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3 years ago

What is the highest occupied molecular orbital in the O2 molecule?

1 answer:

3 0

H2 is known to exist. For dihydrogen, H2, we can identify the frontier molecular orbitals (FMOs). The highest occupied molecular orbital (or HOMO) is the σ (sigma) 1s MO. The lowest unoccupied MO (LUMO) is the σ* (sigma star) 1s MO which is antibonding.

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A surplus (having “extra”) or deficit (having “fewer”) of electrons?

Nikolay [14]

Answer:

Charge

Explanation:

A surplus (having "extra") or deficit (having "fewer") of electrons possessed by an object. A charge can cause attractive or repulsive forces which can be observed in some cases (e.g. pith balls, bits of plastic).

7 0

3 years ago

Calculate the approximate volume of a 0.6000mol sample of gas at 288.15K and a pressure of 1.10atm.

11111nata11111 [884]

Answer:

The volume of the sample of the gas is found to be 12.90 L.

Explanation:

Given pressure of the gas = P = 1.10 atm

Number of moles of gas = n = 0.6000 mole

Temperature = T = 288.15 K

Assuming the volume of the gas to be V liters

The ideal gas equation is shown below

$\textrm{PV} =\textrm{nRT} \\1.10 \textrm{ atm}\times V \textrm{ L} = 0.6000 \textrm{ mole}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 288.5\textrm{K} \\\textrm{V} = 12.90 \textrm{ L}$

Volume occupied by gas = 12.90 L

6 0

4 years ago

How much nano3 is needed to prepare 225 ml of a 1.55 m solution of nano3?

andrezito [222]

Answer:

= 29.64 g NaNO3

Explanation:

Molarity is given by the formula;

Molarity = Moles/Volume in liters

Therefore;

Number of moles = Molarity × Volume in liters

= 1.55 M × 0.225 L

= 0.34875 moles NaNO3

Thus; 0.34875 moles of NaNO3 is needed equivalent to;

= 0.34875 moles × 84.99 g/mol

= 29.64 g

5 0

3 years ago

If two students in a lab are asked to calculate the density of a block with a mass of 31.0 g; they are not told how many digits

Arada [10]

We can see here that the both measurements are accurate. This is the true because the both values are in the correct number of significant figures.

<h3>What is density?</h3>

The term density has to do with the ratio of the mass to the density of the object. We know that the accuracy of a measurement has a lot to do with the number of significant figures in the measurement. In this case, we are told that two students in a lab are asked to calculate the density of a block with a mass of 31.0 g; they are not told how many digits to round their measurements. Student A finds length to be 2 cm width to be 4 cm and height 8 cm. Student B finds the length to be 2.65 cm, width to be 4.20 cm, and height to be 8.35 cm.

For the first student, the density is obtained as;

Density = mass/ volume

= 31.0 g/ 2 cm * 4 cm * 8 cm

= 0.5 g/cm^3

For the second student;

Density = mass/ volume

= 31.0 g/2.65 cm * 4.20 cm * 8.35 cm

= 0.33 g/cm^3

Learn more about density:brainly.com/question/15164682

#SPJ1

3 0

1 year ago

Hydrogen and Methanol have both been proposed as alternatives to hydrocarbon fuels. Write balanced reactions for the complete co

Arlecino [84]

Answer:

The order of energy released per mass is

CH₃OH (-2.268 × 10⁴ kJ/kg) < C₈H₁₈ (-4.826 × 10⁴ kJ/kg) < H₂ (-2.835 × 10⁵ kJ/kg)

Explanation:

In order to calculate the enthalpy of a reaction (ΔH°r) we can use the following expression.

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where

ΔH°f(i) are the enthalpies of formation of reactants and products

ni are the moles of reactants and products

<u>Combustion of hydrogen</u>

H₂(g) + 1/2 O₂(g) ⇒ H₂O(l)

ΔH°r = 2 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(H₂) - 1/2 mol × ΔH°f(O₂)

ΔH°r = 2 mol × (-285.8 kJ/mol) - 1 mol × 0 - 1/2 mol × 0

ΔH°r = -571.6 kJ

571.6 kJ are released when 1 mole of H₂ is burned. The amount of heat released per kilogram is:

$\frac{-571.6kJ}{1molH_{2}} .\frac{1molH_{2}}{2.016gH_{2}} .\frac{10^{3}gH_{2} }{1kgH_{2}} =-2.835 \times 10^{5} kJ/kgH_{2}$

<u>Combustion of methanol</u>

CH₃OH(l) + 3/2 O₂(g) ⇒ CO₂(g) + 2 H₂O(l)

ΔH°r = 1 mol × ΔH°f(CO₂) + 2 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(CH₃OH) - 3/2 mol × ΔH°f(O₂)

ΔH°r = 1 mol × (-393.5 kJ/mol) + 2 mol × (-285.8 kJ/mol) - 1 mol × (-238.4 kJ/mol) - 3/2 mol × 0

ΔH°r = -726.7 kJ

726.7 kJ are released when 1 mole of CH₃OH is burned. The amount of heat released per kilogram is:

$\frac{-726.7kJ}{1molCH_{3}OH} .\frac{1molCH_{3}OH}{32.04gCH_{3}OH} .\frac{10^{3}gCH_{3}OH }{1kgCH_{3}OH} =-2.268 \times 10^{4} kJ/kgCH_{3}OH$

<u>Combustion of octane</u>

C₈H₁₈(l) + 12.5 O₂(g) ⇒ 8 CO₂(g) + 9 H₂O(l)

ΔH°r = 8 mol × ΔH°f(CO₂) + 9 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(C₈H₁₈) - 12.5 mol × ΔH°f(O₂)

ΔH°r = 8 mol × (-393.5 kJ/mol) + 9 mol × (-285.8 kJ/mol) - 1 mol × (-208.4 kJ/mol) - 12.5 mol × 0

ΔH°r = -5511.8 kJ

5511.8 kJ are released when 1 mole of C₈H₁₈ is burned. The amount of heat released per kilogram is:

$\frac{-5511.8kJ}{1molC_{8}H_{18}} .\frac{1molC_{8}H_{18}}{114.2gC_{8}H_{18}} .\frac{10^{3}gC_{8}H_{18} }{1kgC_{8}H_{18}} =-4.826 \times 10^{4} kJ/kgC_{8}H_{18}$

5 0

3 years ago