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ella [17]
3 years ago
12

What is a molecular orbital? Question 3 options: Another method of describing the orbitals of individual atoms A mixing of the a

tomic orbitals from an atom’s inner-most shell with those of the valence shell An area around a molecule that describes a region of space where the electrons that are shared between bonded atoms can reside A mixing of the atomic orbitals from an atom’s valence shell
Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
3 0

Answer:

An area around a molecule that describes a region of space where the electrons that are shared between bonded atoms can reside

Explanation:

A molecular orbital is an area around a molecule that describes a region of space where the electrons that are shared between bonded atoms can reside.

Molecular orbitals are formed by linear combination of atomic orbitals of about the same energy. The number of atomic orbitals that are combined to give molecular orbitals is the same as the number of molecular orbitals formed. The build up of electrons in molecular orbitals also follow the Aufbau principle, Hund's rule and Pauli exclusion principle.

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What pressure (in atmospheres) does a gas exert when at 385 mm of Hg?
Westkost [7]
To find out the pressure in atm. You must divide 385 by 760. So pressure in atmospheres equals 385/760.
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Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon
Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: Cu_2O\text{ and }CuO

  • <u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = \frac{2.26}{2.26}=1

For Oxygen  = \frac{4.55}{2.26}=2.01\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is CO_2

  • <u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = \frac{3.57}{3.57}=1

For Oxygen  = \frac{3.57}{3.57}=1

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is CO

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = \frac{1}{2}\times 1=\frac{1}{2}

Now, taking the ratio of carbon atoms in both the samples, we get:

C_1:C_2=\frac{1}{2}:1=1:2

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0
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The ksp value for lead(ii chloride is 2.4 × 10?4. what is the molar solubility of lead(ii chloride?
charle [14.2K]
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The molar solubility of lead(ii) chloride with ksp value of 2.4 × 10e4 can be solve as: 

Ksp = s2 = 2.4 × 10e4
s2 = 2.4 × 10e4
s = √(2.4 × 10e4)
s =  154.9 mol/L
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