A molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
The bonding orbital, which would be more stable and encourages the bonding of the two H atoms into 
, is the orbital that is located in a less energetic state than just the electron shells of the separate atoms. The antibonding orbital, which has higher energy but is less stable, resists bonding when it is occupied.
An asterisk (sigma*) is placed next to the corresponding kind of molecular orbital to indicate an antibonding orbital. The antibonding orbital known as * would be connected to sigma orbitals, as well as antibonding pi orbitals are known as 
* orbitals.
Therefore, molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
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Hence, the correct answer will be option (b)
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I believe it is <span>d. the bonds of both the reactants and the products are formed.</span>
Answer:
572 g
Explanation:
Molar mass is the mass of 1 mol of an element or compound
molar mass of Li₂SO₄ is the sum of the products of the molar masses of the elements by the number of atoms in the compound
molar masses of each element making up lithium sulphate
Li - 7 g/mol
S - 32 g/mol
O - 16 g/mol
molar mass of Li₂SO₄ - (7 g/mol x 2) + ( 32 g/mol x 1) + ( 16 g/mol x 4 )
molar mass = 110 g/mol
mass of 1 mol of Li₂SO₄ is 110 g
therefore mass of 5.2 mol of Li₂SO₄ is - 110 g/mol x 5.2 mol = 572 g
mass is 572 g
633.97 L
Explanation:
Well use the combined gas law;
P₁V₁T₁ = P₂V₂T₂
We need to change the temperatures into Kelvin;
18.9°C= 292.05 K
5.9°C = 279.05 K
756 * 512 * 292.05 = 639 * V₂ * 279.05
113,044,377.6 = 178,312.95 V₂
V₂ = 113,044,377.6 / 178,312.95
V₂ = 633.97 L
Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
or,
..........(1)
where,
= rate of effusion of nitrogen gas = 
= rate of effusion of sulfur dioxide gas = ?
= molar mass of nitrogen gas = 28 g/mole
= molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:
Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.
