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3 years ago

Difference b\w bonding nd Antibonding molecular orbitals

1 answer:

7 0

Answer: Electrons in bonding orbitals stabilize the molecule because they are between the nuclei. They also have lower energies because they are closer to the nuclei. Antibonding orbitals place less electron density between the nuclei. The nuclear repulsions are greater, so the energy of the molecule increases.

Explanation:

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20 points!! Mark Brainliest!! How do I complete and balance this= Zn(NO3)2 + CaS

FromTheMoon [43]

Answer:

it's balanced form is ZnS+Ca(NO3)2

Explanation:

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4 years ago

What is the molecular weight of potassium ( k2so4)

WINSTONCH [101]

Answer:

174.259 g/mol

Explanation:

The molecular weight of potassium (K2so4) is 174.259 g/mol .

Hope it is helpful to you

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3 years ago

How many atoms are there in 10.0 g of sodium?

Alex73 [517]

Use the grams to moles first. The molar mass of sodium is 22.99g, so 10/22.99 ≈ 0.43 moles. The use moles to atoms conversions. 0.43 x (6.022 x 10^23), which would be about 2.62 x 10^23 atoms in 10g of sodium.

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4 years ago

How many atoms are in a 158.69 g sample of nickel?

Alchen [17]

<h3>Answer:</h3>

1.6283 × 10²⁴ atoms Ni

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Use Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

158.69 g Ni

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Ni - 58.69 g/mol

Step 3: Convert

Set up: $\displaystyle 158.69 \ g \ Ni(\frac{1 \ mol \ Ni}{58.69 \ g \ Ni})(\frac{6.022 \cdot 10^{23} \ atoms \ Ni}{1 \ mol \ Ni})$
Multiply/Divide: $\displaystyle 1.62827 \cdot 10^{24} \ atoms \ Ni$

Step 4: Check

Follow sig fig rules and round. We are given 5 sig figs.

1.62827 × 10²⁴ atoms Ni ≈ 1.6283 × 10²⁴ atoms Ni

8 0

3 years ago

Enter your answer in the provided box. Calculate the number of moles of CrCl, that could be produced from 49.4 g Cr202 according

Mrrafil [7]

Answer:

0.4694 moles of CrCl₃

Explanation:

The balanced equation is:

Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)

The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.

The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:

MCr = 52 g/mol

MCl = 35.5 g/mol

MO = 16 g/mol

So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.

The number of moles is the mass divided by the molar mass, so:

n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.

For the stoichiometry:

1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃

0.2347 mol of Cr₂O₃----------- x

By a simple direct three rule:

x = 0.4694 moles of CrCl₃

6 0

3 years ago

Difference b\w bonding nd Antibonding molecular orbitals​

Difference b\w bonding nd Antibonding molecular orbitals