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3 years ago

What kind of molecular orbital would result when atomic orbitals are combined as indicated?

Chemistry

1 answer:

Romashka-Z-Leto [24]3 years ago

3 0

Provided question is incomplete as it lacks image or diagtram related to the question, however the correct indication is attach with answer :

Answer:

The correct answer is : antibonding molecular orbital

Explanation:

At the point when two atomic orbitals can consolidate directly produce two molecualr orbital, of which one is holding sub-atomic orbital of lower vitality and the other antibonding molecular orbital of higher energy.

Holding molecular orbital would by and large outcomes which nuclear orbitals of same phase joins while antibonding molecular orbital would results when atomic orbitals of various stages can combine.

In the given approach, it is obvious that one of S - orbital is going consolidate directly with opposite phase of p - orbital along their inter nuclear axis. Subsequently it prompts the formation of sigma antibonding molecular orbital.

Thus, the correct answer is : antibonding molecular orbital

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Is iron conductive in water?

Rudik [331]

Answer: yes

Explanation: it is always in a conductive state

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2 years ago

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Calculate how many grams of hydrogen can be burned if 40. liters of oxygen at 200. k and 1.0 atm. show work pls.

Wittaler [7]

The grams of hydrogen gas can be burned if 40. liters of oxygen at 200. k and 1.0 atm is 4.88 grams.

<h3>How do we calculate grams from moles?</h3>

Grams (W) of any substance will be calculated by using their moles (n) through the following equation:

n = W/M, where

M = molar mass

And moles of the gas will be calculated by using the ideal gas equation as:

PV = nRT, where

P = pressure = 1atm

V = volume = 40L

n = moles = ?

R = universal gas constant = 0.082 L.atm / K.mol

T = temperature = 200K

On putting these values on the above equation, we get

n = (1)(40) / (0.082)(200) = 2.439 = 2.44 moles

Now grams of hydrogen gas will be calculated by using the first equation as:

W = (2.44mol)(2g/mol) = 4.88g

Hence required mass of hydrogen gas is 4.88g.

To know more about ideal gas equation, visit the below link:

brainly.com/question/15046679

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7 0

3 years ago

What is the amplitude of the wave above? *

erica [24]

The correct answer is 15

8 0

3 years ago

Answer these please ASAP need help no idea how to do these

STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass = g

Cl₂:

Number of moles = Mass / molar masa

Number of moles = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol = Mass / 2 g/mol

Mass = 16 g

P₄:

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles = 1.6 g /48 g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles = 100 g / 100 g/mol

Number of moles = 1 mol

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO → 2Fe + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

Fe : F₂O₃

2 : 1

1.78 : 1/2×1.78 = 0.89 mol

Mass of F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂ → 2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

Al : AlI₃

2 : 2

0.05 : 0.05

I₂ : AlI₃

3 : 2

0.1 : 2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂ → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

Pb : O₂

2 : 1

0.03 : 1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass = 0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂ → 2PbO

6 0

3 years ago

The pH of a solution is measured as 8.3. What is the hydrogen ion concentration of the solution?

castortr0y [4]

The logarithmic of the reciprocal of hydrogen - ion concentrate in grams atoms per litre

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3 years ago