Atomic Number of Lithium is 3, so it has 3 electrons in its neutral state. Also, Li₂ will have 6 electrons. But the chemical formula we are given has a negative charge on it (i.e Li₂⁻) so there is an additional electron (RED) present on this compound. So, the total number of electrons are 7. The
MOT diagram for this compound is shown below. According to diagram we are having 4 electrons in Bonding Molecular Orbitals (
BMO) and 3 electrons in Anti-Bonding Molecular Orbitals (
ABMO). Bond Order is calculated as,
Bond Order = (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2
Bond Order = (4 - 3) ÷ 2
Bond Order = 1 ÷ 2
Or,
Bond Order = 1/2Or,
Bond Order = 0.5
The mole fraction of helium is 0.33.
The mole fraction can be calculated with the aid of dividing the number of moles of 1 factor of an answer by means of the overall number of moles of all the additives of a solution. it's far noted that the sum of the mole fraction of all the components in the solution should be same to one.
Mole fraction is a unit of attention. within the answer, the relative amount of solute and solvents are measured by the mole fraction and it is represented. The mole fraction is the wide variety of moles of a specific thing within the solution divided by way of the entire wide variety of moles.
Mole fraction represents the quantity of molecules of a selected thing in a combination divided by using the total variety of moles within the given mixture. it is a manner of expressing the concentration of a solution. consequently, the sum of mole fraction of all of the additives is always equal to one.
Mole of neon = 1
mole of helium = 2
mole of argon = 3
Total mole = 1 + 2 + 3 = 6
Mole fraction = mole of helium / total mole
= 2 / 6 = 0.33
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The answer is D
Explanation: A RED FLOWER WITH ONE PART COLORED YELLOW
Boyle Law says “the pressure of fixed amount of ideal gas which is at constant temperature is
inversely proportional to its volume".<span>
P = 1/V
<span>Where, P is pressure of the ideal gas and V is volume of the ideal gas.</span>
<span>For two situations, this law can be added as;
P</span>₁V₁ = P₂V₂<span>
</span><span>14 lb/in² x V₁ = 70 lb/in² x 500 mL</span><span>
</span><span>V₁ =
2500 mL</span><span>
Hence, the needed volume of atmospheric air = 2500
mL
<span>Here, we made two </span>assumptions. They are,
1. The
atmospheric air acts as ideal gas.
2.
Temperature is a constant.
<span>We didn't convert the units to SI units since
converting volume and pressure are products of two numbers, they will cut off. </span></span></span>
Na3P is the formula if that helps