.4, or 2384. 8 mL.

We shall now calculate the mass (m) of kerosene:

ρ=mVm\s=ρ×Vm\s=0.749g/mL×2384.mLm\s=1786.2g

We shall now discover the heat that 1786 generated.

Two grams of kerosene

Kerosene's molar mass is 170.33 g/mol.

The mass of two moles of kerosene is equal to 2*170.33*340.66g.

1.50104kJ of heat are generated by 340.66 g of kerosene.

1786 produced heat.

Kerosene 2 grams = 1.50 104 kJ 340.66 1786.2 g = 78650.3 kJ

Learn more about enthalpy here-

brainly.com/question/13996238

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1 year ago
After fertilization, female cones become very:<br> sticky<br> light<br> O hard
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Answer:

light

Explanation:

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8 0
3 years ago
A tank at is filled with of sulfur tetrafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal g
dangina [55]

The question is incomplete, the complete question is:

A 7.00 L tank at 21.4^oC is filled with 5.43 g of sulfur hexafluoride gas and 14.2 g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas. Round each of your answers to significant digits.

<u>Answer:</u> The mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

  • <u>For sulfur hexafluoride:</u>

Given mass of sulfur hexafluoride = 5.43 g

Molar mass of sulfur hexafluoride = 146.06 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfur hexafluoride}=\frac{5.43g}{146.06g/mol}=0.0372mol

  • <u>For sulfur tetrafluoride:</u>

Given mass of sulfur tetrafluoride = 14.2 g

Molar mass of sulfur tetrafluoride = 108.07 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfur tetrafluoride }=\frac{14.2g}{108.07g/mol}=0.1314mol

Total moles of gas in the tank = [0.0372+ 0.1314] mol = 0.1686 mol

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

\chi_A=\frac{n_A}{n_A+n_B} .....(2)

where n is the number of moles

Putting values in equation 2, we get:

\chi_{SF_6}=\frac{0.0372}{0.1686}=0.221

\chi_{SF_4}=\frac{0.1314}{0.1686}=0.779

Hence, the mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

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Tems11 [23]

I believe it is because, Lead has a higher boiling point than zinc. by the time Lead melts and becomes a liquid, the temperature is way higher than the boiling point of Zinc therefore, Zinc becomes a gas.
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