Answer:
Bohr's model of the hydrogen atom is based on three postulates:
1) An electron moves around the nucleus in a circular orbit,
2) An electron's angular momentum in the orbit is quantised,
3) The change in an electron's energy as it makes a quantum jump from one orbit to another is always accompanied by the emission or absorption of a photon. Bohr's model is semi-classical because it combines the classical concept of electron orbit (postulate 1) with the new concept of quantisation ( postulates 2 and ).
Answer:
The mass of SO2 will be equal to the sum of the mass of S and O2.
Explanation:
This can be explained by the <em>Law of Conservation of Mass</em>. This law states that mass can neither be created nor destroyed. Knowing this, we can say that the reactants of a chemical reaction must be equal to the products.
In this case, the reactants Sulfur (S) and Oxygen (O2) must equal the mass of the product Sulfur Dioxide (SO2). Therefore, the statement <em>"The mass of SO2 will be equal to the sum of the mass of S and O2" </em>is correct.
mol of Na2CO3 = 2.36 x 10⁻⁴
<h3>Further explanation</h3>
Given
Mass : 0.025 g of Na2CO3
Required
moles
Solution
The mole is the number of particles contained in a substance
1 mol = 6.02.10²³
Moles can also be determined from the amount of substance mass and its molar mass :
mol = mass : molar mass
mass = mol x molar mass
Input the value :
mol = mass : MW Na2CO3
mol = 0.025 g : 106 g/mol
mol = 2.36 x 10⁻⁴
Answer:
Balanced equation is related to the law of conservation of mass as it states that mass can neither be created nor be destroyed in a chemical reaction so the total no. of atoms in a reaction should remain same. in a balanced chemical equation also total no. of atoms of each elements remain same.
Answer : The heat of combustion of n-propanol is 0.554 kJ/mol
Explanation :
First we have to calculate the moles of n-propanol.
Molar mass of n-propanol = 60.09 g/mole
Now we have to calculate the heat of combustion of n-propanol.
As, 0.0166 mole of n-propanol liberated heat of combustion = -33.4 kJ
So, 1 mole of n-propanol liberated heat of combustion = 0.0166 × (-33.4 kJ)
= 0.554 kJ/mol
Therefore, the heat of combustion of n-propanol is 0.554 kJ/mol
