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Andrej [43]
1 year ago
13

Does vb theory indicate that the diatomic molecule he2 is a viable species? rationalize your answer

Chemistry
1 answer:
iragen [17]1 year ago
5 0

VB theory does not indicate that the diatomic molecule He₂ is a viable species.

Helium has atomic number 2, it has 2 protons and 2 electrons.

Electron configuration of helium atom: ₂He 1s².

Helim (He) is a noble gas.

Noble gases (group 18) are in group 18: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). They have very low chemical reactivity.

Noble gases have very stable electron configuration and does not need to gain electrons, only when they gain energy.

Valence bond (VB) theory explains chemical bonding.

Helium atom is stable and does not form molecule.

More about valence bond (VB) theory: brainly.com/question/23129240

#SPJ4

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25cm3 of gas at 1 atm has a temperature of 20 degree celsis .
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Answer:

1.28 atm

Explanation:

To solve this problem, you need to use the gas laws, more specifically the Combined Gas Law. It is P1V1/T1 = P2V2/T2. Simply plug your values in. But be careful! Make sure you convert your 20 degree C and 28 deg C to Kelvin, as that it the only temperature scale the Gas Laws work with. Upon plugging in your values, you get approximately 1.28 atm.

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3 years ago
The sulfur content of an ore is determined gravimetrically by reacting the ore with concentrated nitric acid and potassium chlor
katen-ka-za [31]

Answer:

13.92 %

Explanation:

Mass of BaSO_4 = 12.5221 g

Molar mass of BaSO_4 = 233.43 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{12.5221\ g}{233.43\ g/mol}

Moles of BaSO_4 = 0.0536 moles

According to the given reaction,

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1 mole of BaSO_4 is formed from 1 mole of SO_4^{2-}

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0.0536 moles of BaSO_4 is formed from 0.0536 moles of SO_4^{2-}

Moles of SO_4^{2-} = 0.0536 moles

Moles of sulfur in 1 mole SO_4^{2-} = 1 mole

Moles of sulfur in 0.0536 mole SO_4^{2-} = 0.0536 mole

Molar mass of sulfur = 32.065 g/mol

Mass = Moles * Molar mass = 0.0536 * 32.065 g = 1.7187 g

Mass of ore = 12.3430 g

Mass % = \frac{Mass\ of\ Sulfur}{Mass_{ore}}\times 100 = \frac{1.7187}{12.3430}\times 100 = 13.92 %

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